MATH1061 — Week 4 Notes

Lecture 10 — Divisibility

Pre-work: Video 013 (Divisibility). First lecture with Dr Katie Clinch. See divisibility-and-factorisation for the reference definitions, the Unique Factorisation Theorem and the standard proofs.

Learning goals

  1. Understand the definition of “is divisible by” and how to formulate it mathematically.
  2. Understand and apply the Unique Factorisation Theorem.
  3. Gain fluency in proof techniques (direct proof, proof by contradiction and proof by contraposition).

Important note: in the definition of “\(n\) is divisible by \(d\)”, \(n\) and \(d\) are integers and \(d \neq 0\).

\[d \mid n \iff \frac{n}{d} \in \mathbb{Z}\]

All of these say the same thing: \(d\) divides \(n\); \(d\) is a divisor of \(n\); \(d\) is a factor of \(n\); \(n\) is a multiple of \(d\).

Student questions and comments

In the proof from the video that every integer greater than 1 can be written as a product of primes — why must \(r\) and \(s\) be products of primes? Revisited below.

Activity 1 — finding counterexamples

Which collections of values for \(a\), \(b\) and \(m\) give a counterexample disproving \(\forall a, b, m \in \mathbb{Z}\), if \(m \mid (a+b)\) then \(m \mid (a-b)\)? (Multiple answer.)

Recall the conditional is false only when the hypothesis is T and the conclusion is F:

\(m \mid (a+b)\) \(m \mid (a-b)\) \(m\mid(a+b) \to m\mid(a-b)\)
(a) \(m=5, a=20, b=10\) \(5 \mid 30\) T \(5 \mid 10\) T T
(b) \(m=5, a=21, b=13\) \(5 \mid 34\) F \(5 \mid 8\) F T
(c) \(m=5, a=27, b=7\) \(5 \mid 34\) F \(5 \mid 20\) T T
(d) \(m=5, a=19, b=-9\) \(5 \mid 10\) T \(5 \mid 28\) F F
(e) \(m=5, a=23, b=12\) \(5 \mid 35\) T \(5 \mid 11\) F F
(f) \(m=5, a=-10, b=-3\) \(5 \mid -13\) F \(5 \mid -7\) F T

So (d) and (e) provide counterexamples.

Activity 2 — prove or disprove

(a) \(\forall m \in \mathbb{Z}\), \(6m(2m + 10m^2)\) is divisible by 4. — True.

Direct proof. Suppose \(m \in \mathbb{Z}\). Then \[6m(2m + 10m^2) = 2(3m) \cdot 2(m + 5m^2) = 4\big(3m(m + 5m^2)\big).\] Since \(m \in \mathbb{Z}\) we have that \(3m(m+5m^2)\) is also an integer. Thus \(4 \mid 6m(2m+10m^2)\). \(\square\)

Saying why the other factor is an integer is the step that’s easy to skip and shouldn’t be.

(b) \(\forall a, b, c \in \mathbb{Z}\), if \(a \mid (b+c)\) then \(a \mid b\) or \(a \mid c\). — False.

Counterexample. Let \(a = 3\), \(b = 4\), \(c = 2\). Then \(a \mid (b+c)\) because \(3 \mid 6\), but \(a \nmid b\) and \(a \nmid c\) because \(3 \nmid 4\) and \(3 \nmid 2\). Thus the statement is false. \(\square\)

(Recall \(a \mid b \iff \frac{b}{a} \in \mathbb{Z}\) — so \(10 \nmid 5\), since \(\frac{5}{10} \notin \mathbb{Z}\). Watch the order.)

(c) \(\forall a, b, c \in \mathbb{Z}\), if \(a\) is a multiple of \(c\), then \(ab\) is a multiple of \(c\). — True.

Direct proof. Let \(a, b, c \in \mathbb{Z}\) and suppose \(c \mid a\). ← initial hypotheses Then \(\exists k \in \mathbb{Z}\) such that \(a = ck\). Multiplying both sides by \(b\) gives \(ab = ckb = c(kb)\). Since \(k, b \in \mathbb{Z}\) we have that \(kb \in \mathbb{Z}\). Thus \(c \mid ab\). ← conclusion \(\square\)

Every integer \(n > 1\) is a product of primes

Theorem (from the video): every integer \(n > 1\) can be written as a product of primes.

Proof (by contradiction). Suppose the theorem is false. Then there exists an integer \(n > 1\) that is not a product of primes.

Choose the smallest such \(n\). A smallest such \(n\) exists by the well-ordering principle (covered in a later lecture). We choose this \(n\) so that we can use the fact that every integer less than \(n\) (but greater than 1) can be written as a product of primes.

Either \(n\) is prime or \(n\) is composite — splitting into two cases using the fact that every integer \(n > 1\) is one or the other, not both.

Case 1: if \(n\) is prime, then \(n\) is a product of primes (\(n = p\)) — a contradiction already.

Case 2: if \(n\) is composite, then \(n = rs\) for some positive integers \(r\) and \(s\) where \(r \neq 1\) and \(s \neq 1\). This implies \(1 < r < n\) and \(1 < s < n\). Because we chose \(n\) to be the smallest integer greater than 1 that is not a product of primes, both \(r\) and \(s\) — which are smaller than \(n\) — must be products of primes: \(r = p_1p_2\cdots p_k\) and \(s = q_1q_2 \cdots q_\ell\). Therefore \(n = rs = p_1 \cdots p_k q_1 \cdots q_\ell\) is a product of primes also.

So regardless of whether \(n\) is prime or composite, \(n\) is a product of primes. This contradicts our choice of \(n\), and hence the theorem is proved. \(\square\)

That answers the student question above: \(r\) and \(s\) must be products of primes precisely because \(n\) was chosen to be the smallest counterexample, so nothing smaller than \(n\) can be one.

Unique Factorisation Theorem for the integers

Given any integer \(n > 1\), the standard factored form of \(n\) is \[n = p_1^{e_1} p_2^{e_2} \cdots p_k^{e_k}\] where \(k\) is a positive integer, \(p_1, p_2, \dots, p_k\) are prime numbers, \(e_1, \dots, e_k\) are positive integers, and \(p_1 < p_2 < \cdots < p_k\).

Q3 from the pre-work questions

What is the unique prime factorisation of 6975 in standard factored form?

Try dividing by primes \(2, 3, 5, 7, 11, \dots\), keeping in mind you only have to go as far as \(\sqrt{6975} = 83.51\dots\) due to the lemma below.

\[6975 = 3 \cdot 2325 = 3 \cdot 3 \cdot 775 = 3 \cdot 3 \cdot 5 \cdot 155 = 3 \cdot 3 \cdot 5 \cdot 5 \cdot 31 = 3^2 \cdot 5^2 \cdot 31\]

The prime factorisation of a positive divisor of 6975 has 0, 1 or 2 threes; 0, 1 or 2 fives; and 0 or 1 thirty-ones. Thus there are \(3 \cdot 3 \cdot 2 = 18\) positive divisors.

Lemma: for every integer \(n > 1\), if no integer \(x\) in the range \(1 < x \leq \sqrt{n}\) divides \(n\), then \(n\) is prime. (Optional activity — proof by contraposition given in divisibility-and-factorisation.)

Activity 3 — working with the factored form

(a) Standard factored form of 1400: \[1400 = 2(700) = 2 \cdot 2(350) = 2 \cdot 2(35)(10) = 2^2(5 \cdot 7)(2 \cdot 5) = 2^3 \cdot 5^2 \cdot 7^1\] — note the primes in increasing order.

(b) \(1400^2 = (2^3 \cdot 5^2 \cdot 7^1)^2 = 2^6 \cdot 5^4 \cdot 7^2\) — raising to a power multiplies the exponents.

(c) \(1400^3 = 1400 \cdot 1400^2 = (2^3 \cdot 5^2 \cdot 7^1)(2^6 \cdot 5^4 \cdot 7^2) = 2^9 \cdot 5^6 \cdot 7^3\) — multiplying adds the exponents.

(d) Number of positive divisors of 1400: \((3+1)(2+1)(1+1) = 4 \cdot 3 \cdot 2 = 24\).

(e) Positive divisors of 1400 divisible by 8: all factors of the form \(2^3 \cdot 5^* \cdot 7^*\) —

\(7^0\) \(7^1\)
\(5^0\) 8 56
\(5^1\) 40 280
\(5^2\) 200 1400

(f) Smallest positive integer \(n\) such that \(1400n\) is a perfect square: \(1400n\) is a perfect square if \(\exists t \in \mathbb{Z}\) with \(1400n = t^2\). Then \[t^2 = 2^3 \cdot 5^2 \cdot 7^1 \cdot n.\] Split each power in half, bigger powers on the left: \(t = 2^2 \cdot 5^1 \cdot 7^1\) paired against \(2^1 \cdot 5^1 \cdot 7^0 \cdot n\), so \(n = 2 \cdot 7 = 14\).

Activity 4 — a lemma by contradiction

Lemma: for any integer \(a\) and any prime number \(p\), if \(p \mid a\) then \(p \nmid (a+1)\). Hint: once you’ve set up the proof, write \((a+1) - a\) in terms of \(p\).

Proof (by contradiction). Let \(a \in \mathbb{Z}\), let \(p\) be prime, and suppose \(p \mid a\) and — for a contradiction — assume \(p \mid (a+1)\). Since \(p \mid a\) there exists \(k \in \mathbb{Z}\) such that \(a = pk\). Since \(p \mid (a+1)\) there exists \(\ell \in \mathbb{Z}\) such that \(a + 1 = p\ell\). So \[1 = (a+1) - a = p\ell - pk = p(\ell - k).\] Since \(\ell, k \in \mathbb{Z}\) we have \(\ell - k \in \mathbb{Z}\). Thus \(1 = p(\ell - k)\) means \(p \mid 1\) — which is a contradiction, because \(p \geq 2\) for any prime, whereas \(p \mid 1\) would force \(p = \pm 1\). So the assumption was false: \(p \nmid (a+1)\). \(\square\)

The technique worth keeping: rewrite the divisibility hypotheses as equations, then combine them so the unwanted quantity appears on its own.

Theorem: there are an infinite number of primes

Proof (by contradiction). Suppose the number of primes is finite. Then some prime \(p\) is the largest, and we can list the primes in ascending order: \(2, 3, 5, 7, 11, \dots, p\).

Let \(N\) be 1 more than the product of all the primes: \(N = (2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdots p) + 1\).

Then \(N > 1\), so by the Unique Factorisation Theorem \(N\) can be written as a product of primes, and hence is divisible by some prime \(q\). Because \(q\) is prime, it must equal one of the primes in our list.

Thus \(q\) divides \(2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdots p\) — that is, \(q \mid (N-1)\) — and so, by the lemma above, \(q\) does not divide \((2 \cdot 3 \cdots p) + 1 = N\).

Hence \(q\) divides \(N\) and \(q\) does not divide \(N\): a contradiction. We conclude the original statement must be true. \(\square\)

See also

Lecture 12 — Modular Arithmetic and the Euclidean Algorithm

Pre-work: Video 015 (The Euclidean Algorithm). See gcd-lcm-and-euclidean-algorithm for the reference definitions, worked runs of the algorithm and the gcd/lcm proofs.

Learning goals

  1. Continue to gain fluency in applying modular arithmetic.
  2. Understand the definitions of \(\gcd(a,b)\) and \(\operatorname{lcm}(a,b)\) for integers \(a\) and \(b\).
  3. Be able to apply the Euclidean algorithm to find the greatest common divisor of two integers.

Student questions and comments

Is \(\gcd(a,b) = \gcd(b,a)\)? Yes.

How do you do gcd and lcm with a mix of positive and negative integers? Covered below — take absolute values.

Where are gcd, lcm and the Euclidean algorithm used? Used all the time in arithmetic (adding fractions, etc.); used to determine if certain types of equations have solutions; applications in cryptography.

Recap on mod

For \(n, d \in \mathbb{Z}\), \(d > 0\): \(n \bmod d = r\) indicates the remainder when \(n\) is divided by \(d\) is \(r\), with \(0 \leq r < d\). For \(m, n, d \in \mathbb{Z}\), \(d > 0\): \(m \equiv n \pmod d\) indicates \(m\) and \(n\) have the same remainder when divided by \(d\) — equivalently \(d \mid (m-n)\). See modular-arithmetic.

Poll question 1

If \(m \in \mathbb{Z}\) and \(m \bmod 6 = 4\), what is the remainder when \(35m\) is divided by 6?

C. 2. Since \(35 \bmod 6 = 5\) and \(m \equiv 4\), we get \(35m \equiv 5 \cdot 4 \equiv 20 \equiv 2 \pmod 6\).

Poll question 2

If \(m \in \mathbb{Z}\) and \(m \bmod 6 = 4\), what is the remainder when \(3m^2 + 2m + 1\) is divided by 6?

D. 3. \(3m^2 + 2m + 1 \equiv 3(4)^2 + 2(4) + 1 \equiv 3(16) + 8 + 1 \equiv 48 + 9 \pmod 6\), and \(48 \equiv 0 \pmod 6\) (since \(6 \mid 48\)), so this is \(\equiv 9 \equiv 3 \pmod 6\).

\(\gcd(a,b)\)

Definition: for integers \(a\) and \(b\), not both zero, the greatest common divisor of \(a\) and \(b\), denoted \(\gcd(a,b)\), is the integer \(d\) for which:

  • \(d \mid a\) and \(d \mid b\) — \(d\) is a divisor of both \(a\) and \(b\); and
  • \(\forall c \in \mathbb{Z}\), if \(c \mid a\) and \(c \mid b\) then \(c \leq d\) — every other divisor of both \(a\) and \(b\) is less than \(d\).

(Also written \(\operatorname{hcf}(a,b)\).) The order of \(a\) and \(b\) does not matter.

What if one, or both, of \(a\) and \(b\) are negative? The divisors of 15 are \(1, 3, 5, 15, -1, -3, -5, -15\) — and these are also the divisors of \(-15\). The divisors of an integer \(a\) are the same as the divisors of \(-a\). Thus \(\gcd(-15, 36) = \gcd(15,36)\), and in general

\[\gcd(a,b) = \gcd(|a|, |b|)\]

Note that \(\gcd(0,b) = b\) for all positive integers \(b\) — this is the base case the algorithm below terminates on.

\(\operatorname{lcm}(a,b)\)

Definition: for non-zero integers \(a\) and \(b\), the lowest (least) common multiple of \(a\) and \(b\), denoted \(\operatorname{lcm}(a,b)\), is the smallest integer \(n\) for which:

  • \(n > 0\) — every other positive multiple of both \(a\) and \(b\) is greater than \(n\); and
  • \(a \mid n\) and \(b \mid n\) — \(n\) is a multiple of both \(a\) and \(b\).

Order does not matter. The multiples of 5 are \(\dots, -10, -5, 0, 5, 10, 15, \dots\) — the same as the multiples of \(-5\). So \(\operatorname{lcm}(-5, 36) = \operatorname{lcm}(5, 36)\), and in general

\[\operatorname{lcm}(a,b) = \operatorname{lcm}(|a|, |b|)\]

Relationship between gcd and lcm

Use the unique prime factorisation of each integer (see divisibility-and-factorisation) — step 1: factorise; step 2: take minimum exponents for gcd, maximum for lcm.

\[30 = 2 \cdot 3 \cdot 5 \qquad 72 = 2^3 \cdot 3^2\] \[\gcd(30,72) = 2 \cdot 3 = 6 \qquad \operatorname{lcm}(30,72) = 2^3 \cdot 3^2 \cdot 5 = 360\]

Lemma: if \(a, b \in \mathbb{Z}^+\), then \(\gcd(a,b) \cdot \operatorname{lcm}(a,b) = ab\).

Checking: \(\gcd(30,72) \cdot \operatorname{lcm}(30,72) = 6 \cdot 360 = 30 \cdot 72\), since \((2 \cdot 3)(2^3 \cdot 3^2 \cdot 5) = (2 \cdot 3 \cdot 5)(2^3 \cdot 3^2)\).

More generally, for non-zero integers \(a\) and \(b\): \(\gcd(a,b) \cdot \operatorname{lcm}(a,b) = |ab|\).

Activity 1

\(\gcd(63, -105)\) and \(\operatorname{lcm}(63, -105)\):

  • Step 1: \(63 = 3^2 \cdot 7\) and \(-105 = -3 \cdot 5 \cdot 7\).
  • Step 2: \(\gcd(63,-105) = 3 \cdot 7 = 21\); \(\operatorname{lcm}(63,-105) = 3^2 \cdot 5 \cdot 7 = 315\).

Activity 2

Let \(p\) and \(q\) be distinct primes (\(p \neq q\)). Then \(\gcd(p,q) = 1\) and \(\operatorname{lcm}(p,q) = pq\).

The Euclidean algorithm

To determine \(\gcd(a,b)\) for \(a, b \in \mathbb{Z}^+\), based on the fact:

if \(a, b \in \mathbb{Z}\) are not both zero and \(q, r \in \mathbb{Z}\) satisfy \(a = bq + r\), then \(\gcd(a,b) = \gcd(b,r)\).

The proof of this fact is basically showing that every divisor of \(a\) and \(b\) is also a divisor of \(b\) and \(r\), and vice versa:

  • Suppose \(c \mid a\) and \(c \mid b\). Then \(c \mid (a - bq)\) and hence \(c \mid r\).
  • Suppose \(d \mid b\) and \(d \mid r\). Then \(d \mid (bq + r)\) and hence \(d \mid a\).

Divide the big by the small, then repeat on (small, remainder).

Example: \(\gcd(1232, 5859)\).

Division Consequence
\(5859 = 1232 \cdot 4 + 931\) \(\gcd(5859, 1232) = \gcd(1232, 931)\)
\(1232 = 931 \cdot 1 + 301\) \(\gcd(1232, 931) = \gcd(931, 301)\)
\(931 = 301 \cdot 3 + 28\) \(\gcd(931, 301) = \gcd(301, 28)\)
\(301 = 28 \cdot 10 + 21\) \(\gcd(301, 28) = \gcd(28, 21)\)
\(28 = 21 \cdot 1 + 7\) \(\gcd(28, 21) = \gcd(21, 7)\)
\(21 = 7 \cdot 3 + 0\) \(\gcd(21, 7) = \gcd(7, 0) = 7\)

Therefore \(\gcd(5859, 1232) = 7\).

Activity 3

Use the Euclidean Algorithm to determine \(\gcd(4131, 2431)\).

\(a\) \(=\) \(b \times q\) \(+ \; r\)
4131 \(=\) \(2431 \times 1\) \(+ \; 1700\)
2431 \(=\) \(1700 \times 1\) \(+ \; 731\)
1700 \(=\) \(731 \times 2\) \(+ \; 238\)
731 \(=\) \(238 \times 3\) \(+ \; 17\)
238 \(=\) \(17 \times 14\) \(+ \; 0\)

\[\gcd(4131, 2431) = \cdots = \gcd(238, 17) = \gcd(17, 0) = \mathbf{17}\]

Activity 4 — an “if and only if” proof

Prove that for all positive integers \(a\) and \(b\), \(a \mid b\) if and only if \(\operatorname{lcm}(a,b) = b\).

Not a counterexample; a direct proof of a biconditional.

Method 1: prove each direction separately — the recommended method.

(\(\Rightarrow\)) Suppose \(a, b \in \mathbb{Z}^+\) and suppose \(a \mid b\). Then \(\exists k \in \mathbb{Z}\) such that \(b = ka\). So \(\operatorname{lcm}(a,b) = \operatorname{lcm}(a, ka)\). Since \(a \mid ka\) and \(ka \mid ka\), and no smaller \(n > 0\) has \(ka \mid n\), we have \(\operatorname{lcm}(a, ka) = ka\). Therefore \(\operatorname{lcm}(a,b) = b\) (since \(b = ka\)).

(\(\Leftarrow\)) Suppose \(a, b \in \mathbb{Z}^+\) and suppose \(\operatorname{lcm}(a,b) = b\). \(\dots\) Therefore \(a \mid b\). \(\square\)

This direction was left blank on the slides — the completed argument (via \(g = \gcd(a,b)\), \(a = gm\), \(b = gn\) with \(\gcd(m,n) = 1\), so \(\operatorname{lcm}(a,b) = gmn\)) is written out in gcd-lcm-and-euclidean-algorithm.

Method 2: prove both directions simultaneously — be careful! Quicker, but easy to make a logical error, because every step must genuinely be reversible:

Suppose \(a, b \in \mathbb{Z}^+\) and suppose \(a \mid b\). This is true iff \(\exists k \in \mathbb{Z}\) such that \(b = ka\), which holds iff \(\operatorname{lcm}(a,b) = \operatorname{lcm}(a,ka)\), which holds iff \(a \mid ka\) and \(ka \mid ka\) and no smaller \(n > 0\) has \(ka \mid n\), which holds iff \(\operatorname{lcm}(a,ka) = ka\), which holds iff \(\operatorname{lcm}(a,b) = b\). \(\square\)

Each “iff” is doing real work here — if any one of them is only an implication, the proof silently collapses to a single direction.

Challenge activity — reversing the Euclidean algorithm

Non-assessed content.

Theorem: for \(a, b, c \in \mathbb{Z}\), if \(\gcd(a,b) \mid c\) then there exist integers \(x\) and \(y\) such that \(ax + by = c\).

Given \(ax + by = c\), use the Euclidean algorithm to find \(\gcd(a,b)\), then run the steps in reverse to find \(x\) and \(y\).

Example: find integers \(x, y\) satisfying \(1232x + 5859y = 14\).

Step 1 — use the E.A. to find \(\gcd(1232, 5859) = 7\):

\[5859 = 1232 \cdot 4 + 931 \quad \text{(Eqn 1)}\] \[1232 = 931 \cdot 1 + 301 \quad \text{(Eqn 2)}\] \[931 = 301 \cdot 3 + 28 \quad \text{(Eqn 3)}\] \[301 = 28 \cdot 10 + 21 \quad \text{(Eqn 4)}\] \[28 = 21 \cdot 1 + 7 \quad \text{(Eqn 5)}\] \[21 = 7 \cdot 3 + 0\]

Step 2 — work backwards to get \(1232w + 5859z = 7\):

From Eqn 5: \(7 = 28 - 21\). Use Eqn 4 to replace the 21 and collect like terms: \(7 = 28 - (301 - 28 \cdot 10) = 28 \cdot 11 - 301\). Use Eqn 3 to replace the 28: \(7 = (931 - 301 \cdot 3)\cdot 11 - 301 = 931 \cdot 11 - 301 \cdot 34\). Use Eqn 2 to replace the 301: \(7 = 931 \cdot 11 - (1232 - 931)\cdot 34 = 931 \cdot 45 - 1232 \cdot 34\). Use Eqn 1 to replace the 931: \(7 = (5859 - 1232 \cdot 4)\cdot 45 - 1232 \cdot 34 = 5859 \cdot 45 - 1232 \cdot 214\).

Step 3 — multiply through to get \(1232x + 5859y = 14\):

Therefore \(7 = 1232(-214) + 5859(45)\), so \(14 = 1232(-428) + 5859(90)\).

See also

Lecture 11 — Modular Arithmetic

Pre-work: Video 014 (Modular Arithmetic). See modular-arithmetic for the reference definitions, the quotient-remainder theorem and the congruence lemma this lecture builds on.

Learning goals

  1. Understand the proof that there are an infinite number of primes. (Covered in 2026-08-17-divisibility.)
  2. Understand the definition of floor and ceiling.
  3. Understand the quotient-remainder theorem.
  4. Gain fluency in applying modular arithmetic.

Student questions and comments

How does the quotient-remainder theorem work with negative integers? The remainder is still required to satisfy \(0 \leq r < d\), so it stays non-negative: \(-19 = 6 \cdot (-4) + 5\), giving \(-19 \bmod 6 = 5\).

What is the difference between using mod with an equal sign and with an equivalence sign? See “\(\bmod\) versus \(\equiv\)” below — this is the notational trap of the topic.

Floor and ceiling

For the floor you move left on the number line, so \(\lfloor -3.2 \rfloor = -4\). For the ceiling you move right, so \(\lceil -3.2 \rceil = -3\). (And \(\lfloor 3.2 \rfloor = 3\), \(\lceil 3.2 \rceil = 4\).) Watch out for the negatives.

To use floor or ceiling in a proof, it is often helpful to write an expression as an integer plus an amount between 0 and 1. For example, in the video proof that “if \(n\) is an odd integer, then \(\lceil \frac{n}{2}\rceil = \lfloor \frac{n}{2} \rfloor + 1\)”, we wrote

\[\frac{n}{2} = \frac{2k+1}{2} = k + \frac{1}{2}\]

for some integer \(k\) — and \(k + \frac{1}{2}\) sits between \(k\) and \(k+1\), so the floor is \(k\) and the ceiling is \(k+1\). The same trick handles \(\big\lfloor (p^3 + 4p^2 + 5) + \frac{1}{4} \big\rfloor\) where \(p \in \mathbb{Z}\).

Quotient-remainder theorem and “mod”

Given any integer \(n\) and positive integer \(d\), there exist unique integers \(q\) and \(r\) such that \(n = dq + r\) and \(0 \leq r < d\).

The quotient is \(q = \lfloor \frac{n}{d} \rfloor\) and the remainder is \(r = n - d\lfloor \frac{n}{d} \rfloor\). We define \(n \bmod d = r\).

With \(d = 6\): \(53 = 6 \cdot 8 + 5\), \(20 = 6 \cdot 3 + 2\), \(-19 = 6 \cdot (-4) + 5\).

Two equivalent ways of defining \(m \equiv n \pmod d\)

  • Definition 1: \(m\) and \(n\) are congruent modulo \(d\) iff they have the same remainder according to the quotient-remainder theorem.
  • Definition 2: \(m \equiv n \pmod d\) iff \(d \mid (m-n)\).

In the examples above, 53 and \(-19\) are congruent modulo 6: \(53 \equiv -19 \pmod 6\). Checking with Definition 2: does \(6 \mid \big(53 - (-19)\big)\)? We have \(53 + 19 = 72\) and \(6 \mid 72\), so yes.

However 53 and 20 have different remainders on division by 6, so \(53 \not\equiv 20 \pmod 6\).

Take care with notation

  • For \(n, d \in \mathbb{Z}\), \(d > 0\): \(n \bmod d = r\) indicates that the remainder when \(n\) is divided by \(d\) is \(r\), with \(0 \leq r < d\).
  • For \(m, n, d \in \mathbb{Z}\), \(d > 0\): \(m \equiv n \pmod d\) indicates that \(m\) and \(n\) have the same remainder when divided by \(d\) — equivalently \(d \mid (m-n)\).

E.g. \(5 \bmod 4 = 1\) and \(9 \bmod 4 = 1\), so \(9 \equiv 5 \pmod 4\) — but \(9 \bmod 4 \neq 5\).

Q4 from the pre-class multiple choice questions

True or false? \(-50 \equiv 22 \pmod 8\). True, both ways:

  • Definition 1: \(-50 = -48 - 2 = -56 + 6 = 8(-7) + 6\), so \(-50 \bmod 8 = 6\); and \(22 = 16 + 6\), so \(22 \bmod 8 = 6\). Same remainder.
  • Definition 2: \(22 - (-50) = 72 = 8 \times 9\), so \(8 \mid \big(22-(-50)\big)\).

Activity 1

  1. \(\left\lfloor \frac{-32}{5} \right\rfloor\) and \(\left\lceil \frac{-32}{5} \right\rceil\):
  • Fraction approach: \(\lfloor \frac{-30}{5} - \frac{2}{5} \rfloor = \lfloor -6 - \frac{2}{5} \rfloor = -7\) (smaller than \(-6\)).
  • Decimal approach: \(\lfloor -6.4 \rfloor = \lfloor -6 - 0.4 \rfloor = -7\).

So \(\left\lfloor \frac{-32}{5} \right\rfloor = -7\) and \(\left\lceil \frac{-32}{5} \right\rceil = -6\) (add 1).

  1. Quotient and remainder for \(n = -32\), \(d = 5\): \(-32 = -7 \cdot 5 + 3\), so \(q = -7\) and \(r = 3\).

  2. \(-32 \bmod 5 = 3\).

  3. Is \(-32 \equiv 32 \pmod 5\)? By Definition 2: \(32 - (-32) = 64\), and \(5 \nmid 64\), so no — \(32 \not\equiv -32 \pmod 5\).

  4. Integers between 10 and 20 congruent to \(-32\) modulo 5: those leaving remainder 3, i.e. 13 and 18 (stepping by 5 from 3: \(\dots, 13, 18, 23, \dots\)).

Activity 2

(a) Prove that for any odd integer \(n\), \(n^2 \bmod 4 = 1\).

Direct proof. Suppose \(n\) is odd. Then \(\exists k \in \mathbb{Z}\) such that \(n = 2k+1\). So \[n^2 = (2k+1)^2 = 4k^2 + 4k + 1 = 4(k^2 + k) + 1.\] Since \(k^2 + k \in \mathbb{Z}\), this gives by the quotient-remainder theorem: \(n^2 \bmod 4 = 1\). \(\square\)

(b) Use this fact to show that for any odd integer \(n\), \(\left\lfloor \frac{n^2}{4} \right\rfloor = \frac{n^2 - 1}{4}\).

Method 1 — show LHS \(=\) RHS using \(n = 2k+1\):

\[\text{LHS} = \left\lfloor \frac{(2k+1)^2}{4} \right\rfloor = \left\lfloor \frac{4k^2+4k+1}{4} \right\rfloor = \left\lfloor (k^2+k) + \frac{1}{4} \right\rfloor = k^2 + k\] \[\text{RHS} = \frac{(2k+1)^2 - 1}{4} = \frac{4k^2+4k+1-1}{4} = k^2 + k\]

so LHS \(=\) RHS.

Method 2 — use the rule \(r = n - d\lfloor \frac{n}{d} \rfloor\) with number \(n^2\), divisor 4, remainder 1:

\[1 = n^2 - 4\left\lfloor \frac{n^2}{4} \right\rfloor \implies 4\left\lfloor \frac{n^2}{4} \right\rfloor = n^2 - 1 \implies \left\lfloor \frac{n^2}{4}\right\rfloor = \frac{n^2-1}{4}\]

Modular arithmetic — the lemma from the video

Lemma: if \(a \equiv b \pmod d\) and \(m \equiv n \pmod d\), then \[a + m \equiv b + n \pmod d \qquad \text{and} \qquad am \equiv bn \pmod d\]

For example, \(8 \equiv 2 \pmod 3\) and \(46 \equiv 1 \pmod 3\), so \(8 \cdot 46 \equiv 2 \cdot 1 \equiv 2 \pmod 3\), and \(8 + 46 \equiv 2 + 1 \equiv 0 \pmod 3\).

Activity 3

Suppose \(x \bmod 6 = 4\) and \(y \bmod 6 = 3\).

(a) \((x+y) \bmod 6\): \[x + y \equiv 4 + 3 \equiv 7 \equiv 1 \pmod 6\] so the answer is 1.

(b) \((5x^2 + 4y) \bmod 6\): \[5x^2 + 4y \equiv 5(4)^2 + 4(3) \equiv 5(16) + 12 \equiv 5(16) \equiv 5(4) \equiv 20 \equiv 2 \pmod 6\] applying the lemma (addition and multiplication) at each step — \(12 \equiv 0\) and \(16 \equiv 4 \pmod 6\). The answer is 2.

Reducing at every step is what keeps the numbers small.

Remainders and divisibility tests

Modular arithmetic explains the common divisibility tests.

A positive integer is divisible by 3 if and only if the sum of its digits is divisible by 3.

Example: 254 is not divisible by 3 — the sum of its digits is \(2+5+4 = 11\), not divisible by 3.

\[254 \equiv 2 \cdot 100 + 5 \cdot 10 + 4 \cdot 1 \equiv 2 \cdot 1 + 5 \cdot 1 + 4 \cdot 1 \equiv 2 + 5 + 4 \equiv 11 \equiv 2 \pmod 3\]

so the remainder is 2 (confirming with the quotient-remainder theorem: \(254 = 3 \cdot 84 + 2\)).

Example: 3252 is divisible by 3, because \(3+2+5+2 = 12\) is.

\[3252 \equiv 3 \cdot 1 + 2 \cdot 1 + 5 \cdot 1 + 2 \cdot 1 \equiv 12 \equiv 0 \pmod 3 \implies 3 \mid 3252\]

The test works because \(\forall a \in \mathbb{Z}^+\), \(10^a \equiv 1 \pmod 3\). To see \(10^2 \equiv 1 \pmod 3\), use the lemma with \(a = m = 10\) and \(b = n = 1\); to see \(10^3 \equiv 1 \pmod 3\), use the lemma with \(a = 10^2\), \(m = 10\), \(b = n = 1\); and so on.

See also

Reference material

Divisibility & Unique Factorisation

Definition

\[d \mid n \iff d \neq 0 \text{ and } n = dk \text{ for some integer } k\]

Both \(n\) and \(d\) must be integers, and \(d \neq 0\) — this side condition is part of the definition, not an afterthought. Equivalently, \(d \mid n \iff \frac{n}{d} \in \mathbb{Z}\).

All of these say the same thing:

  • \(d\) divides \(n\)
  • \(d\) is a divisor of \(n\)
  • \(d\) is a factor of \(n\)
  • \(n\) is a multiple of \(d\)

\(d \nmid n\) means \(d\) does not divide \(n\). Note that \(d \mid n\) is a statement (it has a truth value) — \(2 \mid 6\) is true, \(10 \nmid 5\) is true since \(\frac{5}{10} \notin \mathbb{Z}\).

Proving and disproving divisibility statements

The pattern for a direct proof: unfold \(d \mid n\) into \(n = dk\), do algebra, then re-fold into the definition by exhibiting the integer witness.

\(\forall a, b, c \in \mathbb{Z}\), if \(c \mid a\) then \(c \mid ab\).

Let \(a, b, c \in \mathbb{Z}\) and suppose \(c \mid a\). Then \(\exists k \in \mathbb{Z}\) such that \(a = ck\). Multiplying both sides by \(b\): \(ab = ckb = c(kb)\). Since \(k, b \in \mathbb{Z}\) we have \(kb \in \mathbb{Z}\). Thus \(c \mid ab\). \(\square\)

\(\forall m \in \mathbb{Z}\), \(6m(2m + 10m^2)\) is divisible by 4.

Let \(m \in \mathbb{Z}\). Then \(6m(2m + 10m^2) = 2(3m) \cdot 2(m + 5m^2) = 4 \big(3m(m + 5m^2)\big)\). Since \(m \in \mathbb{Z}\), \(3m(m + 5m^2) \in \mathbb{Z}\). Thus \(4 \mid 6m(2m + 10m^2)\). \(\square\)

\(\forall a, b, c \in \mathbb{Z}\), if \(a \mid (b + c)\) then \(a \mid b\) or \(a \mid c\) — false. Counterexample \(a = 3, b = 4, c = 2\): \(a \mid (b+c)\) since \(3 \mid 6\), but \(3 \nmid 4\) and \(3 \nmid 2\).

\(\forall a, b, m \in \mathbb{Z}\), if \(m \mid (a+b)\) then \(m \mid (a-b)\) — false. A counterexample needs the hypothesis true and the conclusion false:

\(m \mid (a+b)\) \(m \mid (a-b)\) conditional
\(m=5, a=20, b=10\) \(5 \mid 30\) T \(5 \mid 10\) T T
\(m=5, a=21, b=13\) \(5 \mid 34\) F \(5 \mid 8\) F T
\(m=5, a=27, b=7\) \(5 \mid 34\) F \(5 \mid 20\) T T
\(m=5, a=19, b=-9\) \(5 \mid 10\) T \(5 \mid 28\) F F
\(m=5, a=23, b=12\) \(5 \mid 35\) T \(5 \mid 11\) F F
\(m=5, a=-10, b=-3\) \(5 \mid -13\) F \(5 \mid -7\) F T

Only the two rows with hypothesis T and conclusion F are counterexamples.

\(\forall m, n, q \in \mathbb{Z}\), if \(m \mid n\) and \(n \mid q\) then \(m^2 \mid nq\) — true.

Let \(m, n, q\) be non-zero integers with \(m \mid n\) and \(n \mid q\). By definition there exist \(a, b \in \mathbb{Z}\) with \(n = ma\) and \(q = nb\). Then \[nq = (ma)(nb) = ma \cdot (mab) = m^2(a^2b).\] Since \(a^2 b \in \mathbb{Z}\), we conclude \(m^2 \mid nq\). \(\square\)

If \(k\) is even and \(m\) is odd, then \((k+2)^2 - (m-3)^2\) is divisible by 4.

Let \(k = 2a\) and \(m = 2b+1\) for some \(a, b \in \mathbb{Z}\). Then \((k+2)^2 = (2a+2)^2 = 4(a+1)^2\) and \((m-3)^2 = (2b-2)^2 = 4(b-1)^2\), so \[(k+2)^2 - (m-3)^2 = 4\big((a+1)^2 - (b-1)^2\big),\] and \((a+1)^2 - (b-1)^2 \in \mathbb{Z}\). \(\square\)

\(\forall c, d, e \in \mathbb{Z}\), if \(c \mid d\) and \(c \nmid e\), then \(c \nmid (d+e)\).

Proof by contradiction. Suppose there exist integers \(c, d, e\) with \(c \mid d\), \(c \nmid e\) and \(c \mid (d+e)\). Then \(d = cx\) for some \(x \in \mathbb{Z}\), and \(d + e = cy\) for some \(y \in \mathbb{Z}\). Hence \(cx + e = cy\), so \(e = c(y - x)\) with \(y - x \in \mathbb{Z}\) — so \(c \mid e\), a contradiction. \(\square\)

Every integer \(n > 1\) is a product of primes

Proof by contradiction. Suppose the theorem is false. Then there exists an integer \(n > 1\) that is not a product of primes. Choose the smallest such \(n\) — such a smallest one exists by the well-ordering principle. Choosing the smallest is what lets us assume every integer strictly between \(1\) and \(n\) is a product of primes.

Every integer \(n > 1\) is either prime or composite (not both).

Case 1: \(n\) is prime. Then \(n\) is a product of primes (namely itself) — contradiction.

Case 2: \(n\) is composite. Then \(n = rs\) for positive integers \(r, s\) with \(r \neq 1\) and \(s \neq 1\), which forces \(1 < r < n\) and \(1 < s < n\). Because \(n\) was chosen smallest, both \(r\) and \(s\) are products of primes, say \(r = p_1 p_2 \cdots p_k\) and \(s = q_1 q_2 \cdots q_\ell\). Therefore \(n = rs = p_1 \cdots p_k q_1 \cdots q_\ell\) is a product of primes too.

Either way \(n\) is a product of primes, contradicting our choice of \(n\). \(\square\)

Unique Factorisation Theorem

Provided in the exam. This theorem and the Quotient-Remainder Theorem are both stated on the MATH1061/MATH7861 Examination Formula Page, so they can be cited rather than restated from memory — see also logical-equivalence-laws for the rest of what that page carries.

Given any integer \(n > 1\), the standard factored form of \(n\) is \[n = p_1^{e_1} p_2^{e_2} \cdots p_k^{e_k}\] where \(k \in \mathbb{Z}^+\), the \(p_i\) are primes, the \(e_i\) are positive integers, and \(p_1 < p_2 < \cdots < p_k\).

Both the primes and their exponents are unique — this is what makes the factored form a canonical fingerprint of an integer, and it’s the engine behind divisor counting, gcd and lcm (see gcd-lcm-and-euclidean-algorithm).

Counting divisors from the factorisation

Any positive divisor of \(n = p_1^{e_1} \cdots p_k^{e_k}\) has the form \(p_1^{a_1} \cdots p_k^{a_k}\) with \(0 \leq a_i \leq e_i\), so there are

\[(e_1 + 1)(e_2 + 1) \cdots (e_k + 1)\]

positive divisors.

\(6975 = 3^2 \cdot 5^2 \cdot 31\), found by trial division: \(6975 = 3 \cdot 2325 = 3 \cdot 3 \cdot 775 = 3 \cdot 3 \cdot 5 \cdot 155 = 3^2 \cdot 5^2 \cdot 31\). A divisor has \(0\), \(1\) or \(2\) threes; \(0\), \(1\) or \(2\) fives; and \(0\) or \(1\) thirty-ones — so \(3 \cdot 3 \cdot 2 = 18\) positive divisors.

\(27720 = 2^3 \cdot 3^2 \cdot 5 \cdot 7 \cdot 11\).

\(1400 = 2^3 \cdot 5^2 \cdot 7\), worked through:

  • \(1400 = 2(700) = 2 \cdot 2(350) = 2 \cdot 2(35)(10) = 2^2 (5 \cdot 7)(2 \cdot 5) = 2^3 \cdot 5^2 \cdot 7^1\) — note the primes are written in increasing order.
  • \(1400^2 = (2^3 \cdot 5^2 \cdot 7)^2 = 2^6 \cdot 5^4 \cdot 7^2\) — raising to a power multiplies the exponents.
  • \(1400^3 = 1400 \cdot 1400^2 = 2^9 \cdot 5^6 \cdot 7^3\) — multiplying adds the exponents.
  • Number of positive divisors of \(1400\): \((3+1)(2+1)(1+1) = 24\).
  • Divisors of \(1400\) divisible by \(8\): all factors of the form \(2^3 \cdot 5^* \cdot 7^*\), i.e. \(8, 40, 200, 56, 280, 1400\).
  • Smallest \(n\) with \(1400n\) a perfect square: \(1400n = t^2\) needs every exponent even. \(t^2 = 2^3 \cdot 5^2 \cdot 7^1 \cdot n\), and splitting each power in half (bigger halves on the left) gives \(t = 2^2 \cdot 5^1 \cdot 7^1\) against \(2^1 \cdot 5^1 \cdot 7^0 \cdot n\), so \(n = 2 \cdot 7 = 14\).

Primality testing

Lemma: for every integer \(n > 1\), if no integer \(x\) in the range \(1 < x \leq \sqrt{n}\) divides \(n\), then \(n\) is prime.

This is why trial division only has to go as far as \(\sqrt{n}\) — for \(6975\), only up to \(\sqrt{6975} = 83.51\dots\)

Proof by contraposition. The contrapositive is: \(\forall n \in \mathbb{Z}^{\geq 2}\), if \(n\) is not prime, then there exists an integer \(x\) with \(1 < x \leq \sqrt{n}\) such that \(x \mid n\).

Suppose \(n \in \mathbb{Z}^{\geq 2}\) and \(n\) is not prime. Then \(n\) is composite, so by definition \(\exists r, s \in \mathbb{Z}^+\) with \(n = rs\) and \(1 < r < n\) and \(1 < s < n\).

Case 1: \(r = s\). Then \(r = s = \sqrt{n}\).

Case 2: \(r \neq s\). Then one of \(r, s\) is less than \(\sqrt{n}\) and the other greater — if both were greater we’d have \(rs > n\), and if both were less we’d have \(rs < n\).

In either case there is an \(x \in \{r, s\}\) with \(1 < x \leq \sqrt{n}\) and \(x \mid n\). \(\square\)

There are infinitely many primes

Lemma: for any integer \(a\) and any prime \(p\), if \(p \mid a\) then \(p \nmid (a+1)\).

Proof by contradiction. Let \(a \in \mathbb{Z}\), let \(p\) be prime, suppose \(p \mid a\), and assume for a contradiction that \(p \mid (a+1)\). Since \(p \mid a\) there exists \(k \in \mathbb{Z}\) with \(a = pk\); since \(p \mid (a+1)\) there exists \(\ell \in \mathbb{Z}\) with \(a + 1 = p\ell\). So \[1 = (a+1) - a = p\ell - pk = p(\ell - k),\] and \(\ell - k \in \mathbb{Z}\), so \(p \mid 1\). But \(p \geq 2\) for any prime, whereas \(p \mid 1\) forces \(p = \pm 1\) — a contradiction. \(\square\)

Theorem: there are an infinite number of primes.

Proof by contradiction. Suppose the number of primes is finite. Then some prime \(p\) is the largest, and we can list every prime in ascending order: \(2, 3, 5, 7, 11, \dots, p\).

Let \(N = (2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdots p) + 1\).

Then \(N > 1\), so by the Unique Factorisation Theorem \(N\) is a product of primes and hence divisible by some prime \(q\). Because \(q\) is prime it must be one of \(2, 3, 5, 7, 11, \dots, p\).

So \(q\) divides \(2 \cdot 3 \cdot 5 \cdots p\) — that is, \(q \mid (N - 1)\) — and hence by the lemma \(q \nmid \big((2 \cdot 3 \cdot 5 \cdots p) + 1\big) = N\).

So \(q\) divides \(N\) and \(q\) does not divide \(N\): a contradiction. Therefore the original statement is true. \(\square\)

See also

gcd, lcm & the Euclidean Algorithm

Greatest common divisor

For integers \(a\) and \(b\), not both zero, the greatest common divisor \(\gcd(a,b)\) (also written \(\operatorname{hcf}(a,b)\)) is the integer \(d\) for which:

  • \(d \mid a\) and \(d \mid b\) — \(d\) is a divisor of both; and
  • \(\forall c \in \mathbb{Z}\), if \(c \mid a\) and \(c \mid b\) then \(c \leq d\) — every other common divisor is smaller.

Order doesn’t matter: \(\gcd(a,b) = \gcd(b,a)\).

Negatives. The divisors of \(15\) are \(1, 3, 5, 15, -1, -3, -5, -15\) — exactly the divisors of \(-15\). Since an integer and its negation have the same divisors, \(\gcd(-15, 36) = \gcd(15, 36)\), and in general

\[\gcd(a,b) = \gcd(|a|, |b|)\]

Note \(\gcd(0, b) = b\) for every positive integer \(b\) — this is the base case the Euclidean algorithm terminates on.

Lowest common multiple

For non-zero integers \(a\) and \(b\), the lowest common multiple \(\operatorname{lcm}(a,b)\) is the smallest integer \(n\) for which:

  • \(n > 0\); and
  • \(a \mid n\) and \(b \mid n\) — \(n\) is a multiple of both.

Again \(\operatorname{lcm}(a,b) = \operatorname{lcm}(b,a)\), and since the multiples of \(a\) are the multiples of \(-a\) (\(\dots, -10, -5, 0, 5, 10, 15, \dots\) for both \(5\) and \(-5\)),

\[\operatorname{lcm}(a,b) = \operatorname{lcm}(|a|, |b|)\]

Computing both from the prime factorisation

Using the standard factored form (see divisibility-and-factorisation): take the minimum exponent of each prime for the gcd, the maximum for the lcm.

\[30 = 2 \cdot 3 \cdot 5 \qquad 72 = 2^3 \cdot 3^2\] \[\gcd(30,72) = 2 \cdot 3 = 6 \qquad \operatorname{lcm}(30,72) = 2^3 \cdot 3^2 \cdot 5 = 360\]

Lemma: if \(a, b \in \mathbb{Z}^+\) then \[\gcd(a,b) \cdot \operatorname{lcm}(a,b) = ab\]

Checking: \(\gcd(30,72) \cdot \operatorname{lcm}(30,72) = 6 \cdot 360 = 2160 = 30 \cdot 72\). Each prime’s min and max exponents together account for both original exponents, which is why it works.

More generally, for non-zero integers \(a\) and \(b\): \(\gcd(a,b) \cdot \operatorname{lcm}(a,b) = |ab|\).

Worked

  • \(\gcd(63, -105)\) and \(\operatorname{lcm}(63,-105)\): factor \(63 = 3^2 \cdot 7\) and \(-105 = -3 \cdot 5 \cdot 7\), so \(\gcd = 3 \cdot 7 = 21\) and \(\operatorname{lcm} = 3^2 \cdot 5 \cdot 7 = 315\).
  • For distinct primes \(p, q\): \(\gcd(p,q) = 1\) and \(\operatorname{lcm}(p,q) = pq\).
  • \(\operatorname{lcm}(12,18)\): \(12 = 2^2 \cdot 3\), \(18 = 2 \cdot 3^2\), so \(\operatorname{lcm} = 2^2 \cdot 3^2 = 36\).
  • \(\operatorname{lcm}(2^2 \cdot 3 \cdot 5,\; 2^3 \cdot 3^2) = 2^3 \cdot 3^2 \cdot 5 = 360\).
  • \(\operatorname{lcm}(2800, 6125)\): \(2800 = 2^4 \cdot 5^2 \cdot 7\) and \(6125 = 5^3 \cdot 7^2\), so \(\operatorname{lcm} = 2^4 \cdot 5^3 \cdot 7^2 = 98\,000\).

The Euclidean algorithm

Factorising large numbers is slow; the Euclidean algorithm finds \(\gcd(a,b)\) for \(a, b \in \mathbb{Z}^+\) without it. It rests on this fact:

If \(a, b \in \mathbb{Z}\) are not both zero and \(q, r \in \mathbb{Z}\) satisfy \(a = bq + r\), then \(\gcd(a,b) = \gcd(b,r)\).

Why: suppose \(c \mid a\) and \(c \mid b\); then \(c \mid (a - bq)\) and hence \(c \mid r\). Conversely suppose \(d \mid b\) and \(d \mid r\); then \(d \mid (bq + r)\) and hence \(d \mid a\). So the two pairs have exactly the same common divisors, and therefore the same greatest one.

The method: divide the bigger by the smaller, keep the remainder, repeat — the pair shrinks each round until the remainder is 0, and then \(\gcd(x, 0) = x\).

\(\gcd(5859, 1232)\):

Division Consequence
\(5859 = 1232 \cdot 4 + 931\) \(\gcd(5859,1232) = \gcd(1232,931)\)
\(1232 = 931 \cdot 1 + 301\) \(\gcd(1232,931) = \gcd(931,301)\)
\(931 = 301 \cdot 3 + 28\) \(\gcd(931,301) = \gcd(301,28)\)
\(301 = 28 \cdot 10 + 21\) \(\gcd(301,28) = \gcd(28,21)\)
\(28 = 21 \cdot 1 + 7\) \(\gcd(28,21) = \gcd(21,7)\)
\(21 = 7 \cdot 3 + 0\) \(\gcd(21,7) = \gcd(7,0) = 7\)

So \(\gcd(5859, 1232) = 7\).

\(\gcd(4131, 2431)\):

\[4131 = 2431 \cdot 1 + 1700, \quad 2431 = 1700 \cdot 1 + 731, \quad 1700 = 731 \cdot 2 + 238,\] \[731 = 238 \cdot 3 + 17, \quad 238 = 17 \cdot 14 + 0\]

so \(\gcd(4131,2431) = \gcd(17,0) = 17\).

\(\gcd(14, 3003)\): \(3003 = 14 \cdot 214 + 7\), then \(14 = 7 \cdot 2 + 0\), so the gcd is \(7\).

\(\gcd(931, 301)\): \(931 = 301 \cdot 3 + 28\), \(301 = 28 \cdot 10 + 21\), \(28 = 21 \cdot 1 + 7\), \(21 = 7 \cdot 3 + 0\), so \(\gcd = 7\). Using that, \(7 \mid 931\) and \(931 = 7 \cdot 133 = 7 \cdot 7 \cdot 19 = 7^2 \cdot 19\), which has \((2+1)(1+1) = 6\) positive divisors.

\(\gcd(116, 88)\): \(116 = 88 \cdot 1 + 28\), \(88 = 28 \cdot 3 + 4\), \(28 = 4 \cdot 7 + 0\), so \(\gcd = 4\); hence \(\operatorname{lcm}(116,88) = \frac{116 \cdot 88}{4} = 2552\).

As pseudocode

Input: A, B (integers with A > B >= 0)
a := A; b := B; r := B
while b != 0:
    r := a mod b
    a := b
    b := r
gcd := a
Output: gcd

Trace table for \(A = 1001\), \(B = 871\):

0 1 2 3 4 5
\(a\) 1001 871 130 91 39 13
\(b\) 871 130 91 39 13 0
\(r\) 871 130 91 39 13 0
gcd 13

Proofs about gcd and lcm

For all positive integers \(a\) and \(b\), \(a \mid b\) if and only if \(\operatorname{lcm}(a,b) = b\).

A biconditional can be proved either way round, but proving each direction separately is the recommended method — proving both simultaneously by chaining “iff”s is quicker but easy to get logically wrong (every link in the chain must genuinely be reversible).

(\(\Rightarrow\)) Suppose \(a, b \in \mathbb{Z}^+\) and \(a \mid b\). Then \(b = ka\) for some \(k \in \mathbb{Z}\). So \(\operatorname{lcm}(a,b) = \operatorname{lcm}(a, ka)\). Since \(a \mid ka\) and \(ka \mid ka\), and no smaller \(n > 0\) has \(ka \mid n\), we get \(\operatorname{lcm}(a, ka) = ka = b\).

(\(\Leftarrow\)) Suppose \(\operatorname{lcm}(a,b) = b\). Writing \(g = \gcd(a,b)\) so that \(a = gm\) and \(b = gn\) with \(\gcd(m,n) = 1\), we have \(\operatorname{lcm}(a,b) = gmn\). Then \(gmn = gn\) forces \(m = 1\), so \(a = g\) and therefore \(a \mid b\). \(\square\)

For all positive integers \(a\) and \(b\), \(\gcd(a,b) = \operatorname{lcm}(a,b)\) if and only if \(a = b\).

(\(\Leftarrow\)) If \(a = b\) then both equal \(a\).

(\(\Rightarrow\)) Assume \(\gcd(a,b) = \operatorname{lcm}(a,b)\) and set \(g = \gcd(a,b)\), so \(a = gm\) and \(b = gn\) for positive integers \(m, n\) with \(\gcd(m,n) = 1\). Since \(gm\) and \(gn\) share no prime factors beyond those in \(g\), the lcm uses all primes from both: \(\operatorname{lcm}(a,b) = gmn\). From \(g = gmn\) we get \(mn = 1\), so \(m = n = 1\) and \(a = b = g\). \(\square\)

Extended Euclidean algorithm (non-assessed)

Theorem: for \(a, b, c \in \mathbb{Z}\), if \(\gcd(a,b) \mid c\) then there exist integers \(x\) and \(y\) with \(ax + by = c\).

Running the Euclidean algorithm’s steps in reverse finds them. For \(1232x + 5859y = 14\):

  1. Run the algorithm (as above) to get \(\gcd(1232, 5859) = 7\).
  2. Work backwards from the second-last equation, substituting each remainder in turn:

\[7 = 28 - 21\] \[= 28 - (301 - 28 \cdot 10) = 28 \cdot 11 - 301\] \[= (931 - 301 \cdot 3) \cdot 11 - 301 = 931 \cdot 11 - 301 \cdot 34\] \[= 931 \cdot 45 - 1232 \cdot 34\] \[= (5859 - 1232 \cdot 4) \cdot 45 - 1232 \cdot 34 = 5859 \cdot 45 - 1232 \cdot 214\]

  1. Scale to the target: \(7 = 1232(-214) + 5859(45)\), so \(14 = 1232(-428) + 5859(90)\).

This content is not examinable, but it’s where gcd earns its keep in practice — solving linear Diophantine equations, and modular inverses in cryptography.

Where this gets used

Adding fractions and other everyday arithmetic; deciding whether certain equations have integer solutions; applications in cryptography.

See also

Modular Arithmetic, Floor & the Quotient-Remainder Theorem

Floor and ceiling

For a real number \(x\):

  • \(\lfloor x \rfloor\) — the floor: move left on the number line to the nearest integer.
  • \(\lceil x \rceil\) — the ceiling: move right on the number line to the nearest integer.

So \(\lfloor -3.2 \rfloor = -4\) and \(\lceil -3.2 \rceil = -3\); \(\lfloor 3.2 \rfloor = 3\) and \(\lceil 3.2 \rceil = 4\). Negative numbers are where this goes wrong most often — floor of \(-3.2\) is \(-4\), not \(-3\).

To use floor or ceiling in a proof, it is usually easiest to rewrite the expression as an integer plus an amount between 0 and 1. For instance, in proving “if \(n\) is an odd integer then \(\lceil \frac{n}{2} \rceil = \lfloor \frac{n}{2} \rfloor + 1\)”, write

\[\frac{n}{2} = \frac{2k+1}{2} = k + \frac{1}{2}\]

for some integer \(k\); then \(k + \frac{1}{2}\) sits strictly between \(k\) and \(k+1\), so the floor is \(k\) and the ceiling is \(k+1\). The same move handles things like \(\big\lfloor (p^3 + 4p^2 + 5) + \frac{1}{4} \big\rfloor\) where \(p \in \mathbb{Z}\).

Two ways to compute \(\lfloor \frac{-32}{5} \rfloor\):

  • Fraction approach: \(\lfloor \frac{-32}{5} \rfloor = \lfloor \frac{-30}{5} - \frac{2}{5} \rfloor = \lfloor -6 - \frac{2}{5} \rfloor = -7\) (something smaller than \(-6\)).
  • Decimal approach: \(\lfloor -6.4 \rfloor = \lfloor -6 - 0.4 \rfloor = -7\).

And \(\lceil \frac{-32}{5} \rceil = -6\).

\(\forall x \in \mathbb{R}, \lfloor x^2 \rfloor = \lfloor x \rfloor^2\) is false — take \(x = \frac{3}{2}\): \(\lfloor \frac{9}{4} \rfloor = 2\) but \(\lfloor \frac{3}{2} \rfloor^2 = 1^2 = 1\).

The quotient-remainder theorem

Given any integer \(n\) and positive integer \(d\), there exist unique integers \(q\) and \(r\) such that

\[n = dq + r \quad \text{and} \quad 0 \leq r < d\]

\(q\) is the quotient (\(n \operatorname{div} d\)) and \(r\) is the remainder (\(n \bmod d\)). Explicitly:

\[q = \left\lfloor \frac{n}{d} \right\rfloor \qquad r = n - d\left\lfloor \frac{n}{d} \right\rfloor\]

With \(d = 6\):

\(n\) \(n = dq + r\) \(q\) \(r\)
\(53\) \(6 \cdot 8 + 5\) \(8\) \(5\)
\(20\) \(6 \cdot 3 + 2\) \(3\) \(2\)
\(-19\) \(6 \cdot (-4) + 5\) \(-4\) \(5\)

Negative \(n\) is the case to watch: the remainder is still required to be non-negative, so \(-19 \bmod 6 = 5\), not \(-1\). Likewise \(-95 = 11(-9) + 4\), so \(-95 \operatorname{div} 11 = -9\) and \(-95 \bmod 11 = 4\). And \(58 = 11 \cdot 5 + 3\), so \(58 \operatorname{div} 11 = 5\), \(58 \bmod 11 = 3\).

\(\bmod\) versus \(\equiv\) — take care with notation

These are different statements and the difference is examinable.

  • \(n \bmod d = r\) — an equality between numbers: the remainder when \(n\) is divided by \(d\) is \(r\), with \(0 \leq r < d\).
  • \(m \equiv n \pmod{d}\) — a relation between \(m\) and \(n\): they leave the same remainder when divided by \(d\).

Example: \(5 \bmod 4 = 1\) and \(9 \bmod 4 = 1\), so \(9 \equiv 5 \pmod 4\) — but \(9 \bmod 4 \neq 5\).

Two equivalent definitions of congruence

For \(m, n, d \in \mathbb{Z}\) with \(d > 0\):

  • Definition 1: \(m \equiv n \pmod d\) iff \(m\) and \(n\) have the same remainder under the quotient-remainder theorem.
  • Definition 2: \(m \equiv n \pmod d\) iff \(d \mid (m - n)\).

Definition 2 is usually the faster one to check, and the one to reach for in proofs (see divisibility-and-factorisation).

Worked both ways — is \(-50 \equiv 22 \pmod 8\)?

  • Definition 1: \(-50 = -56 + 6 = 8(-7) + 6\), so \(-50 \bmod 8 = 6\); and \(22 = 16 + 6\), so \(22 \bmod 8 = 6\). Same remainder, so yes.
  • Definition 2: \(22 - (-50) = 72 = 8 \cdot 9\), so \(8 \mid \big(22 - (-50)\big)\), so yes.

Similarly \(53 \equiv -19 \pmod 6\) (both leave remainder 5; and \(53 - (-19) = 72\) is divisible by 6), while \(53 \not\equiv 20 \pmod 6\).

Arithmetic with congruences

Lemma: if \(a \equiv b \pmod d\) and \(m \equiv n \pmod d\), then \[a + m \equiv b + n \pmod d \qquad \text{and} \qquad am \equiv bn \pmod d\]

That is, you may substitute any number for a congruent one before adding or multiplying — which is what makes modular arithmetic cheap.

For example, \(8 \equiv 2 \pmod 3\) and \(46 \equiv 1 \pmod 3\), so \(8 \cdot 46 \equiv 2 \cdot 1 \equiv 2 \pmod 3\) and \(8 + 46 \equiv 2 + 1 \equiv 0 \pmod 3\).

Worked: suppose \(x \bmod 6 = 4\) and \(y \bmod 6 = 3\).

  • \((x + y) \bmod 6\): \(x + y \equiv 4 + 3 \equiv 7 \equiv 1 \pmod 6\), so the answer is \(1\).
  • \((5x^2 + 4y) \bmod 6\): \(5x^2 + 4y \equiv 5(4)^2 + 4(3) \equiv 5(16) + 12 \equiv 5(4) \equiv 20 \equiv 2 \pmod 6\) — applying the lemma at each step.

Worked: if \(m \bmod 6 = 4\), what is \(35m \bmod 6\)? Since \(35 \bmod 6 = 5\), \(35m \equiv 5 \cdot 4 \equiv 20 \equiv 2 \pmod 6\). And \(3m^2 + 2m + 1 \equiv 3(16) + 8 + 1 \equiv 48 + 9 \equiv 0 + 9 \equiv 3 \pmod 6\).

Worked: an integer \(n\) leaves remainder 4 on division by 7 — what is the remainder of \(3n\)? Write \(n = 7k + 4\); then \(3n = 21k + 12 = 7(3k) + 5\), so the remainder is \(5\).

Worked: \(m\) leaves remainder 9 on division by 11 — what remainder does \(4m\) leave on division by 22? Write \(m = 11k + 9\); then \(4m = 44k + 36 = 22(2k+1) + 14\), so the remainder is \(14\). (Note the modulus changed, so you cannot just reduce mod 11.)

Divisibility tests

Modular arithmetic explains the familiar tests. A positive integer is divisible by 3 if and only if the sum of its digits is divisible by 3, because \(10 \equiv 1 \pmod 3\) and hence \(10^k \equiv 1 \pmod 3\) for every \(k \in \mathbb{Z}^+\) (apply the multiplication half of the lemma repeatedly: \(10^2 \equiv 1\) from \(a = m = 10\), \(b = n = 1\); then \(10^3 \equiv 1\) from \(a = 10^2\), \(m = 10\); and so on).

\[254 \equiv 2 \cdot 100 + 5 \cdot 10 + 4 \cdot 1 \equiv 2 + 5 + 4 \equiv 11 \equiv 2 \pmod 3\]

so \(254\) is not divisible by 3 and leaves remainder 2 (checking: \(254 = 3 \cdot 84 + 2\)). Whereas

\[3252 \equiv 3 + 2 + 5 + 2 \equiv 12 \equiv 0 \pmod 3\]

so \(3 \mid 3252\).

Worked proofs

For any odd integer \(n\), \(n^2 \bmod 4 = 1\).

Suppose \(n\) is odd, so \(n = 2k+1\) for some \(k \in \mathbb{Z}\). Then \[n^2 = (2k+1)^2 = 4k^2 + 4k + 1 = 4(k^2 + k) + 1.\] Since \(k^2 + k \in \mathbb{Z}\) and \(0 \leq 1 < 4\), the quotient-remainder theorem gives \(n^2 \bmod 4 = 1\). \(\square\)

Hence for any odd integer \(n\), \(\left\lfloor \frac{n^2}{4} \right\rfloor = \frac{n^2 - 1}{4}\). Two routes:

  • Method 1 — expand both sides. With \(n = 2k+1\): LHS \(= \lfloor \frac{4k^2+4k+1}{4} \rfloor = \lfloor (k^2+k) + \frac{1}{4} \rfloor = k^2 + k\). RHS \(= \frac{(2k+1)^2 - 1}{4} = \frac{4k^2+4k}{4} = k^2 + k\). Equal.
  • Method 2 — use \(r = n - d\lfloor \frac{n}{d} \rfloor\). With number \(n^2\), divisor \(4\), remainder \(1\): \(1 = n^2 - 4\lfloor \frac{n^2}{4} \rfloor\), so \(4\lfloor \frac{n^2}{4} \rfloor = n^2 - 1\) and \(\lfloor \frac{n^2}{4} \rfloor = \frac{n^2-1}{4}\).

For all odd integers \(n\), \(\left\lfloor \frac{n^2}{4} \right\rfloor = \left(\frac{n-1}{2}\right)\left(\frac{n+1}{2}\right)\) — true, since with \(n = 2k+1\) the right-hand side is \(\frac{2k}{2} \cdot \frac{2k+2}{2} = k(k+1)\), matching the \(k^2 + k\) above.

If \(y\) is any integer, then \(y^2\) divided by 5 leaves remainder 0, 1 or 4 — never 2 or 3.

Fix \(y \in \mathbb{Z}\). By the quotient-remainder theorem, \(y = 5m + r\) for some \(m \in \mathbb{Z}\) and \(r \in \{0,1,2,3,4\}\). Then \[y^2 = 25m^2 + 10mr + r^2 \equiv r^2 \pmod 5.\] Checking each case: \(0^2 \equiv 0\), \(1^2 \equiv 1\), \(2^2 \equiv 4\), \(3^2 \equiv 9 \equiv 4\), \(4^2 \equiv 16 \equiv 1 \pmod 5\). So \(y^2 \equiv 0, 1\) or \(4 \pmod 5\). \(\square\)

This “split into cases by remainder” move is the standard way to prove something about every integer using modular arithmetic.

See also

Practice Problems — Applied Class Source

This is the single problem set used in every Applied Class for the whole semester — there is no separate per-week handout. 46 sections, one per pre-work video, each with the relevant textbook pages, a set of practice problems and full worked solutions. Everything is reproduced here in full, so you never need to open the PDF.

Applied Classes start in Week 2 and run to Week 13 — twelve of them, matching the “best 8 of 12” Applied Classes component worth 30% (see math1061). Each week’s class applies the previous week’s lecture content, so the sections in play are roughly one week behind the lecture column below.

How each section is laid out

  1. Relevant textbook pages for Epp 4th edition and 5th (metric) edition
  2. Practice problems
  3. Solutions — full worked solutions, not just answers
  4. Occasionally a non-assessable Application block (see the table below)

Errata

The source document carries an errata page listing corrections made during semester. Last update: 11 August 2026 — on page 195, the solution to 1(b) was corrected (the note in brackets had previously copied the note from 1(a)); that correction is already applied in §45 below. If something looks wrong, post on the Ed Discussion board.

Section index

Every section is reproduced in full below. The Notes column links to the concept note covering that material where one exists; the rest are still ahead in the semester.

§ Topic Lecture (week) 4th ed. 5th ed. Notes
1 Logical Form L2 (wk 1) 023–029 037–043 logical-connectives
2 Logical Equivalence L2 (wk 1) 030–038 043–053 logical-equivalence-laws
3 Conditional Statements L3 (wk 1) 039–050 053–066 conditional-statements
4 Valid and Invalid Arguments L4 (wk 2) 051–063 066–079 valid-argument-forms
5 Methods for Determining Validity L5 (wk 2) 051–063 066–079 determining-argument-validity
6 Quantified Statements L5 (wk 2) 096–108 108–121 quantified-statements
7 Negation of Quantified Statements L6 (wk 2) 108–117 122–131 quantified-statements
8 Statements with Multiple Quantifiers L6 (wk 2) 117–131 131–146 quantified-statements
9 Direct Proofs and Counterexamples L7 (wk 3) 145–163 160–182 proof-techniques
10 Proof by Contradiction L8 (wk 3) 198–207 218–228 proof-techniques
11 Proof by Contraposition L9 (wk 3) 198–207 218–228 proof-techniques
12 Rational Numbers L9 (wk 3) 163–170 183–190 rational-and-irrational-numbers
13 Divisibility L10 (wk 4) 170–179 190–199 divisibility-and-factorisation
14 Modular Arithmetic L11 (wk 4) 482–485 528–531 modular-arithmetic
15 The Euclidean Algorithm L12 (wk 4) 220–224 250–254 gcd-lcm-and-euclidean-algorithm
16 Sequences L13 (wk 5) 227–244 258–275
17 Mathematical Induction L13 (wk 5) 244–268 275–301
18 Strong Induction and the Well Ordering Principle L14 (wk 5) 268–279 301–314
19 Recursive Definitions L15 (wk 6) 290–304 325–340
20 Solving Recurrence Relations L16 (wk 6) 304–328 340–364
21 Set Theory Definitions L17 (wk 6) 336–351 377–391
22 More Definitions and Examples of Sets L17 (wk 6) 336–351 377–391
23 Properties of Sets L18 (wk 7) 352–366 391–407
24 Functions Defined on General Sets L18 (wk 7) 383–396 425–439
25 one-to-one, Onto, and Inverse Functions L19 (wk 7) 397–416 439–461
26 Composition of Functions L20 (wk 7) 416–427 461–472
27 Cardinalities L21 (wk 8) 428–441 473–486
28 Countable and Uncountable Sets L22 (wk 8) 428–441 473–486
29 Relations on Sets L23 (wk 8) 442–449 487–494
30 Reflexivity, Symmetry, Transitivity L23 (wk 8) 449–459 495–505
31 Equivalence Relations L24 (wk 9) 459–478 505–523
32 Partial Order Relations L25 (wk 9) 498–515 546–563
33 Definitions and Examples of Groups L26 (wk 9) — —
34 Elementary Properties of Groups L27 (wk 10) — —
35 Group Isomorphisms not scheduled — —
36 Definitions and Examples of Fields L28 (wk 10) — —
37 Introduction to Counting L29 (wk 11) 516–553 564–604
38 Counting Selections L30 (wk 11) 516–553 564–604
39 Introduction to Probability L31 (wk 11) 516–553 564–604
40 Binomial Coefficients L31 (wk 11) 565–591 617–641
41 Inclusion Exclusion L32 (wk 12) 545–549 595–599
42 The Pigeonhole Principle L33 (wk 12) 554–565 604–616
43 Introduction to Graph Theory L33 (wk 12) 625–660 677–697
44 Walks, Trails and Circuits L34 (wk 12) 625–660 677–697
45 Matrix Representations of Graphs L35 (wk 13) 661–675 698–712
46 Trees L36 (wk 13) 683–701 720–742

§35 Group Isomorphisms has no lecture in the published schedule — the Week 10 row jumps from L27 (V34) to L28 (V36). The section itself carries no practice problems and is marked optional and will not be assessed, so the gap is deliberate.

§28 Countable and Uncountable Sets carries its own note in the source: this content will not be assessed in MATH1061. It still has a full set of problems and solutions.

Sections 33–36 (groups and fields) list no textbook page ranges.

Non-assessable application blocks

Eight sections carry an extra block of problems, not assessable for MATH1061, mostly drawn from K. Rosen, Discrete Mathematics and its Applications, 7th ed. (1991), with two from Epp. They connect the pure material to computer science, and are reproduced in full alongside their sections.

After § Block Source
2 Applications to Computer Science — bit strings and logic circuits Rosen pp. 20–24
5 Applications of Valid and Invalid Arguments Rosen pp. 48–68
9 Applications of Proofs to Computer Science — resolution and automated theorem proving Rosen pp. 74–80
15 Trace Tables for Algorithms — the Division and Euclidean algorithms Epp §4.8 (4th ed.) / §4.10 (5th ed.)
21 Computer Representations of Sets Rosen pp. 134–138
28 Computability of Functions Rosen pp. 170–177
31 Identifiers — C identifiers as an equivalence-relation example Rosen pp. 608–618
37 Runtime — counting basic instructions Epp 5th ed. pp. 787–799

A note on the diagrams: the source PDF draws logic circuits, graphs and Hasse diagrams as figures. Those are reproduced here as explicit gate netlists and edge lists, which carry the same information in a form that renders everywhere.

§1 — Logical Form

Epp 4th ed. pp. 023–029 · 5th ed. (metric) pp. 037–043. Lecture 2 (week 1). Notes: logical-connectives.

Problems

  1. Write a truth table for the statement form \(p \wedge (\sim q \vee r)\).

  2. Suppose you know that \((\sim p \wedge q) \vee p\) is false. What can you conclude about the truth values of each of the two variables?

  3. Let \(p\) be the statement “DATAENDFLAG is off,” \(q\) the statement “ERROR equals 0,” and \(r\) the statement “SUM is less than 1,000.” Express the following sentences in symbolic notation.

    1. DATAENDFLAG is off, ERROR equals 0, and SUM is less than 1,000.

    2. DATAENDFLAG is off but ERROR is not equal to 0.

    3. DATAENDFLAG is off; however, ERROR is not 0 or SUM is greater than or equal to 1,000.

    4. DATAENDFLAG is on and ERROR equals 0 but SUM is greater than or equal to 1,000.

    5. Either DATAENDFLAG is on or it is the case that both ERROR equals 0 and SUM is less than 1,000.

Solutions

  1. The truth table is

    \(p\) \(q\) \(r\) \(\sim q\) \(\sim q \vee r\) \(p \wedge (\sim q \vee r)\)
    T T T F T T
    T T F F F F
    T F T T T T
    T F F T T T
    F T T F T F
    F T F F F F
    F F T T T F
    F F F T T F

  2. We know \((\sim p \wedge q) \vee p\) is false. A disjunction is false only when both parts are false, so we must have

    \[p \text{ is false} \quad \text{and} \quad (\sim p \wedge q) \text{ is false.}\]

    From \(p\) being false, it follows that \(\sim p\) is true. Now \((\sim p \wedge q)\) can be false only if \(q\) is false (since \(\sim p\) is already true). Hence \(q\) is false.

    Therefore \(p\) is false and \(q\) is false.

    1. \(p \wedge q \wedge r\).

    2. \(p \wedge \sim q\).

    3. \(p \wedge (\sim q \vee \sim r)\).

    4. \(\sim p \wedge q \wedge \sim r\).

    5. \(\sim p \vee (q \wedge r)\).

§2 — Logical Equivalence

Epp 4th ed. pp. 030–038 · 5th ed. (metric) pp. 043–053. Lecture 2 (week 1). Notes: logical-equivalence-laws.

Problems

    1. Use the laws of logical equivalence to show that \(p \wedge q \equiv \; \sim (\sim p \vee \sim q)\).

    2. Use the laws of logical equivalence to show that \(\sim (p \vee \sim q) \vee (\sim p \wedge \sim q) \equiv \; \sim p\).

  2. For each of the following, write down a truth table for the statement, and determine whether the statement is a tautology, a contradiction, or neither.

    1. \(((p \wedge q) \vee (q \wedge r)) \vee \sim q\)

    2. \((\sim p \vee q) \vee (p \wedge \sim q)\)

  3. Use De Morgan’s laws to write sentences equivalent to the following:

    1. It is not true that I am studying Computer Science and I am studying Engineering.

    2. I am not going to the movies this weekend or I am not going swimming this weekend.

  4. Let \(\oplus\) denote exclusive or, which means ‘… or … but not both’. The truth table for \(p \oplus q\) is as follows:

    \(p\) \(q\) \(p \oplus q\)
    T T F
    T F T
    F T T
    F F F

    1. Use a truth table to show that \((p \vee q) \wedge \sim p \equiv q \wedge \sim p\).

    2. Use a truth table to show that \((p \oplus q) \wedge r \equiv (p \wedge r) \oplus (q \wedge r)\).

    3. Use the laws of logical equivalence and the equivalence \(p \oplus q \equiv (p \vee q) \wedge \sim (p \wedge q)\) (which is the definition of \(\oplus\)) to show that

    \[(p \oplus q) \wedge r \equiv (p \wedge r) \oplus (q \wedge r).\]

Solutions

    1. We have

    \[\begin{aligned} \sim\big(\sim p \vee \sim q\big) &\equiv \; \sim(\sim p) \wedge \sim(\sim q) && \text{by De Morgan's law} \\ &\equiv p \wedge q && \text{by the double negative law (used twice).} \end{aligned}\]

    1. We have

    \[\begin{aligned} \sim (p \vee \sim q) \vee (\sim p \wedge \sim q) &\equiv (\sim p \wedge \sim \sim q) \vee (\sim p \wedge \sim q) && \text{by De Morgan's law} \\ &\equiv (\sim p \wedge q) \vee (\sim p \wedge \sim q) && \text{by the double negative law} \\ &\equiv \; \sim p \wedge (q \vee \sim q) && \text{by the distributive law} \\ &\equiv \; \sim p \wedge \mathbf{t} && \text{by the negation law} \\ &\equiv \; \sim p && \text{by the identity law.} \end{aligned}\]

    1. We have
    \(p\) \(q\) \(r\) \(p \wedge q\) \(q \wedge r\) \((p \wedge q) \vee (q \wedge r)\) \(\sim q\) \(\big((p \wedge q) \vee (q \wedge r)\big) \vee \sim q\)
    T T T T T T F T
    T T F T F T F T
    T F T F F F T T
    T F F F F F T T
    F T T F T T F T
    F T F F F F F F
    F F T F F F T T
    F F F F F F T T

    Thus the statement is neither a tautology nor a contradiction.

    1. We have
    \(p\) \(q\) \(\sim p\) \(\sim p \vee q\) \(\sim q\) \(p \wedge \sim q\) \((\sim p \vee q) \vee (p \wedge \sim q)\)
    T T F T F F T
    T F F F T T T
    F T T T F F T
    F F T T T F T

    Hence the statement is a tautology.

    1. Let \(C\) be the statement “I am studying Computer Science” and \(E\) be the statement “I am studying Engineering.” The given statement is \(\sim (C \wedge E)\). By De Morgan’s law,

    \[\sim (C \wedge E) \equiv (\sim C) \vee (\sim E).\]

    So an equivalent statement is: “I am not studying Computer Science or I am not studying Engineering.”

    1. Let \(M\) be the statement “I am going to the movies this weekend” and \(S\) be the statement “I am going swimming this weekend.” The given sentence is \((\sim M) \vee (\sim S)\). By De Morgan’s law (in reverse),

    \[(\sim M) \vee (\sim S) \equiv \; \sim (M \wedge S).\]

    So an equivalent statement is: “It is not the case that I am both going to the movies this weekend and going swimming this weekend.”

    1. We have
    \(p\) \(q\) \(p \vee q\) \(\sim p\) \((p \vee q) \wedge \sim p\) \(q \wedge \sim p\)
    T T T F F F
    T F T F F F
    F T T T T T
    F F F T F F

    Since the last two columns agree on every row, we have \((p \vee q) \wedge \sim p \equiv q \wedge \sim p\).

    1. We have
    \(p\) \(q\) \(r\) \(p \oplus q\) \((p \oplus q) \wedge r\) \(p \wedge r\) \(q \wedge r\) \((p \wedge r) \oplus (q \wedge r)\)
    T T T F F T T F
    T T F F F F F F
    T F T T T T F T
    T F F T F F F F
    F T T T T F T T
    F T F T F F F F
    F F T F F F F F
    F F F F F F F F

    The columns for \((p \oplus q) \wedge r\) and \((p \wedge r) \oplus (q \wedge r)\) match on all rows, hence \((p \oplus q) \wedge r \equiv (p \wedge r) \oplus (q \wedge r)\).

    1. To save space, multiple laws have been applied in some steps. There are many different ways to apply the laws to prove this, so your proof may look different to this one. We have

    \[\begin{aligned} &(p \wedge r) \oplus (q \wedge r) \\ &\equiv \big((p \wedge r) \vee (q \wedge r)\big) \wedge \sim \big((p \wedge r) \wedge (q \wedge r)\big) && \text{by the given equivalence (def of } \oplus) \\ &\equiv (p \vee q) \wedge r \wedge \sim \big((p \wedge r) \wedge (q \wedge r)\big) && \text{by distributive and commutative laws (for } \wedge r) \\ &\equiv (p \vee q) \wedge r \wedge \sim \big((p \wedge q) \wedge (r \wedge r)\big) && \text{by associative and commutative laws (for } \wedge) \\ &\equiv (p \vee q) \wedge r \wedge \sim \big((p \wedge q) \wedge r\big) && \text{by idempotent law} \\ &\equiv (p \vee q) \wedge r \wedge \big(\sim (p \wedge q) \vee \sim r\big) && \text{by De Morgan's law} \\ &\equiv (p \vee q) \wedge \Big(\big(r \wedge \sim (p \wedge q)\big) \vee (r \wedge \sim r)\Big) && \text{by distributive law (for } r \wedge) \\ &\equiv (p \vee q) \wedge \Big(\big(r \wedge \sim (p \wedge q)\big) \vee \mathbf{c}\Big) && \text{by negation law} \\ &\equiv (p \vee q) \wedge r \wedge \sim (p \wedge q) && \text{by identity law} \\ &\equiv (p \vee q) \wedge \sim (p \wedge q) \wedge r && \text{by associative and commutative laws (for } \wedge) \\ &\equiv (p \oplus q) \wedge r && \text{by the given equivalence (def of } \oplus) \end{aligned}\]

Application — Applications to Computer Science

Not assessable for MATH1061. Taken from K. Rosen, Discrete Mathematics and its Applications, 7th edition, 1991, pages 20–24.

Information is often represented using binary or bit strings, which are sequences consisting of zeros and ones. The length of a string is the total number of bits in it — for example \(0110110\) is a string of length 7. The logical operators \(\wedge\), \(\vee\) and \(\oplus\) extend to binary strings. The operators AND, OR and XOR are:

\(p\) \(q\) \(p \vee q\) \(p \wedge q\) \(p \oplus q\)
0 0 0 0 0
0 1 1 0 1
1 0 1 0 1
1 1 1 1 0

Propositional logic also applies to the design of computer hardware. A digital logic circuit receives input signals (either 0, the wire being off, or 1, on) and produces output signals which are also bits. Complicated digital logic circuits can be constructed from three basic gates: the OR gate, the NOT gate and the AND gate.

A combinational circuit uses these gates and obeys the following rules.

  1. Two input wires cannot be combined without a gate.
  2. A single input wire can be split partway and directed into two separate gates.
  3. An output wire from a gate can be used as input for another gate.
  4. No output wire from a gate can eventually feed back into that gate.

For example, the logical expression \((p \wedge \sim q) \vee \sim r\) has the combinational circuit

  • \(\text{NOT}(q) \to n_1\)
  • \(\text{AND}(p,\, n_1) \to a_1\)
  • \(\text{NOT}(r) \to n_2\)
  • \(\text{OR}(a_1,\, n_2) \to\) output \((p \wedge \sim q) \vee \sim r\)
Application problems

  1. Find the bitwise OR, bitwise AND and bitwise XOR of each of these pairs of binary strings.

    1. \(1011110\), \(0100001\)

    2. \(11110000\), \(10101010\)

  2. Construct a combinational circuit using the NOT, OR and AND gates that produces the outputs:

    1. \((\sim p \vee \sim r) \wedge q\)

    2. \((p \wedge \sim r) \vee (\sim q \wedge r)\)

  3. Find the output of the following combinational circuit, simplify the expression using logical laws and construct a new circuit for the simplified expression. The circuit is

    • \(\text{NOT}(p) \to n_1\)
    • \(\text{NOT}(q) \to n_2\)
    • \(\text{OR}(p,\, n_2) \to o_1\)
    • \(\text{OR}(n_1,\, n_2) \to o_2\)
    • \(\text{AND}(o_1,\, o_2) \to a_1\)
    • \(\text{OR}(a_1,\, r) \to\) output

    (both \(p\) and \(q\) have their wires split, feeding two gates each).

  4. The logical binary operation NAND, denoted \(\mid\), is defined by

    \[p \mid q \equiv \; \sim (p \wedge q).\]

    1. Write down the truth table for \(\sim (p \mid q)\). Show all working.

    2. Prove, using truth tables or otherwise, that \((p \mid p) \equiv (\sim p)\). Show all working.

    3. Prove, using truth tables or otherwise, that \(((p \mid p) \mid (q \mid q)) \equiv p \vee q\). Show all working.

Application solutions

    1. \[1011110 \text{ OR } 0100001 = 1111111,\] \[1011110 \text{ AND } 0100001 = 0000000,\] \[1011110 \text{ XOR } 0100001 = 1111111.\]

    2. \[11110000 \text{ OR } 10101010 = 11111010,\] \[11110000 \text{ AND } 10101010 = 10100000,\] \[11110000 \text{ XOR } 10101010 = 01011010.\]

    1. The circuit is
    • \(\text{NOT}(p) \to n_1\)
    • \(\text{NOT}(r) \to n_2\)
    • \(\text{OR}(n_1,\, n_2) \to o_1\)
    • \(\text{AND}(o_1,\, q) \to\) output \((\sim p \vee \sim r) \wedge q\)
    1. The circuit is
    • \(\text{NOT}(r) \to n_1\)
    • \(\text{AND}(p,\, n_1) \to a_1\)
    • \(\text{NOT}(q) \to n_2\)
    • \(\text{AND}(n_2,\, r) \to a_2\)
    • \(\text{OR}(a_1,\, a_2) \to\) output \((p \wedge \sim r) \vee (\sim q \wedge r)\)

  3. The given digital logic circuit has output \(\big((p \vee \sim q) \wedge (\sim p \vee \sim q)\big) \vee r\). Using laws of logical equivalence we have

    \[\begin{aligned} \big((p \vee \sim q) \wedge (\sim p \vee \sim q)\big) \vee r &\equiv \big(\sim q \vee (p \wedge \sim p)\big) \vee r && \text{by distributive and commutative laws} \\ &\equiv (\sim q \vee \mathbf{c}) \vee r && \text{by negation law} \\ &\equiv \; \sim q \vee r && \text{by identity law} \end{aligned}\]

    The simplification is \(\sim q \vee r\). The simplified logic circuit is

    • \(\text{NOT}(q) \to n_1\)
    • \(\text{OR}(r,\, n_1) \to\) output \(\sim q \vee r\)
    1. We have
    \(p\) \(q\) \(p \wedge q\) \(p \mid q\) \(\sim (p \mid q)\)
    T T T F T
    T F F T F
    F T F T F
    F F F T F

    Hence \(\sim (p \mid q)\) has the same truth values as \(p \wedge q\) (it is true exactly when both \(p\) and \(q\) are true).

    1. Since \(p \mid p \equiv \; \sim (p \wedge p)\), we have

    \[p \mid p \equiv \; \sim (p \wedge p) \equiv \; \sim p\]

    because \(p \wedge p \equiv p\) (idempotent law). Alternatively, by truth table:

    \(p\) \(p \wedge p\) \(p \mid p\) \(\sim p\)
    T T F F
    F F T T

    The last two columns agree, so \(p \mid p \equiv \; \sim p\).

    1. First note from part (b) that \(p \mid p \equiv \; \sim p\) and \(q \mid q \equiv \; \sim q\). Therefore

    \[\begin{aligned} (p \mid p) \mid (q \mid q) &\equiv (\sim p) \mid (\sim q) && \text{by part (b)} \\ &\equiv \; \sim\big((\sim p) \wedge (\sim q)\big) && \text{by def of NAND} \\ &\equiv p \vee q && \text{by De Morgan's law.} \end{aligned}\]

    Alternatively, by truth table:

    \(p\) \(q\) \(p \mid p\) \(q \mid q\) \((p \mid p) \mid (q \mid q)\) \(p \vee q\)
    T T F F T T
    T F F T T T
    F T T F T T
    F F T T F F

    The last two columns match, hence \((p \mid p) \mid (q \mid q) \equiv p \vee q\).

§3 — Conditional Statements

Epp 4th ed. pp. 039–050 · 5th ed. (metric) pp. 053–066. Lecture 3 (week 1). Notes: conditional-statements.

Problems

  1. Let \(p\), \(q\) and \(r\) be statements. Use the laws of logical equivalence to show that

    \[p \to (q \vee r) \equiv (p \wedge \sim q) \to r.\]

  2. Let \(p\) and \(q\) be statement variables. Are the following statements logically equivalent? Justify your answer.

    • Statement A: \((p \to q) \wedge \sim q\)
    • Statement B: \(\sim (p \wedge q)\)

  3. Write each of the following statements in the form ‘if … then …’.

    1. A sufficient condition for the warranty to be good is that you bought the computer less than a year ago.

    2. Jane gets seasick whenever she is on a boat.

  4. For each of the following, write down a truth table for the statement, and determine whether the statement is a tautology, a contradiction, or neither.

    1. \((\sim p \wedge (p \to q)) \to \sim q\)

    2. \((p \to (q \vee r)) \leftrightarrow ((p \wedge \sim q) \to r)\)

  5. Negate the following two statements.

    1. If it rains, then Sue takes her umbrella.

    2. The cakes burn if the oven temperature is too high.

  6. Recall “a sufficient condition for \(s\) is \(r\)” means \(r\) is a sufficient condition for \(s\), and that “a necessary condition for \(s\) is \(r\)” means \(r\) is a necessary condition for \(s\). Rewrite the following statements in ‘if … then …’ form:

    1. A sufficient condition for John’s team to win the championship is that it wins the rest of its games.

    2. A necessary condition for this computer program to be correct is that it not produce error messages during translation.

Solutions

  1. We have

    \[\begin{aligned} p \to (q \vee r) &\equiv \; \sim p \vee (q \vee r) && \text{by definition of implication} \\ &\equiv (\sim p \vee q) \vee r && \text{by the associative law} \\ &\equiv \; \sim (p \wedge \sim q) \vee r && \text{by De Morgan's law and double negative law} \\ &\equiv (p \wedge \sim q) \to r && \text{by definition of implication.} \end{aligned}\]

  2. Let \(A\) be the statement \((p \to q) \wedge \sim q\) and let \(B\) be the statement \(\sim (p \wedge q)\). Then

    \[\begin{aligned} A &\equiv (\sim p \vee q) \wedge \sim q && \text{by definition of implication} \\ &\equiv (\sim p \wedge \sim q) \vee (q \wedge \sim q) && \text{by distributive and commutative laws} \\ &\equiv (\sim p \wedge \sim q) \vee \mathbf{c} && \text{by negation law} \\ &\equiv \; \sim p \wedge \sim q && \text{by identity law,} \end{aligned}\]

    while

    \[B \equiv \; \sim (p \wedge q) \equiv \; \sim p \vee \sim q \qquad \text{by De Morgan's law.}\]

    Thus Statement A and Statement B are not logically equivalent because, for example, if \(p\) is false and \(q\) is true, then \(A\) is false and \(B\) is true. Hence \(A \not\equiv B\).

    1. If you bought the computer less than a year ago, then the warranty is good.

    2. If Jane is on a boat, then Jane gets seasick.

    1. Solution Option 1. The truth table for the statement is
    \(p\) \(q\) \(\sim p\) \(p \to q\) \(\sim p \wedge (p \to q)\) \(\sim q\) \((\sim p \wedge (p \to q)) \to \sim q\)
    T T F T F F T
    T F F F F T T
    F T T T T F F
    F F T T T T T

    The statement is neither a tautology nor a contradiction.

    Solution Option 2. The same truth table, written under the parts of the statement:

    \(p\) \(q\) \(\sim p\) \(\wedge\) \((p \to q)\) \(\to\) \(\sim q\)
    T T F F T T F
    T F F F F T T
    F T T T T F F
    F F T T T T T
    order ① ④ ② ⑤ ③

    The small numbers show the order in which the columns were completed. Column ④ comes from comparing columns ① and ②. Column ⑤ comes from comparing columns ③ and ④, and shows the truth values for the entire statement. This shows that this statement is neither a tautology nor a contradiction. (You don’t have to number the columns — this is just to help your understanding!)

    1. Solution Option 1. The truth tables for the LHS and RHS are:
    \(p\) \(q\) \(r\) \(q \vee r\) \(p \to (q \vee r)\) \(\sim q\) \(p \wedge \sim q\) \((p \wedge \sim q) \to r\)
    T T T T T F F T
    T T F T T F F T
    T F T T T T T T
    T F F F F T T F
    F T T T T F F T
    F T F T T F F T
    F F T T T T F T
    F F F F T T F T

    Therefore the truth table for the statement is:

    \(p\) \(q\) \(r\) \(\big(p \to (q \vee r)\big) \leftrightarrow \big((p \wedge \sim q) \to r\big)\)
    T T T T
    T T F T
    T F T T
    T F F T
    F T T T
    F T F T
    F F T T
    F F F T

    The statement is a tautology.

    Solution Option 2. Another way to represent the truth table for \((p \to (q \vee r)) \leftrightarrow ((p \wedge \sim q) \to r)\):

    \(p\) \(q\) \(r\) \(p \to\) \((q \vee r)\) \(\leftrightarrow\) \((p \wedge \sim q)\) \(\to r\)
    T T T T T T F T
    T T F T T T F T
    T F T T T T T T
    T F F F F T T F
    F T T T T T F T
    F T F T T T F T
    F F T T T T F T
    F F F T F T F T
    order ② ① ⑤ ③ ④

    Here column ⑤ comes from comparing columns ② and ④. We see (from column ⑤) that this is a tautology.

    1. Let \(R\) be the statement “it rains” and \(U\) be the statement “Sue takes her umbrella.” The given statement is \(R \to U\). Its negation is

    \[R \wedge \sim U,\]

    i.e. “It rains and Sue does not take her umbrella.”

    1. Let \(H\) be the statement “the oven temperature is too high” and \(B\) be the statement “the cakes burn.” The given statement is \(H \to B\). Its negation is

    \[H \wedge \sim B,\]

    i.e. “The oven temperature is too high and the cakes do not burn.”

    1. If John’s team wins the rest of its games, then John’s team will win the championship.

    2. If this computer program is correct, then it does not produce error messages during translation.

§4 — Valid and Invalid Arguments

Epp 4th ed. pp. 051–063 · 5th ed. (metric) pp. 066–079. Lecture 4 (week 2). Notes: valid-argument-forms.

Problems

  1. State truth values for the statement variables \(p\) and \(q\) that demonstrate that the following argument is invalid, and justify your answer.

    Premise 1: \(q\) Premise 2: \(p \to q\) Conclusion: \(p\)

  2. Find values of the statement variables \(a\), \(b\) and \(c\) that show that the following argument is invalid.

    1. \(a \to b\)
    2. \(\sim a \to c\)
    3. \(\sim b \to c\) \(\therefore\ b\)

  3. Use a truth table to show that the following argument forms are valid.

    1. Elimination:
    1. \(p \vee q\)
    2. \(\sim q\) \(\therefore\ p\)
    1. Generalisation:
    1. \(p\) \(\therefore\ p \vee q\)

  4. Use a truth table to show that the following argument is invalid.

    1. \((p \wedge q) \to \sim r\)
    2. \(p \vee \sim q\)
    3. \(\sim q \to p\) \(\therefore\ \sim r\)

Solutions

  1. To show the argument is invalid, we must find truth values for \(p\) and \(q\) such that the premises are true and the conclusion is false.

    Suppose \(p\) is false and \(q\) is true. Then Premise 1 (\(q\)) is true and Premise 2 (\(p \to q\)) is true (by definition of implication). However, the conclusion (\(p\)) is false. Therefore the argument is invalid.

  2. To show the argument is invalid, we must find truth values for \(a\), \(b\) and \(c\) such that the premises are true and the conclusion is false.

    Let \(a\) be false, \(b\) be false, and \(c\) be true. By definition of implication, all three premises are true, but the conclusion (\(b\)) is false, so the argument is invalid.

    \(p\) \(q\) premise 1: \(p \vee q\) premise 2: \(\sim q\) conclusion: \(p\)
    T T T F T
    T F T T T
    F T T F F
    F F F T F

    There is only one row of the truth table in which all premises are true (the second row), and here the conclusion is true. Thus, whenever all the premises are true, the conclusion is true, so the argument is valid.

    \(p\) \(q\) premise 1: \(p\) conclusion: \(p \vee q\)
    T T T T
    T F T T
    F T F T
    F F F F

    There are two rows for which all premises (here just the one premise) are true — the first two rows — and for each of these rows the conclusion is true. Thus, whenever all the premises are true, the conclusion is true, so the argument is valid.

  4. We compute the truth table:

    \(p\) \(q\) \(r\) \((p \wedge q) \to \sim r\) \(p \vee \sim q\) \(\sim q \to p\) Prem. 1 \(\wedge\) Prem. 2 \(\wedge\) Prem. 3 \(\sim r\)
    T T T F T T F F
    T T F T T T T T
    T F T T T T T F
    T F F T T T T T
    F T T T F T F F
    F T F T F T F T
    F F T T T F F F
    F F F T T F F T

    There are three rows for which all the premises are true, however for one of these rows the conclusion is false. In particular, taking \(p\) to be true, \(q\) to be false, and \(r\) to be true makes all the premises true but the conclusion false, so the argument is invalid.

§5 — Methods for Determining Validity

Epp 4th ed. pp. 051–063 · 5th ed. (metric) pp. 066–079. Lecture 5 (week 2). Notes: determining-argument-validity.

Problems

  1. Determine whether the following arguments are valid. If the argument is valid, prove it using rules of inference. If the argument is invalid, demonstrate it with a choice of truth values for the variables.

    \(p \to q\) \(q \to p\) \(\therefore\ p \vee q\)

    \(p\) \(p \to q\) \(\sim q \vee r\) \(\therefore\ r\)

    \(p \vee q\) \(p \to \sim q\) \(p \to r\) \(\therefore\ r\)

  2. Write the following argument in symbolic form. Then determine whether the argument is valid or not.

    An error message is received if the output is flagged. The system is running normally only if the output is not flagged. An error message is received. Therefore, the system is not running normally.

    Please use the following variables:

    \(p\) the output is flagged
    \(q\) an error message is received
    \(r\) the system is running normally

  3. Give truth values for the variables \(p\), \(q\), \(r\), \(s\) and \(w\) which demonstrate that the following argument is invalid.

    \[\big[(p \vee q) \wedge (q \to r) \wedge ((p \wedge s) \to w) \wedge (\sim r) \wedge (q \to s)\big] \to w\]

Solutions

    1. The argument is invalid. When \(p\) and \(q\) are both false, all the premises are true, but the conclusion is false. One could arrive at this example by computing the following truth table:
    \(p\) \(q\) \(p \to q\) \(q \to p\) \(p \vee q\)
    T T T T T
    T F F T T
    F T T F T
    F F T T F

    (In the last row, the premises are true, but the conclusion is false.)

    1. The argument is valid, and we prove this with rules of inference:
    1. \(p\) Premise
    2. \(p \to q\) Premise
    3. \(\sim q \vee r\) Premise
    4. \(q\) from 1 and 2 by Modus Ponens
    5. \(r\) from 3 and 4 by elimination (and the double negative law)

    Therefore the argument is valid.

    1. Suppose the argument is invalid. Then it is possible to find truth values for the statement variables that make all the premises true and the conclusion false. For the conclusion to be false, we have \(r\) is false. Since \(r\) is false but the third premise (\(p \to r\)) is true, we must have \(p\) is false. Since \(p\) is false, this means the second premise (\(p \to \sim q\)) is also true. Now since \(p\) is false but the first premise (\(p \vee q\)) is true, we must have \(q\) is true.

    Thus, with these truth values for the three variables, we have that all the premises are true, but the conclusion is false. So the argument is invalid.

  2. The argument has the symbolic form:

    1. \(p \to q\) (An error message is received if the output is flagged.)
    2. \(r \to \sim p\) (The system is running normally only if the output is not flagged.)
    3. \(q\) (An error message is received.)
    \(\therefore\) \(\sim r\) (The system is not running normally.)

    Suppose the argument is invalid. Then it is possible to find truth values for the statement variables that make all the premises true and the conclusion false. For the conclusion to be false, we have \(r\) is true. Since \(r\) is true, and premise 2 is true, we must have \(\sim p\) true, that is \(p\) is false. Since premise 3 is true, we have \(q\) is true.

    Thus, with these truth values for the three variables, we have that all the premises are true, but the conclusion is false. So the argument is invalid.

  3. The argument being valid is the same as this overall implication statement being a tautology.

    \[\big[(p \vee q) \wedge (q \to r) \wedge ((p \wedge s) \to w) \wedge (\sim r) \wedge (q \to s)\big] \to w\]

    We can write the argument as follows:

    1. \(p \vee q\)
    2. \(q \to r\)
    3. \((p \wedge s) \to w\)
    4. \(\sim r\)
    5. \(q \to s\) \(\therefore\ w\)

    Suppose the argument is invalid. Then it is possible to find truth values for the statement variables that make all the premises true and the conclusion false. For the conclusion to be false, we have \(w\) is false. Since premise 4 is true we have \(r\) is false. Since premise 2 is true and \(r\) is false, we must have \(q\) is false. Since premise 1 is true and \(q\) is false we have \(p\) is true. Since premise 3 is true and \(w\) is false, we must have \((p \wedge s)\) false, and since \(p\) is true, this means \(s\) is false. These truth values also mean premise 5 is true.

    Thus with these truth values for the variables, we have that all the premises are true, but the conclusion is false. So the argument is invalid.

Application — Applications of Valid and Invalid Arguments

Not assessable for MATH1061. Taken from K. Rosen, Discrete Mathematics and its Applications, 7th edition, 1991, pages 48–68.

Application problems

  1. Translating English sentences into logical expressions is an essential part of specifying both hardware and software systems. Software engineers must take requirements in English and produce precise, unambiguous specifications that can be used as the basis for system development.

    Consider the following predicates and domains:

    Predicates:

    \(F(x, y)\) disk \(x\) has more than \(y\) kilobytes of free space.
    \(S(m)\) mail message \(m\) can be saved.
    \(A(x)\) alert \(x\) is active.
    \(Q(m)\) message \(m\) is queued.
    \(T(m)\) message \(m\) is transmitted.
    \(D(s)\) the diagnostic monitor tracks the status of system \(s\).

    Domains:

    \(M\) the nonempty set of all messages.
    \(X\) the set of all disks.
    \(Y\) the set of all alerts.
    \(S\) the set of all systems.

    Express the system specifications below using quantifiers and logical connectives:

    1. At least one mail message, among the nonempty set of messages, can be saved if there is a disk with more than 10 kilobytes of free space.

    2. Whenever there is an active alert, all queued messages are transmitted.

    3. The diagnostic monitor tracks the status of all systems except the main console.

  2. System specifications should be consistent; that is, they should not contain conflicting requirements that could be used to derive a contradiction. When specifications are not consistent, there would be no way to develop a system that satisfies all specifications. Logically, a system is consistent if it is possible for all the premises to be simultaneously true.

    Use logical expressions and methods of logical deduction to determine whether the following specifications are consistent.

    1. The system is in multiuser state if and only if it is operating normally. If the system is operating normally, the kernel is functioning. The kernel is not functioning or the system is in interrupt mode. If the system is not in multiuser state, then it is in interrupt mode. The system is not in interrupt mode.

    2. The router can send packets to the edge system only if it supports the new address space. For the router to support the new address space it is necessary that the latest software release be installed. The router can send packets to the edge system if the latest software release is installed. The router does not support the new address space.

Application solutions

    1. \((\exists x \in X \text{ s.t. } F(x, 10)) \to (\exists m \in M \text{ s.t. } S(m))\).

    2. \((\exists y \in Y \text{ s.t. } A(y)) \to (\forall m \in M,\ (Q(m) \to T(m)))\).

    3. Let \(c \in S\) denote the main console. \(\forall s \in S,\ (s \neq c \to D(s))\).

    1. Define the following variables:
    \(m\) the system is in multiuser state
    \(n\) the system is operating normally
    \(k\) the kernel is functioning
    \(i\) the system is in interrupt mode

    Now we can express the specifications as follows:

    1. \(m \leftrightarrow n\)
    2. \(n \to k\)
    3. \(\sim k \vee i\)
    4. \(\sim m \to i\)
    5. \(\sim i\)

    Solution Option 1. If these specifications are consistent, then we must be able to find truth values for each of the variables that makes all of the specifications true. Assume the specifications are consistent (so we assume each specification is true). For specification 5 to be true we know \(i\) is false. For specification 4 to be true, with \(i\) false, we know \(\sim m\) is false so \(m\) is true. For premise 1 to be true, with \(m\) true, we have \(n\) is true. For specification 2 to be true, and since \(n\) is true, we have \(k\) is true. However, now \(\sim k\) is false and \(i\) is false, which means specification 3 is false. This contradicts our assumption that all the specifications are true, and so it is impossible for the system to be consistent. Therefore, the system is not consistent.

    Solution Option 2. Suppose the specifications are all true. Using laws of inference, we have

    1. \(m \leftrightarrow n\) specification
    2. \(n \to k\) specification
    3. \(\sim k \vee i\) specification
    4. \(\sim m \to i\) specification
    5. \(\sim i\) specification
    6. \(\sim k\) from 3 and 5 by elimination (and double negative law)
    7. \(\sim n\) from 2 and 6 by Modus Tollens
    8. \((m \to n) \wedge (n \to m)\) from 1 by definition of \(\leftrightarrow\)
    9. \(m \to n\) from 8 by specialisation
    10. \(\sim m\) from 9 and 7 by Modus Tollens
    11. \(i\) from 4 and 10 by Modus Ponens

    But now we have both \(i\) and \(\sim i\) being true, which is a contradiction. This contradicts our assumption that all the specifications are true, and so it is impossible for the system to be consistent. Therefore, the system is not consistent.

    1. Define the following variables:
    \(r\) the router can send packets to the edge system
    \(a\) the router supports the new address space
    \(s\) the latest software release is installed

    Now we can express the specifications as follows:

    1. \(r \to a\)
    2. \(a \to s\)
    3. \(s \to r\)
    4. \(\sim a\)

    Assume the specifications are consistent, so each specification is true. For specification 4 to be true, \(\sim a\) is true, so \(a\) is false. For specification 1 to be true with \(a\) false we must have \(r\) is false. Now, since \(r\) is false and specification 3 is true, we have \(s\) is false. We double-check that specification 2 is true, which it is because \(a\) is false. Thus, we have found truth values of the statement variables that make all of the specifications true, so the system is consistent.

§6 — Quantified Statements

Epp 4th ed. pp. 096–108 · 5th ed. (metric) pp. 108–121. Lecture 5 (week 2). Notes: quantified-statements.

Problems

  1. Rewrite each of the following in the form “\(\forall \underline{\phantom{xx}},\ \underline{\phantom{xx}}\).” Here, let \(D\) be the set of all dinosaurs, and let \(L\) be the set of all logicians.

    1. All dinosaurs are extinct.

    2. Every real number is positive, negative, or zero.

    3. No irrational numbers are integers.

    4. No logicians are lazy.

    5. The number \(2{,}147{,}581{,}953\) is not equal to the square of any integer.

    6. The number \(-1\) is not equal to the square of any real number.

  2. Rewrite each of the following in the form “\(\exists \underline{\phantom{xx}}\) such that \(\underline{\phantom{xx}}\).” Here, let \(X\) be the set of all exercises.

    1. Some exercises have answers.

    2. Some real numbers are rational.

  3. Determine whether the following statements are true or false. Justify your reasoning.

    1. \(\forall x \in \mathbb{R},\ x \geq \frac{1}{x}\).

    2. \(\forall a \in \mathbb{Z},\ \frac{a-1}{a} \notin \mathbb{Z}\).

    3. \(\forall m, n \in \mathbb{Z}^+,\ mn \geq m + n\).

Solutions

    1. \(\forall x \in D,\ x\) is extinct.

    2. \(\forall x \in \mathbb{R},\ (x > 0 \vee x < 0 \vee x = 0)\).

    3. \(\forall x \in \mathbb{R},\ (x \text{ is irrational} \to x \notin \mathbb{Z})\).

    4. \(\forall x \in L,\ x\) is not lazy.

    5. \(\forall x \in \mathbb{Z},\ x^2 \neq 2147581953\).

    6. \(\forall x \in \mathbb{R},\ -1 \neq x^2\).

    1. \(\exists x \in X\) such that \(x\) has an answer.

    2. \(\exists x \in \mathbb{R}\) such that \(x \in \mathbb{Q}\).

    1. False. Counterexample: take \(x = \frac{1}{2}\). Then \(x = \frac{1}{2}\) but \(\frac{1}{x} = 2\), so \(x \geq \frac{1}{x}\) is false.

    2. False. Counterexample: take \(a = 1\). Then \(\frac{a-1}{a} = \frac{0}{1} = 0 \in \mathbb{Z}\).

    3. False. Counterexample: take \(n = m = 1\). Then \(mn = 1\) but \(m + n = 2\).

§7 — Negation of Quantified Statements

Epp 4th ed. pp. 108–117 · 5th ed. (metric) pp. 122–131. Lecture 6 (week 2). Notes: quantified-statements.

Problems

  1. Write the negation of each of the following statements, and then determine whether the original or the negation is true:

    1. \(\forall x \in \mathbb{R}\), if \(x^2 > 9\) then \(x \geq 2\).

    2. \(\forall n \in \mathbb{Z}^+\), \(n\) is either prime or composite.

    3. \(\exists x, y \in \mathbb{R}\) such that \(x \notin \mathbb{Z}\) and \(y \notin \mathbb{Z}\) but \(xy \in \mathbb{Z}\).

  2. Write down the negation of each of the following statements:

    1. All fire engines are red.

    2. Some apples are ripe.

    3. A triangle with equal sides exists.

Solutions

    1. Original: \(\forall x \in \mathbb{R}\), if \(x^2 > 9\) then \(x \geq 2\). Negation: \(\exists x \in \mathbb{R}\) such that \(x^2 > 9\) and \(x < 2\).

    The original statement is false. For instance, take \(x = -4\). Then \(x^2 > 9\) but \(x \not\geq 2\).

    1. Original: \(\forall n \in \mathbb{Z}^+\), \(n\) is either prime or composite. Negation: \(\exists n \in \mathbb{Z}^+\) such that \(n\) is neither prime nor composite.

    The original statement is false. For instance, consider \(n = 1\), which is neither prime nor composite.

    1. Original: \(\exists x, y \in \mathbb{R}\) such that \(x \notin \mathbb{Z}\) and \(y \notin \mathbb{Z}\) but \(xy \in \mathbb{Z}\). Negation: \(\forall x, y \in \mathbb{R}\), if \(x \notin \mathbb{Z}\) and \(y \notin \mathbb{Z}\) then \(xy \notin \mathbb{Z}\).

    The original statement is true. For instance, take \(x = \sqrt{2}\) and \(y = \sqrt{2}\). Then \(x \notin \mathbb{Z}\) and \(y \notin \mathbb{Z}\), but \(xy = 2 \in \mathbb{Z}\).

    1. Original: All fire engines are red. Negation: There exists a fire engine which is not red.

    2. Original: Some apples are ripe. Negation: All apples are not ripe.

    3. Original: A triangle with equal sides exists. Negation: There does not exist a triangle with equal sides.

§8 — Statements with Multiple Quantifiers

Epp 4th ed. pp. 117–131 · 5th ed. (metric) pp. 131–146. Lecture 6 (week 2). Notes: quantified-statements.

Problems

  1. Write down the negation of each of the following statements. Then determine whether the statement is true, or whether its negation is true.

    1. \(\forall x, y \in \mathbb{Z}\), \(\exists z \in \mathbb{Z}\) such that \(z = x - y\).

    2. \(\forall x \in \mathbb{R}\), \(\exists y \in \mathbb{R}\) such that \(x^2 - xy - 2y^2 = 0\).

    3. \(\exists x \in \mathbb{Z}^+\) such that \(\forall y \in \mathbb{Z}^+\), \(x \geq y\).

    4. \(\forall r, s \in \mathbb{R}\), if \(rs \in \mathbb{Q}\), then \(r \in \mathbb{Q}\) and \(s \in \mathbb{Q}\).

  2. For each of the following English sentences:

    1. For any rational number \(x\) and any negative integer \(y\) there exists a real number that is greater than the product of \(x\) and \(y\).

    2. For every real number \(x\), there is a real number \(y\) for which \(3y = x\).

    do the following:

    1. Write the statement in symbolic form.

    2. Write the negation of the statement in symbolic form.

    3. Determine which of the statement or its negation is true.

Solutions

    1. Original: \(\forall x, y \in \mathbb{Z}\), \(\exists z \in \mathbb{Z}\) such that \(z = x - y\). Negation: \(\exists x \in \mathbb{Z}\ \exists y \in \mathbb{Z}\) such that \(\forall z \in \mathbb{Z}\), \(z \neq x - y\).

    The original statement is true (and the negation is false). For any given \(x, y \in \mathbb{Z}\), choose \(z = x - y \in \mathbb{Z}\).

    1. Original: \(\forall x \in \mathbb{R}\), \(\exists y \in \mathbb{R}\) such that \(x^2 - xy - 2y^2 = 0\). Negation: \(\exists x \in \mathbb{R}\) s.t. \(\forall y \in \mathbb{R}\), \(x^2 - xy - 2y^2 \neq 0\).

    The original statement is true (and the negation is false). For any \(x \in \mathbb{R}\), taking \(y = \frac{x}{2}\) gives

    \[x^2 - x\left(\frac{x}{2}\right) - 2\left(\frac{x}{2}\right)^2 = x^2 - \frac{x^2}{2} - 2 \cdot \frac{x^2}{4} = x^2 - \frac{x^2}{2} - \frac{x^2}{2} = 0.\]

    1. Original: \(\exists x \in \mathbb{Z}^+\) such that \(\forall y \in \mathbb{Z}^+\), \(x \geq y\). Negation: \(\forall x \in \mathbb{Z}^+\ \exists y \in \mathbb{Z}^+\) such that \(x < y\).

    The negation is true (and the original is false). For any \(x \in \mathbb{Z}^+\), take \(y = x + 1\). Then \(y\) is an integer satisfying the inequality \(x < y\).

    1. Original: \(\forall r, s \in \mathbb{R}\), if \(rs \in \mathbb{Q}\), then \(r \in \mathbb{Q}\) and \(s \in \mathbb{Q}\). Negation: \(\exists r, s \in \mathbb{R}\) such that \(rs \in \mathbb{Q}\) and (\(r \notin \mathbb{Q}\) or \(s \notin \mathbb{Q}\)).

    The negation is true (and the original is false). Take \(r = \sqrt{2}\) and \(s = \sqrt{2}\). Then \(rs = 2 \in \mathbb{Q}\) but \(r, s \notin \mathbb{Q}\).

      1. Original: \(\forall x \in \mathbb{Q},\ \forall y \in \mathbb{Z}^{<0},\ \exists z \in \mathbb{R}\) s.t. \(z > xy\).

    1. Negation: \(\exists x \in \mathbb{Q}\ \exists y \in \mathbb{Z}^{<0}\) s.t. \(\forall z \in \mathbb{R}\), \(z \leq xy\).

    2. The original statement is true. For any given \(x \in \mathbb{Q}\) and \(y \in \mathbb{Z}^{<0}\), take \(z = xy + 1\). This is a real number satisfying the inequality \(z > xy\).

      1. Original: \(\forall x \in \mathbb{R},\ \exists y \in \mathbb{R}\) s.t. \(3y = x\).

    1. Negation: \(\exists x \in \mathbb{R}\) s.t. \(\forall y \in \mathbb{R}\), \(3y \neq x\).

    2. The original statement is true. Given any \(x \in \mathbb{R}\), choose \(y = \frac{x}{3}\). This is a real number satisfying the equality \(3y = x\).

§9 — Direct Proofs and Counterexamples

Epp 4th ed. pp. 145–163 · 5th ed. (metric) pp. 160–182. Lecture 7 (week 3). Notes: proof-techniques.

Problems

  1. For each of the following statements, either prove the statement or disprove it using a counterexample.

    1. The sum of any pair of even integers is even.

    2. There exist positive integers \(a\), \(b\) and \(c\) such that \(a^2 + b^2 = c^2\).

    3. There is an even integer \(n\) such that \(5n - 4\) is prime.

    4. \(\forall n \in \mathbb{Z}^+\), if \(n \geq 4\) then \(2n^2 - 5n + 2\) is composite.

    5. \(\forall x, y \in \mathbb{R}\), if \(y^2 > x^2\) then \(y > x\).

  2. Find a counterexample that disproves the following statement:

    \[\text{For all } x \in \mathbb{R},\ \lfloor x^2 \rfloor = \lfloor x \rfloor^2.\]

    (Recall that \(\lfloor \cdot \rfloor\) denotes the floor function.)

  3. Find the errors in the following proofs and rewrite the proof correctly. Here is a list of common errors made when writing proofs:

    • Arguing from examples
    • Using the same variable to mean two different things
    • Jumping to the conclusion
    • Assuming what is to be proved
    • Confusion between what is known and what is to be shown
    • Use of the word any instead of some
    • Misuse of the word if
    1. The difference between any odd integer and any even integer is odd.

    Proof. Let \(m\) be any odd integer, then \(m = 2k + 1\) where \(k \in \mathbb{Z}\) and let \(n\) be any even integer, so \(n = 2k\) for any \(k \in \mathbb{Z}\). Then \(m - n = 2k + 1 - 2k = 1\) and 1 is odd so the statement is true.

    1. If \(a\) and \(b\) are even integers, then \(4a + 2b\) is even.

    Proof. We have \(a = 2k\) and \(b = 2j\) for some \(k, j \in \mathbb{Z}\). Then we want to show that \(4a + 2b\) is even, so \[\begin{aligned} 4a + 2b &= 2\ell \\ 4(2k) + 2(2j) &= 2\ell \\ 8k + 4j &= 2\ell \\ 2(4k + 2j) &= 2\ell \end{aligned}\] therefore the statement is true.

    1. \(\forall x \in \mathbb{R}\), \(\exists y \in \mathbb{R}\) such that \(xy = 1\).

    Proof. This is true, for example if \(x = 10\) then let \(y = \frac{1}{10}\) and \(xy = 10 \cdot \frac{1}{10} = 1\). Therefore the statement is true.

    1. \(\forall n \in \mathbb{Z}\), if \(n > 0\) then \(n^2 + 2n + 1\) is composite.

    Proof. Suppose \(n\) is an integer and \(n > 0\), then we want to show that \(n^2 + 2n + 1\) is composite, so this means \[n^2 + 2n + 1 = rs\] where \[1 < r, s < n^2 + 2n + 1\] Clearly neither \(r\) nor \(s\) are equal to 1 so \(n^2 + 2n + 1\) is not prime and hence must be composite.

Solutions

    1. The sum of any pair of even integers is even.

    Proof. Let \(m, n \in \mathbb{Z}\) be even. Then \(m = 2k\) and \(n = 2\ell\) for some \(k, \ell \in \mathbb{Z}\). Hence \[m + n = 2k + 2\ell = 2(k + \ell),\] which is even. Therefore, the sum of any pair of even integers is even. \(\square\)

    1. There exist positive integers \(a\), \(b\) and \(c\) such that \(a^2 + b^2 = c^2\).

    Proof. Consider the positive integers \(a = 3\), \(b = 4\), \(c = 5\). Then \(a^2 + b^2 = 9 + 16 = 25 = c^2\). \(\square\)

    1. There is an even integer \(n\) such that \(5n - 4\) is prime. This statement is false. We prove the negation: \(\forall n \in \mathbb{Z}^{\text{even}}\), \(5n - 4\) is not prime.

    Proof. Suppose that \(n\) is an even integer. Thus \(n = 2k\) for some integer \(k\). Hence \[5n - 4 = 5(2k) - 4 = 10k - 4 = 2(5k - 2).\] By definition, this means that \(5n - 4\) is even. The only even prime is 2. If \(2(5k-2) = 2\), then \(5k - 2 = 1\) and hence \(k = \frac{3}{5}\) which is not an integer. Therefore \(5n - 4\) is not a prime number. \(\square\)

    1. \(\forall n \in \mathbb{Z}^+\), if \(n \geq 4\) then \(2n^2 - 5n + 2\) is composite.

    Proof. Suppose \(n\) is a positive integer and \(n \geq 4\). Let us factor \[2n^2 - 5n + 2 = (2n - 1)(n - 2).\] Since \(n \geq 4\), we have \(2n - 1 \geq 7\) and \(n - 2 \geq 2\), so both factors are integers \(> 1\). Hence the expression is composite for all \(n \geq 4\). \(\square\)

    1. \(\forall x, y \in \mathbb{R}\), if \(y^2 > x^2\) then \(y > x\). This statement is false. Take \(x = 1\) and \(y = -2\). Then \(y^2 = 4 > x^2 = 1\), but \(y = -2 \not> 1 = x\).

  2. Take \(x = \frac{3}{2}\). Then \(\lfloor x^2 \rfloor = \left\lfloor \frac{9}{4} \right\rfloor = 2\), while \(\lfloor x \rfloor^2 = \left\lfloor \frac{3}{2} \right\rfloor^2 = 1^2 = 1\). Hence \(\lfloor x^2 \rfloor \neq \lfloor x \rfloor^2\), so the statement is false.

    1. Statement. The difference between any odd integer and any even integer is odd.

    Proof. Let \(m\) be any odd integer, then \(m = 2k + 1\) where \(k \in \mathbb{Z}\) and let \(n\) be any even integer, so \(n = 2k\) for any \(k \in \mathbb{Z}\). [Error: using \(k\) for both \(m\) and \(n\) means that the statement is only proved when \(m = n + 1\).] Then \(m - n = 2k + 1 - 2k = 1\) and 1 is odd so the statement is true.

    Correct proof. Let \(m\) be any odd integer and let \(n\) be any even integer. Then there exist \(k, \ell \in \mathbb{Z}\) such that \(m = 2k + 1\) and \(n = 2\ell\). Hence \[m - n = (2k + 1) - 2\ell = 2(k - \ell) + 1,\] which is an odd integer by definition. \(\square\)

    1. Statement. If \(a\) and \(b\) are even integers, then \(4a + 2b\) is even. [Error: the proof forgets to state its assumptions.]

    Proof. We have \(a = 2k\) and \(b = 2j\) for some \(k, j \in \mathbb{Z}\). Then we want to show that \(4a + 2b\) is even, so \[\begin{aligned} 4a + 2b &= 2\ell && \textbf{[Error: this line assumes what is to be proved.]} \\ 4(2k) + 2(2j) &= 2\ell \\ 8k + 4j &= 2\ell \\ 2(4k + 2j) &= 2\ell \end{aligned}\] therefore the statement is true.

    Correct proof. Suppose \(a\) and \(b\) are even integers. Then \(a = 2k\) and \(b = 2j\) for some \(k, j \in \mathbb{Z}\). Therefore \[4a + 2b = 4(2k) + 2(2j) = 8k + 4j = 2(4k + 2j).\] Since \(k, j \in \mathbb{Z}\), we have \((4k + 2j) \in \mathbb{Z}\) and so, by definition, \(4a + 2b\) is even. \(\square\)

    1. Statement. \(\forall x \in \mathbb{R}\), \(\exists y \in \mathbb{R}\) such that \(xy = 1\).

    Proof. This is true, for example if \(x = 10\) then let \(y = \frac{1}{10}\) and \(xy = 10 \cdot \frac{1}{10} = 1\). Therefore the statement is true. [Error: the proof argues from a single example, which cannot establish a universal statement. Moreover, the statement is false.]

    Correct disproof. The statement is false. As a counterexample, let \(x = 0\). We know that \(0 \cdot y = 0\) for all \(y \in \mathbb{R}\) and so there is no real number \(y\) such that \(0 \cdot y = 1\). \(\square\)

    1. Statement. \(\forall n \in \mathbb{Z}\), if \(n > 0\) then \(n^2 + 2n + 1\) is composite.

    Proof. Suppose \(n\) is an integer and \(n > 0\), then we want to show that \(n^2 + 2n + 1\) is composite, so this means \[n^2 + 2n + 1 = rs \qquad \textbf{[Error: forgets to define } r, s \textbf{ — e.g. for some } r, s \in \mathbb{Z}\textbf{]}\] where \[1 < r, s < n^2 + 2n + 1\] Clearly neither \(r\) nor \(s\) are equal to 1 so \(n^2 + 2n + 1\) is not prime and hence must be composite. [Error: this is jumping to the conclusion without showing what needs to be shown. The expression we need to show (the definition of composite) has been given but it has not been shown.]

    Correct proof. Let \(n \in \mathbb{Z}\) with \(n > 0\). Then we have \[n^2 + 2n + 1 = (n+1)(n+1).\] Now since \(0 < n < n^2 + 2n\), we must have that \(1 < n + 1 < n^2 + 2n + 1\). So by letting \(r = s = n + 1\) we have shown that \[n^2 + 2n + 1 = rs\] for integers \(1 < r, s < n^2 + 2n + 1\). By definition this means \(n^2 + 2n + 1\) is composite. \(\square\)

Application — Applications of Proofs to Computer Science

Not assessable for MATH1061. Taken from K. Rosen, Discrete Mathematics and its Applications, 7th edition, 1991, pages 74–80.

Computer programs have been developed to automate the task of reasoning and proving theorems. Many of these programs make use of a rule of inference known as resolution. This rule of inference is based on the tautology

\[\big((p \vee q) \wedge (\sim p \vee r)\big) \to (q \vee r) \tag{1}\]

Resolution can be used to build automatic theorem proving systems.

To construct proofs using resolution as the only rule of inference, the hypotheses and conclusion must be expressed as clauses. Clauses are logical expressions which only feature disjunctions of variables or their negations (i.e. expressions which only use \(\vee\), \(\sim\), so they do not use \(\wedge\), \(\to\)). To express any statement as a clause, we can use the laws of logical equivalence.

Application problems

    1. Using a truth table, show that expression (1) is a tautology.

    2. Write the following expressions as clauses.

    1. \(\sim (\sim r \wedge w)\)

    2. \(r \to u\)

    3. \(\sim (u \wedge w)\)

    1. Use resolution only to show that the following argument is valid:

    If it is raining, then Yvette has their umbrella. It is not the case that Yvette has their umbrella and gets wet. It is not the case that Yvette gets wet while it is not raining. Therefore, Yvette does not get wet.

Application solutions
    1. The truth table is
    \(p\) \(q\) \(r\) \(p \vee q\) \(\sim p \vee r\) \((p \vee q) \wedge (\sim p \vee r)\) \(q \vee r\) \(\big((p \vee q) \wedge (\sim p \vee r)\big) \to (q \vee r)\)
    T T T T T T T T
    T T F T F F T T
    T F T T T T T T
    T F F T F F F T
    F T T T T T T T
    F T F T T T T T
    F F T F T F T T
    F F F F T F F T

    Since the last column is always T, the statement is a tautology.

    1. Clause forms:

    1. \(\sim (\sim r \wedge w) \equiv r \vee \sim w\) (by De Morgan’s law and double negative law).

    2. \(r \to u \equiv \; \sim r \vee u\) (by definition of implication).

    3. \(\sim (u \wedge w) \equiv \; \sim u \vee \sim w\) (by De Morgan’s law).

    1. Define the following variables:
    \(r\) it is raining
    \(u\) Yvette has their umbrella
    \(w\) Yvette gets wet

    Then we can express the argument as

    1. \(r \to u\)
    2. \(\sim (u \wedge w)\)
    3. \(\sim (\sim r \wedge w)\) \(\therefore\ \sim w\)

    Using the equivalences in (b), we can rewrite this as

    1. \(\sim r \vee u\)
    2. \(\sim u \vee \sim w\)
    3. \(r \vee \sim w\)
    4. \(\sim r \vee \sim w\) From 1 and 2 by Resolution
    5. \(\sim w\) From 3 and 4 by Resolution.

§10 — Proof by Contradiction

Epp 4th ed. pp. 198–207 · 5th ed. (metric) pp. 218–228. Lecture 8 (week 3). Notes: proof-techniques.

Problems

  1. Prove the following statement using a proof by contradiction:

    \[\text{For integers } a \text{ and } b,\ \text{if } 6a + 3b \text{ is odd, then } b \text{ is odd.}\]

  2. Use a proof by contradiction to prove the following statement:

    \[\text{For all positive integers } x \text{ and } y,\ x^2 - y^2 \neq 1.\]

Solutions

  1. For integers \(a\) and \(b\), if \(6a + 3b\) is odd, then \(b\) is odd.

    Proof. Suppose the statement is false, that is, suppose that there exist integers \(a\) and \(b\) for which \(6a + 3b\) is odd and \(b\) is even.

    Since \(b\) is even, \(b = 2k\) for some integer \(k\). Thus \[6a + 3b = 6a + 3(2k) = 6a + 6k = 2(3a + 3k).\] Since \(a, k \in \mathbb{Z}\), \(3a + 3k \in \mathbb{Z}\) so \(6a + 3b\) is even. This contradicts our assumption that \(6a + 3b\) is odd, so our assumption was incorrect and hence the original statement is true. \(\square\)

  2. For all positive integers \(x\) and \(y\), \(x^2 - y^2 \neq 1\).

    Proof. Suppose, for contradiction, that \(x\) and \(y\) are positive integers satisfying \(x^2 - y^2 = 1\). Now, by factoring, we have \[1 = x^2 - y^2 = (x - y)(x + y).\] Since \(x\) and \(y\) are positive integers, we must have \(x - y = 1\) and \(x + y = 1\). This means \(2x = 2\), so \(x = 1\). But then we must have \(y = 0\) which is not a positive integer; this is a contradiction. Therefore the original statement is true. \(\square\)

§11 — Proof by Contraposition

Epp 4th ed. pp. 198–207 · 5th ed. (metric) pp. 218–228. Lecture 9 (week 3). Notes: proof-techniques.

Problems

  1. Use proof by contraposition to prove the following statement:

    \[\text{For all integers } a \text{ and } b,\ \text{if } (ab)^2 \text{ is odd then } a \text{ is odd and } b \text{ is odd.}\]

  2. Prove the following statement:

    \[\text{For any integer } n,\ n^2 \text{ is odd if and only if } n \text{ is odd.}\]

    Use a direct proof for the reverse implication (\(\Leftarrow\)), and a proof by contraposition for the forward implication (\(\Rightarrow\)).

  3. Use proof by contraposition to prove the following statement:

    \[\text{For all integers } x \text{ and } y,\ \text{if } x^2(y^2 - 2y) \text{ is odd then } x \text{ and } y \text{ are odd.}\]

Solutions

  1. For all integers \(a\) and \(b\), if \((ab)^2\) is odd then \(a\) is odd and \(b\) is odd.

    Proof. We prove the contrapositive statement, which is:

    For all integers \(a\) and \(b\), if \(a\) is even or \(b\) is even, then \((ab)^2\) is even.

    Let \(a\) and \(b\) be integers and suppose \(a\) is even or \(b\) is even. Without loss of generality, suppose \(a\) is even, so \(a = 2k\) for some \(k \in \mathbb{Z}\). Then \(ab = 2kb\) is even, and therefore \((ab)^2 = 4k^2b^2 = 2(2k^2b^2)\). Since \(k, b \in \mathbb{Z}\), we know \(2k^2b^2 \in \mathbb{Z}\), so \((ab)^2\) is even by definition. \(\square\)

  2. For any integer \(n\), \(n^2\) is odd if and only if \(n\) is odd.

    Proof. To prove this “if and only if” statement, we must prove each of the following two statements:

    (\(\Leftarrow\)) For any integer \(n\), if \(n\) is odd, then \(n^2\) is odd. (\(\Rightarrow\)) For any integer \(n\), if \(n^2\) is odd, then \(n\) is odd.

    (\(\Leftarrow\)) Let \(n\) be an integer and suppose \(n\) is odd. Then \(n = 2k + 1\) for some \(k \in \mathbb{Z}\). Hence \(n^2 = (2k+1)^2 = 2(2k^2 + 2k) + 1\). Since \(k \in \mathbb{Z}\), we have \(2k^2 + 2k \in \mathbb{Z}\) and hence \(n^2\) is odd by definition.

    (\(\Rightarrow\)) The contrapositive is: for any integer \(n\), if \(n\) is even, then \(n^2\) is even. Let \(n\) be an integer and suppose \(n\) is even. Then \(n = 2k\) for some \(k \in \mathbb{Z}\). Hence \(n^2 = 4k^2 = 2(2k^2)\). Since \(k \in \mathbb{Z}\), we have \(2k^2 \in \mathbb{Z}\), so \(n^2\) is even by definition.

    Therefore, for all integers \(n\), \(n^2\) is odd if and only if \(n\) is odd. \(\square\)

  3. For all integers \(x\) and \(y\), if \(x^2(y^2 - 2y)\) is odd then \(x\) and \(y\) are odd.

    Proof. We will prove the contrapositive statement, which is:

    For all integers \(x\) and \(y\), if \(x\) is even or \(y\) is even, then \(x^2(y^2 - 2y)\) is even.

    Let \(x, y \in \mathbb{Z}\) and suppose \(x\) or \(y\) is even. We consider the cases of \(x\) even or \(y\) even individually.

    Case 1. Suppose \(x\) is even. Then \(x = 2k\) for some \(k \in \mathbb{Z}\). Hence \[x^2(y^2 - 2y) = (2k)^2(y^2 - 2y) = 2(2k^2)(y^2 - 2y).\] Since \(k, y \in \mathbb{Z}\), we have \((2k^2)(y^2 - 2y) \in \mathbb{Z}\), and hence \(x^2(y^2-2y)\) is even, by definition.

    Case 2. Suppose \(y\) is even. Then \(y = 2\ell\) for some \(\ell \in \mathbb{Z}\). Hence \[x^2(y^2 - 2y) = x^2\big((2\ell)^2 - 2(2\ell)\big) = 2\big(x^2(2\ell^2 - 2\ell)\big).\] Since \(x, \ell \in \mathbb{Z}\), we have \(x^2(2\ell^2 - 2\ell) \in \mathbb{Z}\) and thus \(x^2(y^2-2y)\) is even, by definition. \(\square\)

§12 — Rational Numbers

Epp 4th ed. pp. 163–170 · 5th ed. (metric) pp. 183–190. Lecture 9 (week 3). Notes: rational-and-irrational-numbers.

Problems

  1. Prove the following statement:

    \[\text{For real numbers } m \text{ and } n,\ \text{if } m \text{ is irrational and } n \text{ is rational, then } m - n \text{ is irrational.}\]

  2. Prove the following statement:

    \[\text{For a real number } r,\ \text{if } 2r^2 - 3r \text{ is irrational, then } r \text{ is irrational.}\]

  3. Prove the following statement:

    \[\text{For all real numbers } r, s, n,\ \text{if } r \text{ is irrational, } s \text{ is rational and } n \text{ is a positive integer, then } n(r-s) \text{ is irrational.}\]

  4. Recall that \(\sqrt{2}\) is irrational. Consider the statement

    \[\text{For every rational number } r,\ \text{if } r \neq 0,\ \text{then } r\sqrt{2} \text{ is an irrational number.}\]

    1. Write down the negation of the statement.

    2. Use a proof by contradiction to prove the statement.

Solutions

  1. For real numbers \(m\) and \(n\), if \(m\) is irrational and \(n\) is rational, then \(m - n\) is irrational.

    Proof. Let \(m\) and \(n\) be real numbers and suppose (for a contradiction) that \(m\) is an irrational number and \(n\) is rational, but \(m - n\) is rational. Since \(n\) is rational, we have \(n = \frac{a}{b}\) for some \(a, b \in \mathbb{Z}\) where \(b \neq 0\). Since \(m - n\) is rational, we have \(m - n = \frac{c}{d}\) for some \(c, d \in \mathbb{Z}\) where \(d \neq 0\).

    Notice that \(m = (m-n) + n\). Hence \[m = (m-n) + n = \frac{c}{d} + \frac{a}{b} = \frac{bc + ad}{bd}.\] Since \(a, b, c, d \in \mathbb{Z}\) we have \(bc + ad \in \mathbb{Z}\) and \(bd \in \mathbb{Z}\). Further, since \(b \neq 0\) and \(d \neq 0\), we have \(bd \neq 0\). Thus, by definition, \(m\) is rational. But this contradicts the assumption that \(m\) is irrational. \(\square\)

  2. For a real number \(r\), if \(2r^2 - 3r\) is irrational, then \(r\) is irrational.

    Proof. We prove the contrapositive statement: for a real number \(r\), if \(r\) is rational, then \(2r^2 - 3r\) is rational.

    Let \(r\) be a real number and suppose \(r \in \mathbb{Q}\). Then \(r = \frac{a}{b}\) for some \(a, b \in \mathbb{Z}\) where \(b \neq 0\). So \[2r^2 - 3r = 2\left(\frac{a}{b}\right)^2 - 3\left(\frac{a}{b}\right) = \frac{2a^2 - 3ab}{b^2}.\] Since \(a, b \in \mathbb{Z}\), we have \(2a^2 - 3ab \in \mathbb{Z}\) and \(b^2 \in \mathbb{Z}\). Further, \(b^2 \neq 0\) because \(b \neq 0\). This means \(2r^2 - 3r\) is rational, by definition. \(\square\)

  3. For all real numbers \(r\), \(s\), and \(n\), if \(r\) is irrational, \(s\) is rational and \(n\) is a positive integer, then \(n(r-s)\) is irrational.

    Proof. Suppose, for a contradiction, that the statement is false. Then there exist real numbers \(r\), \(s\) and \(n\) such that \(r\) is an irrational number, \(s\) is rational, \(n\) is a positive integer, but \(n(r-s)\) is rational. Since \(s\) is rational, \(s = \frac{a}{b}\) for some \(a, b \in \mathbb{Z}\) where \(b \neq 0\). Since \(n(r-s)\) is rational, we have \(n(r-s) = \frac{c}{d}\) for some \(c, d \in \mathbb{Z}\) where \(d \neq 0\).

    Now, \[\begin{aligned} r - s &= n(r-s)\left(\frac{1}{n}\right) && \text{noting that } n \neq 0 \\ r - s &= \frac{c}{d} \cdot \frac{1}{n} \\ r &= s + \frac{c}{d} \cdot \frac{1}{n} \\ &= \frac{a}{b} + \frac{c}{dn} \\ &= \frac{adn + bc}{bdn} \end{aligned}\] Since \(a, b, c, d \in \mathbb{Z}\), we have \(adn + bc \in \mathbb{Z}\) and \(bdn \in \mathbb{Z}\). Further, since \(b \neq 0\), \(d \neq 0\) and \(n \neq 0\), we have \(bdn \neq 0\). This means \(r\) is rational, by definition, which is a contradiction. \(\square\)

  4. For every rational number \(r\), if \(r \neq 0\), then \(r\sqrt{2}\) is an irrational number.

    1. Negation: There exists \(r \in \mathbb{Q}\) such that \(r \neq 0\) and \(r\sqrt{2} \in \mathbb{Q}\).

    Proof. Suppose, for a contradiction, that the original statement is false. Then there exists \(r \in \mathbb{Q}\) such that \(r \neq 0\) and \(r\sqrt{2} \in \mathbb{Q}\). Since \(r\) is rational, \(r = \frac{a}{b}\) for some \(a, b \in \mathbb{Z}\) with \(b \neq 0\), and since \(r \neq 0\), we have \(a \neq 0\). Since we have assumed \(r\sqrt{2}\) is rational, \(r\sqrt{2} = \frac{c}{d}\) for some \(c, d \in \mathbb{Z}\) where \(d \neq 0\).

    Now \[\sqrt{2} = \frac{r\sqrt{2}}{r} = \left(r\sqrt{2}\right)\left(\frac{1}{r}\right) = \frac{c}{d} \cdot \frac{b}{a} = \frac{bc}{ad}.\] Since \(a, b, c, d \in \mathbb{Z}\), we know \(bc \in \mathbb{Z}\) and \(ad \in \mathbb{Z}\). Further, since \(a \neq 0\) and \(d \neq 0\), we have \(ad \neq 0\). Thus, by definition, \(\sqrt{2}\) is rational. However, this is a contradiction, as \(\sqrt{2}\) is irrational. Thus, the original statement is true. \(\square\)

§13 — Divisibility

Epp 4th ed. pp. 170–179 · 5th ed. (metric) pp. 190–199. Lecture 10 (week 4). Notes: divisibility-and-factorisation.

Problems

  1. Find the unique factorisation (in standard form) of the integer \(27\,720\).

  2. Prove or disprove the statement: for any integers \(m\), \(n\), and \(q\), if \(m \mid n\) and \(n \mid q\) then \(m^2 \mid nq\).

  3. Prove the following statements.

    1. If \(k\) is any even integer and \(m\) is any odd integer, then \((k+2)^2 - (m-3)^2\) is divisible by 4.

    2. The difference between the cube of two consecutive integers leaves the remainder 1 when it is divided by 6.

  4. Use proof by contradiction to prove this statement:

    \[\text{For all integers } c, d \text{ and } e,\ \text{if } c \mid d \text{ and } c \nmid e,\ \text{then } c \nmid (d+e).\]

Solutions

  1. By repeated division, we have \(27\,720 = 2^3 \cdot 3^2 \cdot 5 \cdot 7 \cdot 11\).

  2. For any integers \(m\), \(n\), and \(q\), if \(m \mid n\) and \(n \mid q\) then \(m^2 \mid nq\). The statement is true, and we will prove it.

    Proof. Let \(m\), \(n\), and \(q\) be non-zero integers. Assume \(m \mid n\) and \(n \mid q\). Then, by definition, there exist integers \(a, b\) such that \[n = ma \quad \text{and} \quad q = nb.\] Then observe \[nq = (ma)(nb) = ma \cdot (mab) = m^2(a^2 b).\] Since \(a^2 b \in \mathbb{Z}\), we conclude that \(m^2 \mid nq\). \(\square\)

    1. If \(k\) is any even integer and \(m\) is any odd integer, then \((k+2)^2 - (m-3)^2\) is divisible by 4.

    Proof. Let \(k\) be even and \(m\) be odd. Then \(k = 2a\) and \(m = 2b + 1\) for some \(a, b \in \mathbb{Z}\). Hence \[(k+2)^2 = (2a+2)^2 = 4(a+1)^2 \quad \text{and} \quad (m-3)^2 = (2b-2)^2 = 4(b-1)^2.\] Therefore \[(k+2)^2 - (m-3)^2 = 4\big[(a+1)^2 - (b-1)^2\big].\] Since \(a, b \in \mathbb{Z}\) we have \((a+1)^2 - (b-1)^2 \in \mathbb{Z}\) so \(4 \mid \big[(k+2)^2 - (m-3)^2\big]\). \(\square\)

    1. The difference between the cube of two consecutive integers leaves the remainder 1 when it is divided by 6.

    Proof. Let \(n\) and \(n+1\) be two consecutive integers. The difference between their cubes is \[(n+1)^3 - n^3 = 3n^2 + 3n + 1 = 3n(n+1) + 1.\] Now whether \(n\) is odd or even (so whatever integer \(n\) is), one of \(n\), \(n+1\) will be odd and the other one will be even. Thus the product \(n(n+1)\) is even.

    Since \(n(n+1)\) is even, write \(n(n+1) = 2t\) for some \(t \in \mathbb{Z}\). Then \[(n+1)^3 - n^3 = 3(2t) + 1 = 6t + 1 \equiv 1 \pmod 6.\] Hence the remainder is 1 upon division by 6. \(\square\)

  4. For all integers \(c\), \(d\) and \(e\), if \(c \mid d\) and \(c \nmid e\), then \(c \nmid (d+e)\).

    Proof. Suppose for a contradiction that the statement is false. Then there exist integers \(c\), \(d\) and \(e\) such that \(c \mid d\) and \(c \nmid e\) and \(c \mid (d+e)\).

    Now \(c \mid d\) means that \(d = cx\) for some integer \(x\). And \(c \mid (d+e)\) means that \(d + e = cy\) for some integer \(y\).

    Hence \(d + e = cx + e = cy\), so \(e = cy - cx = c(y-x)\), and we have \(y - x \in \mathbb{Z}\), because \(x\) and \(y\) are both integers. Thus \(c \mid e\), a contradiction. So the original statement is true. \(\square\)

§14 — Modular Arithmetic

Epp 4th ed. pp. 482–485 · 5th ed. (metric) pp. 528–531. Lecture 11 (week 4). Notes: modular-arithmetic.

Problems

  1. Prove the following statement:

    \[\text{If } y \text{ is any integer, then when } y^2 \text{ is divided by 5, the remainder is always 0, 1 or 4, and never 2 or 3.}\]

    1. Evaluate \(58 \operatorname{div} 11\) and \(58 \bmod 11\).

    2. Find integers \(q\) and \(r\), with \(0 \leqslant r < d\) and \(n = dq + r\), in the case that \(n = -95\) and \(d = 11\).

    1. An integer \(n\), when divided by 7, leaves remainder 4. What is the remainder (between 0 and 6 inclusive) when \(3n\) is divided by 7?

    2. When the integer \(m\) is divided by 11, the remainder 9 is left. What remainder is left when \(4m\) is divided by 22?

  4. Prove or disprove the following statements.

    1. For all real numbers \(x\), \(\lfloor x^2 \rfloor = \lfloor x \rfloor^2\).

    2. For all odd integers \(n\), \(\left\lfloor \dfrac{n^2}{4} \right\rfloor = \left(\dfrac{n-1}{2}\right)\left(\dfrac{n+1}{2}\right)\).

Solutions

  1. If \(y\) is any integer, then when \(y^2\) is divided by 5, the remainder is always 0, 1 or 4, and never 2 or 3.

    Proof. Fix \(y \in \mathbb{Z}\). By the division algorithm, \(y = 5m + r\) for some \(m \in \mathbb{Z}\) and \(r \in \{0, 1, 2, 3, 4\}\). Then \[y^2 = (5m+r)^2 = 25m^2 + 10mr + r^2 \equiv r^2 \pmod 5.\] Let us compute the possible values for \(r^2 \bmod 5\): \[0^2 \equiv 0, \quad 1^2 \equiv 1, \quad 2^2 \equiv 4, \quad 3^2 \equiv 9 \equiv 4, \quad 4^2 \equiv 16 \equiv 1 \pmod 5.\] Thus \(y^2 \equiv 0, 1\), or \(4 \pmod 5\), and never 2 or 3. \(\square\)

    1. Since \(58 = 11 \cdot 5 + 3\) with \(0 \leq 3 < 11\), we have

    \[58 \operatorname{div} 11 = 5 \quad \text{and} \quad 58 \bmod 11 = 3.\]

    1. We want \(-95 = 11q + r\) with \(0 \leq r < 11\). Taking \(q = -9\) gives

    \[11(-9) = -99 \quad \Rightarrow \quad -95 = -99 + 4 = 11(-9) + 4,\]

    so \(q = -9\) and \(r = 4\).

    1. If \(n\) leaves remainder 4 upon division by 7, then \(n = 7k + 4\) for some integer \(k\). Hence

    \[3n = 21k + 12 = 7(3k + 1) + 5,\]

    so the remainder when \(3n\) is divided by 7 is 5. (The source PDF prints this step as \(7(3k) + 5\), which equals \(21k + 5\), not \(21k + 12\) — the quotient is \(3k + 1\), as above. The answer of 5 is unaffected.)

    1. If \(m\) leaves remainder 9 upon division by 11, then \(m = 11k + 9\) for some integer \(k\). Then

    \[4m = 44k + 36 = 22(2k+1) + 14,\]

    so the remainder when \(4m\) is divided by 22 is 14.

    1. For all real numbers \(x\), \(\lfloor x^2 \rfloor = \lfloor x \rfloor^2\). The statement is false. For a counterexample, take \(x = \frac{3}{2}\). Then

    \[\lfloor x^2 \rfloor = \left\lfloor \frac{9}{4} \right\rfloor = 2 \quad \text{but} \quad \lfloor x \rfloor^2 = \left\lfloor \frac{3}{2} \right\rfloor^2 = 1^2 = 1.\]

    1. For all odd integers \(n\), \(\left\lfloor \dfrac{n^2}{4} \right\rfloor = \left(\dfrac{n-1}{2}\right)\left(\dfrac{n+1}{2}\right)\). The statement is true.

    Proof. Let \(n\) be an odd integer, so \(n = 2k+1\) for some integer \(k\). Then \[\frac{n^2}{4} = \frac{(2k+1)^2}{4} = \frac{4k^2 + 4k + 1}{4} = k(k+1) + \frac{1}{4},\] so \(\left\lfloor \frac{n^2}{4} \right\rfloor = k(k+1)\). Moreover, notice \[\left(\frac{n-1}{2}\right)\left(\frac{n+1}{2}\right) = \left(\frac{2k}{2}\right)\left(\frac{2k+2}{2}\right) = k(k+1).\] Hence \(\left\lfloor \frac{n^2}{4} \right\rfloor = \left(\frac{n-1}{2}\right)\left(\frac{n+1}{2}\right)\) for all odd integers \(n\). \(\square\)

§15 — The Euclidean Algorithm

Epp 4th ed. pp. 220–224 · 5th ed. (metric) pp. 250–254. Lecture 12 (week 4). Notes: gcd-lcm-and-euclidean-algorithm.

Problems

  1. Use the Euclidean Algorithm to determine \(\gcd(14, 3003)\).

  2. Compute

    1. \(\operatorname{lcm}(12, 18)\)

    2. \(\operatorname{lcm}(2^2 \cdot 3 \cdot 5,\ 2^3 \cdot 3^2)\)

    3. \(\operatorname{lcm}(2800, 6125)\)

  3. Prove that for all positive integers \(a\) and \(b\), we have

    \[\gcd(a,b) = \operatorname{lcm}(a,b) \iff a = b.\]

  4. Prove that for all positive integers \(a\) and \(b\), we have

    \[a \mid b \iff \operatorname{lcm}(a,b) = b.\]

    1. Use the Euclidean algorithm to determine \(\gcd(931, 301)\). Show your working.

    2. Write down the unique prime factorisation of 931 in standard form.

    3. How many positive divisors does 931 have? (Hint: use part (b).)

    1. Using the Euclidean algorithm, compute \(\gcd(116, 88)\). Show your working.

    2. Using your answer to part (a), compute \(\operatorname{lcm}(116, 88)\).

Solutions

  1. We apply the Euclidean Algorithm:

    \[3003 = 14 \cdot 214 + 7, \qquad 14 = 7 \cdot 2 + 0.\]

    Hence \(\gcd(14, 3003) = 7\).

    1. We factor \(12 = 2^2 \cdot 3\) and \(18 = 2 \cdot 3^2\) so \(\operatorname{lcm}(12, 18) = 2^2 \cdot 3^2 = 36\).

    2. Using the maximum exponents of each prime, we have

    \[\operatorname{lcm}(2^2 \cdot 3 \cdot 5,\ 2^3 \cdot 3^2) = 2^3 \cdot 3^2 \cdot 5 = 360.\]

    1. We factor \(2800 = 2^4 \cdot 5^2 \cdot 7\) and \(6125 = 5^3 \cdot 7^2\). Thus

    \[\operatorname{lcm}(2800, 6125) = 2^4 \cdot 5^3 \cdot 7^2 = 98000.\]

  3. For all positive integers \(a\) and \(b\), \(\gcd(a,b) = \operatorname{lcm}(a,b)\) if and only if \(a = b\).

    Proof. Let \(a\) and \(b\) be positive integers.

    (\(\Longleftarrow\)) It is clear that if \(a = b\) then \(\gcd(a,b) = \operatorname{lcm}(a,b)\).

    (\(\Longrightarrow\)) Assume \(\gcd(a,b) = \operatorname{lcm}(a,b)\). Denoting \(g = \gcd(a,b)\), we have \(a = gm\) and \(b = gn\) for some positive integers \(m\) and \(n\) with \(\gcd(m,n) = 1\).

    Now notice \(gm\) and \(gn\) have no common prime factors beyond those in \(g\), so the least common multiple uses all primes from both \(m\) and \(n\). Therefore \[\operatorname{lcm}(a,b) = gmn.\] Now, we have \(g = gmn\). Therefore \(mn = 1\), and so \(m = n = 1\). This lets us conclude \(a = b\). \(\square\)

  4. For all positive integers \(a\) and \(b\), \(a \mid b\) if and only if \(\operatorname{lcm}(a,b) = b\).

    Proof. Let us denote \(g = \gcd(a,b)\) so that \(a = gm\) and \(b = gn\) for some positive integers \(m\) and \(n\) with \(\gcd(m,n) = 1\). As explained in the previous question, we have \[\operatorname{lcm}(a,b) = gmn.\]

    (\(\Longrightarrow\)) If \(a \mid b\), then \(gm \mid gn\), hence \(m \mid n\). Since \(\gcd(m,n) = 1\), we conclude \(m = 1\). Therefore \(\operatorname{lcm}(a,b) = gmn = gn = b\).

    (\(\Longleftarrow\)) If \(\operatorname{lcm}(a,b) = b\), then \(gmn = gn\) which means \(m = 1\). Therefore \(a = gm = g\) and so \(a \mid b\). \(\square\)

    1. Using the Euclidean Algorithm:

    \[\begin{aligned} 931 &= 301 \cdot 3 + 28 \\ 301 &= 28 \cdot 10 + 21 \\ 28 &= 21 \cdot 1 + 7 \\ 21 &= 7 \cdot 3 + 0. \end{aligned}\]

    Hence \(\gcd(931, 301) = 7\).

    1. Since \(\gcd(931, 301) = 7\), we know \(7 \mid 931\), and we have \(931 = 7 \cdot 133\). The number 133 has prime factorisation \(7 \cdot 19\). Therefore the unique prime factorisation of 931 is \(7^2 \cdot 19\).

    2. We have \(931 = 7^2 \cdot 19^1\) from part (b). Any divisor of 931 must have a prime factorisation of the form \(7^a \cdot 19^b\) where \(a \in \{0,1,2\}\) and \(b \in \{0,1\}\). Thus the number of positive divisors is

    \[(2+1)(1+1) = 6.\]

    (There are 3 choices for the exponent \(a\) and 2 choices for the exponent \(b\).)

    1. Using the Euclidean Algorithm:

    \[\begin{aligned} 116 &= 88 \cdot 1 + 28 \\ 88 &= 28 \cdot 3 + 4 \\ 28 &= 4 \cdot 7 + 0 \end{aligned}\]

    Hence \(\gcd(116, 88) = 4\).

    1. Using \(\gcd(a,b) \cdot \operatorname{lcm}(a,b) = ab\) for \(a, b \in \mathbb{Z}^+\),

    \[\operatorname{lcm}(116, 88) = \frac{116 \cdot 88}{\gcd(116,88)} = \frac{116 \cdot 88}{4} = 116 \cdot 22 = 2552.\]

Application — Trace Tables for Algorithms

Not assessable for MATH1061. For reference, see S. Epp, Discrete Mathematics with Applications, Section 4.8 (4th edition) or Section 4.10 (5th edition), Brooks/Cole, Belmont, CA.

The following pseudo-code describes the Division Algorithm, with annotations to explain what each step is doing.

Input: \(a\) (a non-negative integer), \(d\) (a positive integer). This algorithm will take two inputs, \(a\) and \(d\), and will find the quotient \(a \operatorname{div} d\) and the remainder \(a \bmod d\).

Algorithm body:

r := a          -- defining our variables
q := 0
                -- starting with a, repeatedly subtract lots of d until we reach
                -- a number less than d; each round adds 1 to the quotient count
while r >= d,
    r := r - d
    q := q + 1
end while

Output: \(q\), \(r\). Once we have obtained a number less than \(d\), we are finished. The output \(q\) is the quotient (\(a \operatorname{div} d\)) and the output \(r\) is the remainder (\(a \bmod d\)).

The following pseudo-code describes the Euclidean Algorithm, with annotations to explain what each step is doing.

Input: \(A\), \(B\) (integers with \(A > B \geq 0\)).

Algorithm body:

a := A          -- defining our variables
b := B
r := B
                -- if b /= 0, compute a mod b (call the algorithm above for this
                -- step) and set r to be this value; replace a with b and b with r
                -- and repeat until b = 0
while b /= 0,
    r := a mod b
    a := b
    b := r
end while
gcd := a

Output: \(\text{gcd}\) (a positive integer).

A trace table keeps track of the value of each variable at each step of an algorithm. For example, the trace table for the Division Algorithm with input variables \(a = 19\) and \(d = 4\) is:

Variable / iteration 0 1 2 3 4
\(a\) 19
\(d\) 4
\(r\) 19 15 11 7 3
\(q\) 0 1 2 3 4

Output: \(q = 4\), \(r = 3\).

Application problems

  1. Construct a trace table to trace the action of the Division Algorithm for the following inputs.

    1. \(a = 26\), \(d = 7\)

    2. \(a = 15\), \(d = 6\)

  2. Construct a trace table to trace the action of the Euclidean Algorithm with inputs \(A = 1001\) and \(B = 871\).

Application solutions
    1. Trace table for the Division Algorithm with \(a = 26\), \(d = 7\):
    Variable / iteration 0 1 2 3
    \(a\) 26
    \(d\) 7
    \(r\) 26 19 12 5
    \(q\) 0 1 2 3

    Output: \(q = 3\), \(r = 5\).

    1. Trace table for the Division Algorithm with \(a = 15\), \(d = 6\):
    Variable / iteration 0 1 2
    \(a\) 15
    \(d\) 6
    \(r\) 15 9 3
    \(q\) 0 1 2

    Output: \(q = 2\), \(r = 3\).

  2. Trace table for the Euclidean Algorithm with inputs \(A = 1001\) and \(B = 871\):

    Variable / iteration 0 1 2 3 4 5
    \(A\) 1001
    \(B\) 871
    \(a\) 1001 871 130 91 39 13
    \(b\) 871 130 91 39 13 0
    \(r\) 871 130 91 39 13 0
    \(\text{gcd}\) 13

    Output: \(\text{gcd} = 13\).

§16 — Sequences

Epp 4th ed. pp. 227–244 · 5th ed. (metric) pp. 258–275. Lecture 13 (week 5).

Problems

  1. Write down the first four terms in the sequence defined by \(a_k = \dfrac{k}{10+k}\) for \(k = 1, 2, \dots\).

  2. Consider the sequences \(a_k = 2k+1\) and \(b_k = (k-1)^3 + k + 2\) where \(k = 0, 1, 2, \dots\). Show that the first three terms of these sequences are identical, but that their fourth terms differ.

    1. Find an explicit formula for a sequence \(c_n\) whose first six terms are \(-1, 1, -1, 1, -1, 1\).

    2. Find an explicit formula for a sequence \(d_n\) whose first six terms are \(\frac{1}{4}, \frac{2}{9}, \frac{3}{16}, \frac{4}{25}, \frac{5}{36}, \frac{6}{49}\).

    1. Write \((x^2 - 1) + (2x^3 - 1) + (3x^4 - 1) + (4x^5 - 1) + (5x^6 - 1)\) in summation notation.

    2. Evaluate \(\displaystyle\prod_{i=1}^{5} (-1)^i \cdot i\).

    3. Write \((y-1)(y^2 - 2)(y^3 - 3)(y^4 - 4)\) in product notation.

    4. Evaluate \(\displaystyle\sum_{j=1}^{n} (-1)^j \frac{1}{2^j}\) when \(n = 5\).

Solutions

  1. We have

    \[a_1 = \frac{1}{11}, \qquad a_2 = \frac{2}{12} = \frac{1}{6}, \qquad a_3 = \frac{3}{13}, \qquad a_4 = \frac{4}{14} = \frac{2}{7}.\]

  2. We compute

    \[a_0 = 1,\quad a_1 = 3,\quad a_2 = 5,\quad a_3 = 7;\]

    \[b_0 = (-1)^3 + 0 + 2 = 1, \quad b_1 = 0^3 + 1 + 2 = 3, \quad b_2 = 1^3 + 2 + 2 = 5, \quad b_3 = 2^3 + 3 + 2 = 13.\]

    Hence the first three terms agree (\(a_0 = b_0\), \(a_1 = b_1\) and \(a_2 = b_2\)), but the fourth terms differ (\(a_3 = 7\) while \(b_3 = 13\)).

    1. One explicit formula is

    \[c_n = (-1)^n \qquad n = 1, 2, 3, \dots\]

    1. The denominators are square numbers, so an explicit formula is

    \[d_n = \frac{n}{(n+1)^2} \qquad n = 1, 2, 3, \dots\]

    1. We have

    \[(x^2 - 1) + (2x^3 - 1) + (3x^4 - 1) + (4x^5 - 1) + (5x^6 - 1) = \sum_{k=1}^{5} \left(kx^{k+1} - 1\right).\]

    1. We have

    \[\prod_{i=1}^{5} (-1)^i \cdot i = \left(\prod_{i=1}^{5} (-1)^i\right)\left(\prod_{i=1}^{5} i\right) = (-1)^{1+2+3+4+5} \cdot 5! = (-1)^{15} \cdot 120 = -120.\]

    1. We have

    \[(y-1)(y^2-2)(y^3-3)(y^4-4) = \prod_{k=1}^{4} \left(y^k - k\right).\]

    1. When \(n = 5\), we have

    \[\sum_{j=1}^{5} (-1)^j \frac{1}{2^j} = -\frac{1}{2} + \frac{1}{4} - \frac{1}{8} + \frac{1}{16} - \frac{1}{32} = -\frac{11}{32}.\]

§17 — Mathematical Induction

Epp 4th ed. pp. 244–268 · 5th ed. (metric) pp. 275–301. Lecture 13 (week 5).

Problems

  1. Use mathematical induction to prove the following statements.

    1. For every positive integer \(n\), \(n(n+1)\) is even.

    2. For every integer \(n \geq 2\), \(n^3 - n\) is a multiple of 6.

    3. For all integers \(n \geq 7\), \(3^n < n!\).

    4. For each integer \(n \geq 1\),

    \[\sum_{i=1}^{n} i^3 = \frac{n^2(n+1)^2}{4}.\]

    1. For each integer \(n \geq 0\), \(1 + 4n \leq 5^n\).

Solutions

    1. For every positive integer \(n\), \(n(n+1)\) is even.

    Proof. Let \(P(n)\) be the predicate “\(n(n+1)\) is even”.

    Basis: when \(n = 1\), \(n(n+1) = 2\) which is even, so \(P(1)\) is true.

    (The inductive step is to show that for all integers \(k \geq 1\), \(P(k)\) implies \(P(k+1)\).)

    Inductive hypothesis: suppose that for some integer \(k \geq 1\), \(P(k)\) is true. Thus \(k(k+1) = 2a\) for some integer \(a\).

    Now consider \(n = k+1\). Then \[(k+1)(k+2) = k(k+1) + 2(k+1) = 2a + 2(k+1) = 2(a + k + 1).\] Since \(a\) and \(k\) are integers, \(a + k + 1\) is an integer, so \((k+1)(k+2)\) is even and \(P(k+1)\) is true.

    Thus, by the principle of mathematical induction, \(n(n+1)\) is even for every positive integer \(n\). \(\square\)

    1. For every integer \(n \geq 2\), \(n^3 - n\) is a multiple of 6.

    Proof. Let \(P(n)\) be the predicate \(6 \mid (n^3 - n)\).

    Basis: when \(n = 2\), \(n^3 - n = 8 - 2 = 6\). Since \(6 \mid 6\), \(P(2)\) is true.

    (The inductive step is to show that for all integers \(k \geq 2\), \(P(k)\) implies \(P(k+1)\).)

    Inductive hypothesis: suppose that for some integer \(k \geq 2\), \(P(k)\) is true. Thus \(k^3 - k = 6a\) for some integer \(a\).

    Now consider \(n = k+1\). Then \[(k+1)^3 - (k+1) = k^3 + 3k^2 + 3k + 1 - k - 1 = k^3 - k + 3k^2 + 3k = 6a + 3k(k+1).\] For the integer \(k\), \(k(k+1)\) is even from part (a), so let \(k(k+1) = 2b\) for some integer \(b\). Hence \((k+1)^3 - (k+1) = 6a + 3(2b) = 6(a+b)\). Thus \(6 \mid \big((k+1)^3 - (k+1)\big)\) and so \(P(k+1)\) is true.

    Thus, by the principle of mathematical induction, \(n^3 - n\) is a multiple of 6 for each integer \(n \geq 2\). \(\square\)

    1. For all integers \(n \geq 7\), \(3^n < n!\).

    Proof. Let \(P(n)\) be the predicate \(3^n < n!\).

    Basis: \(P(7)\) is the statement \(3^7 < 7!\). The left-hand side is \(3^7 = 2187\) and the right-hand side is \(7! = 5040\), so \(P(7)\) holds.

    (The inductive step is to show that for all integers \(k \geq 7\), \(P(k)\) implies \(P(k+1)\).)

    Inductive hypothesis: let \(k \geq 7\) be an integer and assume that \(P(k)\) is true; that is \(3^k < k!\). We will now show that \(P(k+1)\) holds, so we will show \(3^{k+1} < (k+1)!\).

    Taking the right-hand side, we have \[\begin{aligned} (k+1)! &= (k+1) \cdot k! \\ &> (k+1) \cdot 3^k && \text{by the inductive hypothesis} \\ &> 3 \cdot 3^k && \text{since } k \geq 7 \\ &= 3^{k+1} \end{aligned}\] Hence \(P(k+1)\) is true.

    Thus by the principle of mathematical induction, for all integers \(n \geq 7\), \(3^n < n!\). \(\square\)

    1. For each integer \(n \geq 1\), \(\displaystyle\sum_{i=1}^{n} i^3 = \frac{n^2(n+1)^2}{4}\).

    Proof. Let \(P(n)\) be the predicate \[\sum_{i=1}^{n} i^3 = \frac{n^2(n+1)^2}{4}.\]

    Basis: consider \(n = 1\). We have \(\sum_{i=1}^{1} i^3 = 1^3 = 1\) and \(\frac{1^2(1+1)^2}{4} = \frac{4}{4} = 1\) so \(P(1)\) is true.

    (The inductive step is to show that for all integers \(k \geq 1\), \(P(k)\) implies \(P(k+1)\).)

    Inductive hypothesis: suppose \(k \geq 1\) is an integer and \(P(k)\) is true. That is, \[\sum_{i=1}^{k} i^3 = \frac{k^2(k+1)^2}{4}.\]

    Now consider \(n = k+1\). \[\begin{aligned} \sum_{i=1}^{k+1} i^3 &= \sum_{i=1}^{k} i^3 + (k+1)^3 && \text{(splitting off the last term)} \\ &= \frac{k^2(k+1)^2}{4} + (k+1)^3 && \text{by the inductive hypothesis} \\ &= \frac{k^2(k+1)^2}{4} + \frac{4(k+1)^3}{4} \\ &= \frac{(k+1)^2}{4}\big(k^2 + 4(k+1)\big) \\ &= \frac{(k+1)^2}{4}\big(k^2 + 4k + 4\big) \\ &= \frac{(k+1)^2}{4}(k+2)^2 \end{aligned}\] Thus \(P(k+1)\) is true.

    Therefore, by the principle of mathematical induction, \(\sum_{i=1}^{n} i^3 = \frac{n^2(n+1)^2}{4}\) for every integer \(n \geq 1\). \(\square\)

    1. For each integer \(n \geq 0\), \(1 + 4n \leq 5^n\).

    Proof. Let \(P(n)\) be the predicate \(1 + 4n \leq 5^n\).

    Basis: when \(n = 0\) we have \(1 + 4 \cdot 0 \leq 5^0\), so \(P(0)\) is true.

    (The inductive step is to show that for all integers \(k \geq 0\), \(P(k)\) implies \(P(k+1)\).)

    Inductive hypothesis: suppose \(k \geq 0\) is an integer and \(P(k)\) is true; that is, \(1 + 4k \leq 5^k\).

    Now consider \(n = k+1\). Starting with the left-hand side of \(P(k+1)\), we have \[\begin{aligned} 1 + 4(k+1) &= 1 + 4k + 4 \\ &\leq 5^k + 4 && \text{by the inductive hypothesis} \\ &\leq 5^k + 4 \cdot 5^k && \text{since } k \geq 0 \text{ so } 5^k \geq 1 \\ &= 5(5^k) \\ &= 5^{k+1} \end{aligned}\] so \(P(k+1)\) is true.

    Thus, by the principle of mathematical induction, \(1 + 4n \leq 5^n\) for every integer \(n \geq 0\). \(\square\)

§18 — Strong Induction and the Well Ordering Principle

Epp 4th ed. pp. 268–279 · 5th ed. (metric) pp. 301–314. Lecture 14 (week 5).

Problems

  1. Consider the statement

    \[\text{For every integer } n \geq 2, \text{ there exist integers } a, b \geq 0 \text{ such that } n = 2a + 3b.\]

    1. Prove the statement using the well-ordering principle.

    2. Prove the statement using strong mathematical induction.

  2. The Lucas sequence \(a_1, a_2, a_3, \dots\) is defined by \(a_1 = 1\), \(a_2 = 3\) and \(a_k = a_{k-1} + a_{k-2}\) for all integers \(k \geqslant 3\). Use strong mathematical induction to prove that \(a_n \leqslant \left(\dfrac{7}{4}\right)^n\) for all \(n \in \mathbb{Z}^+\).

  3. Suppose \(x\) is a real number and \(x + \frac{1}{x}\) is an integer. Use strong mathematical induction to prove \(a_n := x^n + \frac{1}{x^n}\) is an integer for all \(n \geq 0\).

Solutions

  1. For every integer \(n \geq 2\), there exist integers \(a, b \geq 0\) such that \(n = 2a + 3b\).

    Proof (using the well-ordering principle). Consider the set of counterexamples \[S = \{\, n \in \mathbb{Z}^{\geq 2} : \text{there do not exist integers } a, b \geq 0 \text{ with } n = 2a + 3b \,\}.\] If \(S\) is empty, we are done. Otherwise, by the well-ordering principle, \(S\) has a least element. Let us call it \(N\).

    Notice that \(2 \notin S\) because \(2 = 2 \cdot 1 + 3 \cdot 0\), and \(3 \notin S\) because \(3 = 2 \cdot 0 + 3 \cdot 1\). Also \(4 \notin S\) because \(4 = 2 \cdot 2 + 3 \cdot 0\). Hence \(N \geq 5\). By minimality of \(N\), we must have \(N - 2 \notin S\). Therefore there exist integers \(a, b \geq 0\) such that \[N - 2 = 2a + 3b.\] In other words, we have \[N = 2(a+1) + 3b.\] This means \(N \notin S\), contradicting the assumption that \(N \in S\). Hence \(S\) is empty, and the statement holds for all \(n \geq 2\). \(\square\)

    Proof (using strong induction). Let \(P(n)\) be the predicate: “there exist integers \(a, b \geq 0\) such that \(n = 2a + 3b\).”

    Basis: we have \[2 = 2 \cdot 1 + 3 \cdot 0, \qquad 3 = 2 \cdot 0 + 3 \cdot 1, \qquad 4 = 2 \cdot 2 + 3 \cdot 0,\] so \(P(2)\), \(P(3)\), \(P(4)\) are true.

    (The inductive step is to show that for every integer \(k \geq 4\), if \(P(2), P(3), \dots, P(k)\) are all true then \(P(k+1)\) is true.)

    Inductive hypothesis: suppose \(k \geq 4\) is an integer and assume \(P(m)\) is true for every integer \(m = 2, 3, \dots, k\).

    Consider \(n = k+1\). Since \(k \geq 4\), we have \(k - 1 \geq 3\), so by the inductive hypothesis \(P(k-1)\) is true. Thus there exist \(a, b \geq 0\) such that \[k - 1 = 2a + 3b.\] Adding 2 gives \[k + 1 = 2(a+1) + 3b,\] so \(P(k+1)\) holds. \(\square\)

  2. The Lucas sequence \(a_1, a_2, a_3, \dots\) is defined by \(a_1 = 1\), \(a_2 = 3\) and \(a_k = a_{k-1} + a_{k-2}\) for all integers \(k \geqslant 3\). Prove that \(a_n \leqslant \left(\frac{7}{4}\right)^n\) for all \(n \in \mathbb{Z}^+\).

    Proof. Let \(P(n)\) be the predicate \(a_n \leqslant \left(\frac{7}{4}\right)^n\).

    Basis: when \(n = 1\), \(a_1 = 1 \leqslant \left(\frac{7}{4}\right)^1 = 1\frac{3}{4}\) is true, so \(P(1)\) is true. When \(n = 2\), \(P(2)\) holds because \(a_2 = 3 \leqslant \left(\frac{7}{4}\right)^2 = \frac{49}{16} = 3\frac{1}{16}\).

    (The inductive step is to show that for every integer \(k \geq 1\), if \(P(1), \dots, P(k)\) are all true then \(P(k+1)\) is true.)

    Inductive hypothesis: suppose \(k \geq 1\) is an integer and that \(P(m)\) is true for all \(m\) with \(1 \leq m \leq k\).

    Now consider \(P(k+1)\), which we want to prove holds. We have \[\begin{aligned} a_{k+1} &= a_k + a_{k-1} \\ &\leqslant \left(\frac{7}{4}\right)^k + \left(\frac{7}{4}\right)^{k-1} && \text{by the inductive hypothesis} \\ &= \left(\frac{7}{4}\right)^{k-1}\left(\frac{7}{4} + 1\right) \\ &= \left(\frac{7}{4}\right)^{k-1}\left(\frac{11}{4}\right) \\ &\leqslant \left(\frac{7}{4}\right)^{k-1}\left(\frac{49}{16}\right) && \text{noting } \tfrac{11}{4} = \tfrac{44}{16} < \tfrac{49}{16} \\ &= \left(\frac{7}{4}\right)^{k-1}\left(\frac{7}{4}\right)^2 \\ &= \left(\frac{7}{4}\right)^{k+1}. \end{aligned}\] Hence \(P(k+1)\) holds. Therefore, by the principle of strong mathematical induction, \(a_n \leqslant \left(\frac{7}{4}\right)^n\) for all integers \(n \geqslant 1\). \(\square\)

  3. Suppose \(x\) is a real number and \(x + \frac{1}{x}\) is an integer. Prove \(a_n := x^n + \frac{1}{x^n}\) is an integer for all \(n \geq 0\).

    Proof. Suppose \(x\) is a real number and \(x + \frac{1}{x}\) is an integer. Let \(P(n)\) be the predicate “\(a_n\) is an integer.”

    Basis: \(P(0)\) holds since \(a_0 = x^0 + \frac{1}{x^0} = 2\). \(P(1)\) holds since \(a_1 = x + \frac{1}{x}\), which is an integer by assumption.

    (The inductive step is to show that for every integer \(k \geq 1\), if \(P(0), P(1), \dots, P(k)\) are all true then \(P(k+1)\) is true.)

    Inductive hypothesis: suppose \(k \geq 1\) is an integer and \(P(m)\) holds for all integers \(m = 0, 1, \dots, k\).

    We now aim to show that \(P(k+1)\) is true. Notice that \[\begin{aligned} a_1 a_k &= \left(x + \frac{1}{x}\right)\left(x^k + \frac{1}{x^k}\right) \\ &= \left(x^{k+1} + \frac{1}{x^{k+1}}\right) + \left(x^{k-1} + \frac{1}{x^{k-1}}\right) = a_{k+1} + a_{k-1}. \end{aligned}\] Therefore \(a_{k+1} = a_1 a_k - a_{k-1}\). By the inductive hypothesis, \(a_k\) and \(a_{k-1}\) are integers, and \(a_1\) is an integer, so \(a_{k+1}\) is an integer. Hence \(P(k+1)\) holds.

    Therefore, by the principle of strong mathematical induction, \(a_n\) is an integer for all \(n \geq 0\). \(\square\)

§19 — Recursive Definitions

Epp 4th ed. pp. 290–304 · 5th ed. (metric) pp. 325–340. Lecture 15 (week 6).

Problems

  1. Define the sequence \(b_1 = 2\), \(b_2 = 6\) and

    \[b_k = b_{k-1} + b_{k-2} + \gcd(b_{k-1}, b_{k-2})\]

    for all \(k \geq 3\). Use strong induction to prove \(b_n\) is even for all \(n \geq 1\).

  2. A sequence is defined by \(a_1 = 2\) and

    \[a_k = k(a_{k-1} + k - 1)\]

    for all integers \(k \geq 2\). Use mathematical induction to prove \(a_n\) is even for all \(n \geq 1\).

  3. Recursively define the set \(S\) of binary strings as follows.

    1. (Base rule) \(1 \in S\).

    2. (Recursion rule) If \(s \in S\) then \(0s \in S\) and \(1s \in S\) (e.g. if \(s = 1\) then \(0s = 01\)).

    3. (Restriction rule) No other strings lie in \(S\) besides those given by the rules above.

    What are the strings contained in \(S\) arising from the first three applications of the recursive rule?

Solutions

  1. Proof. Let \(P(n)\) be the predicate “\(b_n\) is even”.

    Base case. The base cases are \(n = 1, 2\). We have \(b_1 = 2\) and \(b_2 = 6\) so \(P(1)\) and \(P(2)\) are true.

    Inductive hypothesis. Assume \(P(m)\) is true for all integers \(m\) with \(1 \leq m \leq k\), where \(k \geq 2\) is an integer. That is, we assume \(b_1, b_2, \dots, b_k\) are all even.

    Inductive step. We need to show that \(P(k+1)\) is true, i.e. \(b_{k+1}\) is even. Using the recursive definition, we have \[b_{k+1} = b_k + b_{k-1} + \gcd(b_k, b_{k-1}).\] By the inductive hypothesis, \(b_k\) and \(b_{k-1}\) are even, so \(2 \mid b_k\) and \(2 \mid b_{k-1}\), hence \(2 \mid \gcd(b_k, b_{k-1})\) as well. Therefore each term on the right-hand side is even, so their sum \(b_{k+1}\) is even. Thus \(P(k+1)\) holds.

    Therefore, \(b_n\) is even for all integers \(n \geq 1\). \(\square\)

  2. Proof. Let \(P(n)\) be the predicate “\(a_n\) is even”.

    Base case. The base case is \(n = 1\). We have \(a_1 = 2\), which is even, so \(P(1)\) is true.

    Inductive hypothesis. Assume \(P(k)\) is true for some integer \(k \geq 1\). That is, assume \(a_k\) is even.

    Inductive step. We need to show that \(P(k+1)\) is true, i.e. \(a_{k+1}\) is even. By the recursive definition, we have \[a_{k+1} = (k+1)\big(a_k + (k+1) - 1\big) = (k+1)(a_k + k).\] Note that \(a_k\) is even by the inductive hypothesis. We split into cases:

    • If \(k\) is even, then \(a_k + k\) is even, hence \(a_{k+1}\) is even.
    • If \(k\) is odd, then \(k + 1\) is even, hence \(a_{k+1}\) is even.

    In all cases, \(a_{k+1}\) is even, so \(P(k+1)\) is true.

    Therefore, \(a_n\) is even for all integers \(n \geq 1\). \(\square\)

  3. By the base rule, we only know \(1 \in S\).

    First application of the recursive rule:

    \[01 \in S, \qquad 11 \in S.\]

    Second application of the recursive rule:

    \[001 \in S, \quad 101 \in S, \quad 011 \in S, \quad 111 \in S.\]

    Third application of the recursive rule:

    \[0001 \in S, \quad 1001 \in S, \quad 0101 \in S, \quad 1101 \in S,\] \[0011 \in S, \quad 1011 \in S, \quad 0111 \in S, \quad 1111 \in S.\]

§20 — Solving Recurrence Relations

Epp 4th ed. pp. 304–328 · 5th ed. (metric) pp. 340–364. Lecture 16 (week 6).

Problems

  1. Define a sequence \(a_n = 2^n + 5 \cdot 3^n\) for all integers \(n \geq 0\). Show that

    \[a_n = 5a_{n-1} - 6a_{n-2}\]

    for all integers \(n \geq 2\).

  2. Let \(\{b_n\}_{n \geq 0}\) be the sequence defined by

    \[b_0 = 2, \qquad b_1 = 5, \qquad \text{and} \qquad b_k = 2b_{k-1} - b_{k-2} \text{ for each integer } k \geq 2.\]

    1. Write down the values of \(b_0, b_1, b_2, b_3, b_4, b_5\).

    2. Based on your answers for part (a), guess an explicit formula for this sequence.

    3. Prove that your guess from part (b) is correct.

  3. Let \(T_n\) be a sequence defined by \(T_1 = 1\) and

    \[T_{n+1} = 1 + (T_1 \times T_2 \times T_3 \times \cdots \times T_n)\]

    for all \(n \geq 1\). Prove that \(T_{n+1} = (T_n)^2 - T_n + 1\) for all \(n \geq 2\).

  4. Consider the Fibonacci sequence defined recursively by

    \[\begin{aligned} a_0 &= 0 \\ a_1 &= 1 \\ a_n &= a_{n-2} + a_{n-1} \qquad \text{for all integers } n \geq 2. \end{aligned}\]

    Use strong mathematical induction to prove the explicit formula

    \[a_n = \frac{\phi^n - \psi^n}{\sqrt{5}}\]

    for all integers \(n \geq 0\), where \(\phi = \frac{1+\sqrt{5}}{2}\) and \(\psi = \frac{1-\sqrt{5}}{2}\). (You may use the fact that \(\phi^2 = \phi + 1\) and \(\psi^2 = \psi + 1\).)

Solutions

  1. Define a sequence \(a_n = 2^n + 5 \cdot 3^n\) for all integers \(n \geq 0\). Show that \(a_n = 5a_{n-1} - 6a_{n-2}\) for all integers \(n \geq 2\).

    Proof. For \(n \geq 2\), we have \(n - 2 \geq 0\) so all three terms \(a_n\), \(a_{n-1}\) and \(a_{n-2}\) are defined in the given formula. Then, \[\begin{aligned} 5a_{n-1} - 6a_{n-2} &= 5\left(2^{n-1} + 5 \cdot 3^{n-1}\right) - 6\left(2^{n-2} + 5 \cdot 3^{n-2}\right) \\ &= \left(5 \cdot 2^{n-1} - 6 \cdot 2^{n-2}\right) + \left(25 \cdot 3^{n-1} - 30 \cdot 3^{n-2}\right) \\ &= (10 - 6)2^{n-2} + (75 - 30)3^{n-2} \\ &= 4 \cdot 2^{n-2} + 45 \cdot 3^{n-2} \\ &= 2^n + 5 \cdot 3^n \\ &= a_n. \end{aligned}\] Hence \(a_n = 5a_{n-1} - 6a_{n-2}\) for all integers \(n \geq 2\). \(\square\)

    \[\begin{aligned} b_0 &= 2 \\ b_1 &= 5 \\ b_2 &= 2(5) - 2 = 8 \\ b_3 &= 2(8) - 5 = 11 \\ b_4 &= 2(11) - 8 = 14 \\ b_5 &= 2(14) - 11 = 17 \end{aligned}\]

    1. Guess that \(b_n = 3n + 2\) for each integer \(n \geq 0\).

    Proof. The guess gives \(b_0 = 3(0) + 2 = 2\) and \(b_1 = 3(1) + 2 = 5\) so it gives the correct initial conditions. Now we substitute the guess into the recurrence relation. The LHS of the recurrence relation is \(b_k = 3k + 2\). The RHS of the recurrence relation is \[\begin{aligned} 2b_{k-1} - b_{k-2} &= 2\big(3(k-1) + 2\big) - \big(3(k-2) + 2\big) \\ &= 2(3k - 1) - (3k - 4) \\ &= 6k - 2 - 3k + 4 \\ &= 3k + 2 \end{aligned}\] So the guess also satisfies the recurrence relation. Thus the guess is correct. \(\square\)

    Alternative proof (by strong induction). Let \(P(n)\) be the predicate \(b_n = 3n + 2\).

    Basis: \(b_0 = 2\) from the recursive definition and \(3(0) + 2 = 2\) so \(P(0)\) is true. \(b_1 = 5\) from the recursive definition and \(3(1) + 2 = 5\) so \(P(1)\) is true.

    Inductive hypothesis: suppose \(k \geq 1\) is an integer and \(P(0), P(1), \dots, P(k)\) are true.

    Now \[\begin{aligned} b_{k+1} &= 2b_k - b_{k-1} && \text{by the rec. rel. since } k+1 \geq 2 \\ &= 2(3k+2) - \big(3(k-1) + 2\big) && \text{since } P(k) \text{ and } P(k-1) \text{ are true} \\ &= 6k + 4 - (3k - 1) \\ &= 3k + 5 \\ &= 3(k+1) + 2 \end{aligned}\] So \(P(k+1)\) is true. Therefore, by the principle of (strong) mathematical induction, the guess is correct for this recursively defined sequence. \(\square\)

  3. Let \(T_n\) be a sequence defined by \(T_1 = 1\) and \(T_{n+1} = 1 + (T_1 \times \cdots \times T_n)\) for all \(n \geq 1\). Prove that \(T_{n+1} = (T_n)^2 - T_n + 1\) for all \(n \geq 2\).

    Proof. We can prove this simply by applying the recursive formula twice, once for \(T_{n+1}\) and once for \(T_n\): \[\begin{aligned} T_{n+1} &= 1 + (T_1 \cdot T_2 \cdot T_3 \cdots T_n) && \text{by the recursive definition} \\ &= 1 + T_n \cdot (T_1 \cdot T_2 \cdot T_3 \cdots T_{n-1}) \\ &= 1 + T_n \cdot (T_n - 1) && \text{since } T_n = 1 + (T_1 \cdots T_{n-1}) \text{ by rec. def.} \\ &= 1 + (T_n)^2 - T_n \\ &= (T_n)^2 - T_n + 1. \end{aligned}\] Note that both applications of the recursive definition are valid, since \(n - 1 \geq 1\). \(\square\)

  4. Fibonacci explicit formula.

    Proof. Let \(P(n)\) be the predicate \[a_n = \frac{\phi^n - \psi^n}{\sqrt{5}}.\]

    Basis. The base cases are \(n = 0, 1\), which are true because \[\frac{\phi^0 - \psi^0}{\sqrt{5}} = \frac{1-1}{\sqrt{5}} = 0 = a_0,\] and \[\frac{\phi - \psi}{\sqrt{5}} = \frac{\frac{1+\sqrt{5}}{2} - \frac{1-\sqrt{5}}{2}}{\sqrt{5}} = \frac{\sqrt{5}}{\sqrt{5}} = 1 = a_1.\]

    Inductive hypothesis. Now suppose that for some integer \(m \geq 1\), we have that \(P(i)\) is true for all \(0 \leq i \leq m\). In particular, \(P(m)\) and \(P(m-1)\) are true, that is, \[a_m = \frac{\phi^m - \psi^m}{\sqrt{5}} \qquad \text{and} \qquad a_{m-1} = \frac{\phi^{m-1} - \psi^{m-1}}{\sqrt{5}}.\]

    We need to show that \(P(m+1)\) is true. That is, we aim to show \[a_{m+1} = \frac{\phi^{m+1} - \psi^{m+1}}{\sqrt{5}}.\]

    Starting with the right-hand side of \(P(m+1)\), we have \[\begin{aligned} \frac{\phi^{m+1} - \psi^{m+1}}{\sqrt{5}} &= \frac{\phi^2 \phi^{m-1} - \psi^2 \psi^{m-1}}{\sqrt{5}} \\ &= \frac{(\phi+1)\phi^{m-1} - (\psi+1)\psi^{m-1}}{\sqrt{5}} && \text{since } \phi^2 = \phi+1 \text{ and } \psi^2 = \psi+1 \\ &= \frac{(\phi^m + \phi^{m-1}) - (\psi^m + \psi^{m-1})}{\sqrt{5}} \\ &= \frac{\phi^m - \psi^m}{\sqrt{5}} + \frac{\phi^{m-1} - \psi^{m-1}}{\sqrt{5}} \\ &= a_m + a_{m-1} && \text{by the inductive hypothesis} \\ &= a_{m+1} && \text{by definition.} \end{aligned}\] Hence \(P(m+1)\) is true. \(\square\)

§21 — Set Theory Definitions

Epp 4th ed. pp. 336–351 · 5th ed. (metric) pp. 377–391. Lecture 17 (week 6).

Problems

  1. Define the sets

    \[\begin{aligned} V &= \{x \in \mathbb{Z} \mid 2 \leqslant x \leqslant 11 \text{ and } x \text{ is odd}\}, \\ X &= \{x \in \mathbb{Z} \mid -1 \leqslant x \leqslant 4\}, \\ Y &= \{2, 3, 7, 9\}, \\ Z &= \{x \in \mathbb{Z} \mid x \neq 0\}. \end{aligned}\]

    Write down the following sets, listing their elements and using brackets as appropriate.

    1. \(V \cap X\). (b) \(V \cup Y\). (c) \(X \cap Z\).

    2. \(X \cap (Z - Y)\). (e) \((X \cup Y) \cap V\). (f) \(X \cup (Y \cap V)\).

  2. For each of the following sets, write down its elements and then state the size of the set; that is, find \(|A|\), \(|B|\), \(|C|\) and \(|D|\).

    1. \(A = \{x \in \mathbb{Z} \mid \exists\, i \in \mathbb{Z} \text{ such that } x = 1 + (-1)^i\}\).

    2. \(B = \{y \in \mathbb{Z} \mid -8 < y \leqslant -5\}\).

    3. \(C = \{s \in \mathbb{Z} \mid s^2 + 1 = 0\}\).

    4. \(D = \{t \in \mathbb{Z} \mid t^4 = 1\}\).

  3. Define the sets \(A = \{0,1,2,3,4\}\), \(B = \{3,4,5,6\}\) and \(C = \{1,2,4,5,7\}\). List the elements of the set

    \[R = \big((A - B) \cap (A - C)\big) \cup (B \cap C \cap A).\]

  4. Are the following statements true or false? Justify in one or two sentences.

    1. \(x \in \{x\}\) (b) \(\{x\} \subseteq \{x\}\) (c) \(\{\emptyset\} \in \{\{\emptyset\}\}\)

    2. \(\emptyset \subseteq \{x\}\) (e) \(\emptyset \in \{x\}\)

Solutions

    1. \(V \cap X = \{3\}\)

    2. \(V \cup Y = \{2, 3, 5, 7, 9, 11\}\)

    3. \(X \cap Z = \{-1, 1, 2, 3, 4\}\)

    4. \(X \cap (Z - Y) = \{-1, 1, 4\}\)

    5. \((X \cup Y) \cap V = \{3, 7, 9\}\)

    6. \(X \cup (Y \cap V) = \{-1, 0, 1, 2, 3, 4, 7, 9\}\)

    1. Note that \((-1)^i\) is either \(+1\) or \(-1\) (according as \(i\) is even or odd). So \(x\) is either \(1+1\) or \(1-1\), that is, 2 or 0. So \(A = \{2, 0\}\) and \(|A| = 2\).

    2. \(B = \{-7, -6, -5\}\), so \(|B| = 3\).

    3. If \(s^2 + 1 = 0\), we have \(s^2 = -1\), and there are no such real numbers \(s\), let alone integers! So \(C = \emptyset\) and \(|C| = 0\).

    4. \(D = \{-1, 1\}\), since \(t^4 = 1\) with \(t \in \mathbb{Z}\) implies \(t = \pm 1\). Hence \(|D| = 2\).

  3. We have \(A - B = \{0,1,2\}\) and \(A - C = \{0,3\}\), so \((A-B) \cap (A-C) = \{0\}\). Also \(B \cap C \cap A = \{4\}\). Thus \(R = \{0\} \cup \{4\} = \{0, 4\}\).

    1. True: \(\{x\}\) is the set whose only element is \(x\), so \(x \in \{x\}\).

    2. True: every set is a subset of itself, so \(\{x\} \subseteq \{x\}\).

    3. True: \(\{\{\emptyset\}\}\) has the single element \(\{\emptyset\}\), so \(\{\emptyset\} \in \{\{\emptyset\}\}\).

    4. True: \(\emptyset \subseteq S\) for every set \(S\), in particular \(\emptyset \subseteq \{x\}\).

    5. False (in general): \(\emptyset \in \{x\}\) would require \(\emptyset = x\).

Application — Computer Representations of Sets

Not assessable for MATH1061. Taken from K. Rosen, Discrete Mathematics and its Applications, 7th edition, 1991, pages 134–138.

There are various ways to represent sets using a computer. One such method involves the use of binary strings.

If we fix a finite universal set \(U\) with an arbitrary ordering, for example \(U = \{a_1, a_2, \dots, a_n\}\), we can represent subsets of \(U\) using binary strings of length \(n\). A 1 in the \(i\)th position of the string indicates that element \(a_i\) is in the subset and a 0 indicates it is not.

For example if \(U = \{1,2,3\}\) then the string \(000\) represents the empty set, \(101\) represents the subset \(\{1,3\}\) and \(111\) represents the entire set \(U\).

The set operations union and intersection can be computed at the level of binary strings by utilising the bitwise OR and AND operations. The union of two sets corresponds to the bitwise OR operation on the strings and the intersection corresponds to bitwise AND.

Application problems

  1. Define the universal set \(U = \{1,2,3,4,5,6,7,8,9,10\}\).

    1. Express each of the following subsets of \(U\) using binary strings:

    1. \(\{3,4,5\}\)

    2. \(\{1,3,6,10\}\)

    3. \(\{a \in U \mid a \text{ is prime}\}\).

    1. Using the same \(U\), find the set represented by the following strings:

    1. \(1111001111\)

    2. \(0101111000\)

    3. \(1000000001\)

    1. Use binary strings to compute the following set operations.

    1. \(\{1,3,4,6,10\} \cap \{2,3,6,9,10\}\)

    2. \(\{2,4,6,8\} \cup \{3,4,5\}\)

Application solutions
      1. \(0011100000\)

    1. \(1010010001\)

    2. \(0110101000\) — the primes in \(U\) are \(2, 3, 5, 7\). (The source PDF prints \(1010111000\) here, which decodes to \(\{1,3,5,6,7\}\) and is not the set of primes; the string above is the correct one.)

      1. \(\{1,2,3,4,7,8,9,10\}\)

    1. \(\{2,4,5,6,7\}\)

    2. \(\{1,10\}\)

      1. \(1011010001 \text{ AND } 0110010011 = 0010010001\), so the intersection is \(\{3,6,10\}\).
    1. \(0101010100 \text{ OR } 0011100000 = 0111110100\), so the union is \(\{2,3,4,5,6,8\}\).

§22 — More Definitions and Examples of Sets

Epp 4th ed. pp. 336–351 · 5th ed. (metric) pp. 377–391. Lecture 17 (week 6).

Problems

  1. Let \(A = \{1,2\}\), \(B = \{2,3\}\) and \(C = \{3,4\}\). List the elements of the following sets.

    1. \(A \cap C\)

    2. \((A \cup C) - B\)

    3. \((A \times B) \cap (A \times C)\)

    4. \(\mathcal{P}(B \cap C)\)

    5. \(\{x : (x \subseteq A) \wedge (x \in \mathcal{P}(B))\}\)

  2. Are the following statements true or false?

    1. \(\mathbb{Q} \in \mathbb{R}\).

    2. \(\{1,2\} \subseteq \mathbb{Z}\).

    3. \(\{-3, 0\} \subseteq \mathcal{P}(\mathbb{Z})\).

    4. \(\{(1,2), (0,1)\} \in \mathbb{Z} \times \mathbb{Z}\).

  3. Let \(X = \mathcal{P}(\emptyset)\) and \(Y = \mathcal{P}(\mathcal{P}(\emptyset))\). List the elements of each of the following sets.

    1. \(X\) (b) \(Y\) (c) \(X \cap Y\) (d) \(X \cup Y\)

  4. Define sets \(S = \{3,5\}\) and \(T = \{2,3,7\}\). Are the following statements true or false?

    1. \(S \subseteq T\)

    2. \(\emptyset \in \mathcal{P}(S)\)

    3. \(\emptyset \in T\)

    4. \((3,3,3) \in S \times T\)

Solutions

    1. \(A \cap C = \emptyset\).

    2. \((A \cup C) - B = \{1, 4\}\).

    3. \((A \times B) \cap (A \times C) = \{(1,3), (2,3)\}\).

    4. \(\mathcal{P}(B \cap C) = \{\emptyset, \{3\}\}\).

    5. \(\{x : (x \subseteq A) \wedge (x \in \mathcal{P}(B))\} = \{\emptyset, \{2\}\}\).

    1. False. (\(\mathbb{Q}\) is a subset of \(\mathbb{R}\), not an element of it.)

    2. True.

    3. False. (The elements of \(\mathcal{P}(\mathbb{Z})\) are sets of integers; \(-3\) and \(0\) are integers, not sets.)

    4. False. (It is a subset of \(\mathbb{Z} \times \mathbb{Z}\), not an element — the elements of \(\mathbb{Z} \times \mathbb{Z}\) are ordered pairs.)

    1. \(X = \mathcal{P}(\emptyset) = \{\emptyset\}\).

    2. \(Y = \mathcal{P}(\mathcal{P}(\emptyset)) = \{\emptyset, \{\emptyset\}\}\).

    3. \(X \cap Y = \{\emptyset\}\).

    4. \(X \cup Y = \{\emptyset, \{\emptyset\}\}\).

    1. False. (\(5 \in S\) but \(5 \notin T\).)

    2. True. (The empty set is an element of every power set.)

    3. False. (\(T\)’s elements are \(2\), \(3\) and \(7\).)

    4. False. (The elements of \(S \times T\) are ordered pairs, not triples.)

§23 — Properties of Sets

Epp 4th ed. pp. 352–366 · 5th ed. (metric) pp. 391–407. Lecture 18 (week 7).

Problems

    1. Use the element method to prove that for all sets \(A\) and \(B\)

    \[A \cup (B - A) = A \cup B.\]

    1. Use the element method to prove that for all sets \(X\), \(Y\) and \(Z\),

    \[(X - Z) \cap (Y - Z) \cap (X - Y) = \emptyset.\]

    1. Use set identities to prove that for all sets \(A\), \(B\) and \(C\)

    \[(A - B) - C = A - (B \cup C).\]

  2. Prove or disprove the following:

    1. For all sets \(A\) and \(B\), we have \(\mathcal{P}(A \cup B) = \mathcal{P}(A) \cup \mathcal{P}(B)\).

    2. For all sets \(A\) and \(B\), we have \(\mathcal{P}(A) = \mathcal{P}(B) \iff A = B\).

    3. There exists a set \(A\) such that \(\mathcal{P}(A) = \{\{a\}, \{b\}, \{c\}, \{a,b\}, \{b,c\}, \{a,c\}, \{a,b,c\}\}\).

    4. Let \(A\), \(B\) and \(C\) be sets. If \(B \subseteq C\) then \(A \times B \subseteq A \times C\).

  3. Let \(A\), \(B\) and \(C\) be sets. Prove that

    \[A - (B \cup C) = (A - B) - C.\]

Solutions

    1. For all sets \(A\) and \(B\), \(A \cup (B - A) = A \cup B\).

    Proof. First we show that \(A \cup (B-A) \subseteq A \cup B\).

    Consider an arbitrary element \(x \in A \cup (B-A)\). Then we know \(x \in A\) or \(x \in B - A\). This means \(x \in A\), or \(x \in B\) and \(x \notin A\). If \(x \in A\), then \(x \in A \cup B\). Similarly if \(x \in B - A\), then it must be in \(B\) and so \(x \in A \cup B\). In either case, we have \(x \in A \cup B\), thus \(A \cup (B-A) \subseteq A \cup B\).

    Now we show that \(A \cup B \subseteq A \cup (B-A)\).

    Let \(x \in A \cup B\), then \(x \in A\) or \(x \in B\) or it is in both \(A\) and \(B\). If \(x \in A\), then \(x \in A \cup (B-A)\). If \(x \notin A\) then \(x \in B\) since \(x \in A \cup B\). So we have that \(x \in B\) and \(x \notin A\), hence \(x \in B - A\) and thus \(x \in A \cup (B-A)\). This proves that \(A \cup B \subseteq A \cup (B-A)\), and together this proves that \(A \cup B = A \cup (B-A)\). \(\square\)

    1. For all sets \(X\), \(Y\) and \(Z\), \((X-Z) \cap (Y-Z) \cap (X-Y) = \emptyset\).

    Proof. Assume that \((X-Z) \cap (Y-Z) \cap (X-Y) \neq \emptyset\).

    So let \(x \in (X-Z) \cap (Y-Z) \cap (X-Y)\). Then, in particular, \(x \in Y - Z\) and \(x \in X - Y\). The former implies \(x \in Y\), but the latter implies \(x \notin Y\). No such \(x\) exists. Therefore the set is empty. \(\square\)

    1. For all sets \(A\), \(B\) and \(C\), \((A-B) - C = A - (B \cup C)\).

    Proof. Using \(A - B = A \cap B^c\), \((S - T) = S \cap T^c\), and De Morgan’s law, we have \[(A-B) - C = (A \cap B^c) \cap C^c = A \cap (B^c \cap C^c) = A \cap (B \cup C)^c = A - (B \cup C). \qquad \square\]

    1. For all sets \(A\) and \(B\), \(\mathcal{P}(A \cup B) = \mathcal{P}(A) \cup \mathcal{P}(B)\). False. Counterexample: \(A = \{1\}\), \(B = \{2\}\). Then

    \[\mathcal{P}(A \cup B) = \{\emptyset, \{1\}, \{2\}, \{1,2\}\} \neq \{\emptyset, \{1\}\} \cup \{\emptyset, \{2\}\} = \{\emptyset, \{1\}, \{2\}\}.\]

    1. For all sets \(A\) and \(B\), \(\mathcal{P}(A) = \mathcal{P}(B) \iff A = B\). True.

    (\(\Longleftarrow\)) If \(A = B\) then clearly \(\mathcal{P}(A) = \mathcal{P}(B)\).

    (\(\Longrightarrow\)) We next show that if \(\mathcal{P}(A) = \mathcal{P}(B)\), then \(A = B\). Suppose (for a contradiction) that \(\mathcal{P}(A) = \mathcal{P}(B)\) but \(A \neq B\). Then (without loss of generality), there exists an element \(x\) in \(A\) that is not in \(B\). Therefore, there is an element \(\{x\}\) in \(\mathcal{P}(A)\) that is not in \(\mathcal{P}(B)\). Therefore \(\mathcal{P}(A) \neq \mathcal{P}(B)\) — a contradiction. Thus, for all sets \(A\) and \(B\), if \(\mathcal{P}(A) = \mathcal{P}(B)\) then \(A = B\). \(\square\)

    1. There exists a set \(A\) such that \(\mathcal{P}(A) = \{\{a\}, \{b\}, \{c\}, \{a,b\}, \{b,c\}, \{a,c\}, \{a,b,c\}\}\). False.

    No such set exists. The power set of any set must contain the empty set — that is, \(\emptyset \in \mathcal{P}(A)\), where \(A\) is any set. Also, the power set of any set of size \(n\) has cardinality \(2^n\). The power set above contains 7 elements, and there does not exist any integer \(n\) such that \(2^n = 7\).

    1. Let \(A\), \(B\) and \(C\) be sets. If \(B \subseteq C\) then \(A \times B \subseteq A \times C\). True.

    First note that if \(A = \emptyset\) then \(A \times B = A \times C = \emptyset\) and the statement is true. Also, if \(B = \emptyset\) then \(A \times B = \emptyset\) and \(\emptyset \subseteq A \times C\) for any \(C\), so the statement is true.

    Suppose \(A\), \(B\) and \(C\) are non-empty sets with \(B \subseteq C\). Thus if \(b \in B\), then \(b \in C\). Consider any element \((a,b) \in A \times B\), where \(a \in A\) and \(b \in B\). Since \(b \in B\), \(b \in C\). As such, \(a \in A\) and \(b \in C\) for all \(a\) or \(b\), so \((a,b) \in A \times C\). Therefore \(A \times B \subseteq A \times C\). \(\square\)

  3. For all sets \(A\), \(B\) and \(C\), \(A - (B \cup C) = (A-B) - C\).

    Proof Option 1. For any \(x\), we have \[x \in A - (B \cup C) \iff x \in A \wedge x \notin (B \cup C) \iff x \in A \wedge (x \notin B) \wedge (x \notin C).\] Also, we have \[x \in (A-B) - C \iff x \in (A-B) \wedge x \notin C \iff (x \in A \wedge x \notin B) \wedge x \notin C.\] The statements on the right-hand side are equivalent. Hence \(A - (B \cup C) = (A-B) - C\). \(\square\)

    Proof Option 2. \[\begin{aligned} A - (B \cup C) &= A \cap (B \cup C)^c && \text{by the Set Difference Law} \\ &= A \cap (B^c \cap C^c) && \text{by De Morgan's law} \\ &= (A \cap B^c) \cap C^c && \text{by Associative law} \\ &= (A - B) \cap C^c && \text{by the Set Difference Law} \\ &= (A - B) - C && \text{by the Set Difference Law} \qquad \square \end{aligned}\]

§24 — Functions Defined on General Sets

Epp 4th ed. pp. 383–396 · 5th ed. (metric) pp. 425–439. Lecture 18 (week 7).

Problems

  1. Consider the function \(f : \mathbb{R} \to \mathbb{R}\) defined by \(f(x) = x^2 - 1\).

    1. What is the domain, codomain and range of \(f\)?

    2. What is the image of \(\{x \in \mathbb{R} \mid -1 \leq x \leq 2\}\) under \(f\)?

    3. What is the preimage of \(\{y \in \mathbb{R} \mid 0 \leq y \leq 3\}\) under \(f\)?

  2. Let \(X = \{1,3,5\}\) and \(Y = \{s,t,u,v\}\) be sets. Define \(f : X \to Y\) by the arrow diagram

    \[1 \mapsto v, \qquad 3 \mapsto s, \qquad 5 \mapsto v\]

    (the elements \(t\) and \(u\) of \(Y\) receive no arrows).

    1. Write the domain of \(f\) and the co-domain of \(f\).

    2. Determine \(f(1)\), \(f(3)\), and \(f(5)\).

    3. Determine the range of \(f\).

    4. Is 3 an inverse image of \(s\)? Is 1 an inverse image of \(u\)?

    5. What is the inverse image of \(s\)? of \(u\)? of \(v\)?

    6. Represent \(f\) as a set of ordered pairs.

  3. Let \(D\) be the set of all finite subsets of \(\mathbb{Z}^+\). Consider the function \(T : \mathbb{Z}^+ \to D\) defined by \(T(n) := \{d \mid d \in \mathbb{Z}^+ \text{ and } d \text{ divides } n\}\). Find the following:

    1. \(T(1)\) (b) \(T(15)\) (c) \(T(17)\) (d) \(T(5)\) (e) \(T(18)\) (f) \(T(21)\)

  4. Let \(S\) be the set of all strings of \(a\)’s and \(b\)’s.

    1. Define the function \(f : S \to \mathbb{Z}\) by

    \[f(s) = \begin{cases} \text{the number of } b\text{'s to the left of the left-most } a \text{ in } s, & \text{if } s \text{ contains an } a, \\ 0, & \text{if } s \text{ contains no } a\text{'s.} \end{cases}\]

    Find \(f(aba)\), \(f(bbab)\), and \(f(b)\). What is the range of \(f\)?

    1. Define the function \(g : S \to S\) by

    \[g(s) = \text{the string obtained by writing the characters of } s \text{ in reverse order.}\]

    Find \(g(aba)\), \(g(bbab)\), and \(g(b)\). What is the range of \(g\)?

Solutions

    1. The domain is \(\mathbb{R}\), the codomain is \(\mathbb{R}\), and the range is \([-1, \infty)\).

    2. \(f\big(\{x \in \mathbb{R} \mid -1 \leq x \leq 2\}\big) = [-1, 3]\).

    3. \(f^{-1}\big(\{y \in \mathbb{R} \mid 0 \leq y \leq 3\}\big) = \{x \in \mathbb{R} \mid 1 \leq x^2 \leq 4\} = [-2,-1] \cup [1,2]\).

    1. The domain is \(X = \{1,3,5\}\) and the codomain is \(Y = \{s,t,u,v\}\).

    2. \(f(1) = v\), \(f(3) = s\), \(f(5) = v\).

    3. \(\operatorname{range}(f) = \{s, v\}\).

    4. Yes, 3 is an inverse image of \(s\). No, 1 is not an inverse image of \(u\).

    5. \(f^{-1}(s) = \{3\}\), \(f^{-1}(u) = \emptyset\), \(f^{-1}(v) = \{1, 5\}\).

    6. \(f = \{(1,v), (3,s), (5,v)\}\).

    1. \(T(1) = \{1\}\).

    2. \(T(15) = \{1, 3, 5, 15\}\).

    3. \(T(17) = \{1, 17\}\).

    4. \(T(5) = \{1, 5\}\).

    5. \(T(18) = \{1, 2, 3, 6, 9, 18\}\).

    6. \(T(21) = \{1, 3, 7, 21\}\).

    1. \(f(aba) = 0\), \(f(bbab) = 2\), \(f(b) = 0\), \(\operatorname{range}(f) = \mathbb{Z}^{\geq 0}\).

    2. \(g(aba) = aba\), \(g(bbab) = babb\), \(g(b) = b\), \(\operatorname{range}(g) = S\).

§25 — one-to-one, Onto, and Inverse Functions

Epp 4th ed. pp. 397–416 · 5th ed. (metric) pp. 439–461. Lecture 19 (week 7).

Problems

  1. Consider the function \(f : \mathbb{Z} \to \mathbb{Z}\) defined by \(f(n) = \left\lfloor \dfrac{1-6n}{3} \right\rfloor\).

    1. Is \(f\) one-to-one?

    2. Is \(f\) onto?

    3. Is \(f\) bijective?

    4. Prove or disprove that for all \(m, n \in \mathbb{Z}\), we have \(f(mn) = f(m)f(n)\).

  2. Define the set \(A = \{0,1,2,3,4\}\) and the functions \(f : A \to A\) and \(g : A \to A\) by

    \[f(a) = (a+4)^2 \pmod 5, \qquad g(a) = (a^2 + 3a + 1) \pmod 5.\]

    Is \(f = g\)?

  3. Consider the function \(f : \mathbb{Z}^+ \times \mathbb{Z}^+ \to \mathbb{Z}^+\) defined by \(f((m,n)) = 2^m 3^n\).

    1. Determine \(f((3,1))\).

    2. Determine the pre-image of \(\{36\}\).

    3. Prove \(f\) is one-to-one.

  4. Consider the function \(f : \mathbb{R} \to \mathbb{Z}\) defined by \(f(x) = \lfloor x^2 \rfloor\).

    1. What is the image \(f([-1,1])\)?

    2. What is the preimage \(f^{-1}(\{1\})\)?

    3. Is \(f\) one-to-one?

    4. Is \(f\) onto?

    5. Is \(f\) bijective?

Solutions

  1. First, we note that, for \(n \in \mathbb{Z}\),

    \[f(n) = \left\lfloor \frac{1-6n}{3} \right\rfloor = \left\lfloor -2n + \frac{1}{3} \right\rfloor = -2n\]

    since \(-2n \in \mathbb{Z}\) and \(\frac{1}{3} \in [0,1)\).

    1. Yes. If \(f(m) = f(n)\) then \(-2m = -2n\), hence \(m = n\). Therefore \(f\) is one-to-one.

    2. No. Since \(f(n) = -2n\) we see that \(f\) never maps to any odd integer; for example, there is no integer \(w\) with \(f(w) = 1\). Therefore \(f\) is not onto.

    3. No, \(f\) is not a bijection. (It is injective but not surjective.)

    4. This is false; take \(m = n = 1\). Then \(f(mn) = -2\) but \(f(m)f(n) = 4\).

  2. Yes. First note \(f\) and \(g\) have the same domain and codomain. This means we just need to show \(f(a) = g(a)\) for all \(a \in A\). Notice that we have

    \[(a+4)^2 = a^2 + 8a + 16 \equiv a^2 + 3a + 1 \pmod 5,\]

    so \(f(a) = g(a)\) for all \(a \in A\), and hence \(f = g\).

    1. \(f((3,1)) = 2^3 3^1 = 24\).

    2. \(36 = 2^2 3^2\), so \(f((m,n)) = 36\) if and only if \((m,n) = (2,2)\). Hence \(f^{-1}(\{36\}) = \{(2,2)\}\).

    3. If \(f((m,n)) = f((m',n'))\), then \(2^m 3^n = 2^{m'} 3^{n'}\), so \(2^{m-m'} = 3^{n'-n}\). By unique prime factorisation, this forces \(m - m' = 0\) and \(n' - n = 0\). That is, \(m = m'\) and \(n = n'\), and hence \((m,n) = (m',n')\). So \(f\) is one-to-one.

    1. Note that if \(x \in [-1,1]\) then \(x^2 \in [0,1]\). This means we have \(\lfloor x^2 \rfloor \in \{0,1\}\). Both values occur because \(\lfloor 0^2 \rfloor = 0\) and \(\lfloor 1^2 \rfloor = 1\). Thus \(f([-1,1]) = \{0,1\}\).

    2. \(f^{-1}(\{1\}) = \{x \in \mathbb{R} : 1 \leq x^2 < 2\} = (-\sqrt{2}, -1] \cup [1, \sqrt{2})\).

    3. No, \(f\) is not one-to-one. For example, \(f\!\left(\frac{1}{2}\right) = 0 = f\!\left(-\frac{1}{2}\right)\) but \(\frac{1}{2} \neq -\frac{1}{2}\).

    4. No, \(f\) is not onto since \(f(x) = \lfloor x^2 \rfloor \geq 0\) for all \(x\). This means, for instance, that \(-1\) is not in the image of \(f\).

    5. No, \(f\) is not bijective.

§26 — Composition of Functions

Epp 4th ed. pp. 416–427 · 5th ed. (metric) pp. 461–472. Lecture 20 (week 7).

Problems

  1. Let \(f, g : \mathbb{Z} \to \mathbb{Z}\) be defined by \(f(x) = x^3 - 1\) and \(g(x) = x + 5\).

    1. Determine the functions \((g \circ f)(x)\) and \((f \circ g)(x)\).

    2. Is \(g \circ f\) a bijection?

  2. Define \(f : \mathbb{Q} \to \mathbb{Z}\) by \(f(x) = 2\lfloor x + 1 \rfloor\).

    1. What is the range of \(f\)?

    2. Is \(f\) one-to-one?

    3. Is \(f\) onto?

    4. Define \(g : \mathbb{Z} \to \mathbb{Z}\) by \(g(x) = 2x - 1\). Determine the composition \((g \circ f)(x)\).

    5. Calculate \((g \circ f)\left(-\frac{3}{4}\right)\).

    6. What is the range of \(g \circ f\)?

  3. Let \(f\) and \(g\) be functions. Prove or disprove the following statements:

    1. If \(f\) and \(f \circ g\) are one-to-one, then \(g\) is one-to-one.

    2. If \(f\) and \(f \circ g\) are onto, then \(g\) is onto.

Solutions

    1. We have

    \[(g \circ f)(x) = g(x^3 - 1) = x^3 + 4\]

    and

    \[(f \circ g)(x) = f(x+5) = (x+5)^3 - 1 = x^3 + 15x^2 + 75x + 124.\]

    1. No, \(g \circ f\) is not a bijection. The map \(x \mapsto x^3 + 4\) is one-to-one on \(\mathbb{Z}\), but it is not onto (e.g. 0 is not in its image since \(x^3 = -4\) has no integer solution).

    1. The range is a subset of the even integers because \(\lfloor x+1 \rfloor \in \mathbb{Z}\). If \(2m\) is an even integer then \(f(m-1) = 2\lfloor m-1+1 \rfloor = 2m\), so the range is the set of all even integers.

    2. No, \(f\) is not one-to-one. For example,

    \[f(0) = 2\lfloor 0+1 \rfloor = 2\lfloor 1 \rfloor = 2 \times 1 = 2, \quad \text{and}\] \[f\!\left(\tfrac{1}{2}\right) = 2\left\lfloor \tfrac{1}{2} + 1 \right\rfloor = 2\left\lfloor \tfrac{3}{2} \right\rfloor = 2 \times 1 = 2,\]

    but \(0 \neq \frac{1}{2}\).

    1. No, \(f\) is not onto since \(\operatorname{range}(f) \neq \mathbb{Z}\) (e.g. there is no \(x \in \mathbb{Q}\) such that \(f(x) = 1\) because 1 is odd).

    2. We have

    \[(g \circ f)(x) = g(f(x)) = 2\big(2\lfloor x+1 \rfloor\big) - 1 = 4\lfloor x+1 \rfloor - 1.\]

    1. We have

    \[(g \circ f)\left(-\frac{3}{4}\right) = 4 \cdot 0 - 1 = -1.\]

    1. The range is a subset of \(\{4n - 1 \mid n \in \mathbb{Z}\}\) because \(\lfloor x+1 \rfloor \in \mathbb{Z}\). This is the range because, given an integer \(4n-1\), we have \(f(n-1) = 4\lfloor n-1+1 \rfloor - 1 = 4n-1\). Note that you can also write the range as \(\{n \in \mathbb{Z} \mid n \equiv -1 \pmod 4\}\) which is the same as \(\{n \in \mathbb{Z} \mid n \equiv 3 \pmod 4\}\).
    1. The statement is true.

    Proof. Suppose that \(f\) and \(f \circ g\) are one-to-one functions. Suppose for a contradiction that \(g\) is not one-to-one. Thus there exist two distinct elements in the domain of \(g\), say \(x_1\) and \(x_2\), for which \(g(x_1) = g(x_2)\). Let \(g(x_1) = y = g(x_2)\). Thus \(f(g(x_1)) = f(y) = f(g(x_2))\), but \(x_1 \neq x_2\), which contradicts the fact that \(f \circ g\) is one-to-one. Therefore \(g\) must be one-to-one. \(\square\)

    1. The statement is false.

    Counterexample. Consider the functions \(g : \{1\} \to \{2,3\}\) defined by \(g(1) = 2\), and \(f : \{2,3\} \to \{4\}\) defined by \(f(2) = 4\) and \(f(3) = 4\). Then \(f\) is onto, and \(f \circ g\) is onto (since \(f \circ g : \{1\} \to \{4\}\) and \(f(g(1)) = 4\)). However, \(g\) is not onto. \(\square\)

§27 — Cardinalities

Epp 4th ed. pp. 428–441 · 5th ed. (metric) pp. 473–486. Lecture 21 (week 8).

Problems

  1. Let \(A = \{-2,-1,0,1\}\) and \(B = \{3,4,5,6\}\). Prove that \(|A| = |B|\) by providing a bijection between \(A\) and \(B\).

  2. Prove that \(|\{1,2,3\}| \leq |\mathbb{Z}|\) by providing an appropriate injective (one-to-one) function.

  3. Consider the intervals

    \[[6,12] = \{x \in \mathbb{R} \mid 6 \leq x \leq 12\} \quad \text{and} \quad (6,12) = \{x \in \mathbb{R} \mid 6 < x < 12\}.\]

    Prove that \(|[6,12]| \leq |(6,12)|\) by providing an appropriate injective (one-to-one) function.

  4. Let \(A = \{x \in \mathbb{R} \mid -1 \leq x \leq 1\}\) and \(B = \{x \in \mathbb{R} \mid 3 \leq x \leq 7\}\). Prove that \(|A| = |B|\) by providing a bijection between \(A\) and \(B\).

Solutions

  1. Define the function \(f : A \to B\) by

    \[f(-2) = 3, \qquad f(-1) = 4, \qquad f(0) = 5, \qquad f(1) = 6.\]

    Then \(f\) is bijective, hence \(|A| = |B|\).

  2. Define the function \(g : \{1,2,3\} \to \mathbb{Z}\) by \(g(1) = 1\), \(g(2) = 2\), and \(g(3) = 3\). Then \(g\) is injective (one-to-one), so \(|\{1,2,3\}| \leq |\mathbb{Z}|\).

  3. Define \(h : [6,12] \to (6,12)\) by

    \[h(x) = \frac{x}{2} + 4.\]

    For each \(x \in [6,12]\) we have

    \[\begin{aligned} \frac{6}{2} + 4 \leq\ &h(x) \leq \frac{12}{2} + 4 \\ 7 \leq\ &h(x) \leq 10. \end{aligned}\]

    Thus \(h\) is indeed a function from \([6,12]\) to \((6,12)\), and \(h\) has range \([7,10] \subseteq (6,12)\).

    Suppose \(x, y \in [6,12]\) and \(h(x) = h(y)\). Then \(\frac{x}{2} + 4 = \frac{y}{2} + 4\), so \(x = y\). Therefore \(h\) is injective (one-to-one).

    Thus \(|[6,12]| \leq |(6,12)|\).

  4. Let \(A = \{x \in \mathbb{R} \mid -1 \leq x \leq 1\}\) and \(B = \{x \in \mathbb{R} \mid 3 \leq x \leq 7\}\).

    Consider the function \(f : A \to B\) where \(f(x) = 2x + 5\). Since \(x \geq -1\), \(f(x) = 2x+5 \geq 2(-1)+5 = 3\), and since \(x \leq 1\), \(f(x) = 2x+5 \leq 2(1)+5 = 7\). Also each value of \(x\) gives a unique value of \(2x+5\). Thus \(f\) is a function with the correct domain and codomain.

    We now prove that \(f\) is one-to-one. Suppose that \(x_1, x_2 \in A\) and \(f(x_1) = f(x_2)\). Thus \(2x_1 + 5 = 2x_2 + 5\), and so \(x_1 = x_2\). Thus \(f\) is one-to-one.

    We now prove that \(f\) is onto. Suppose \(y \in [3,7]\). Let \(x = \frac{y-5}{2}\). Since \(y \leq 7\), we have \(x \leq 1\) and since \(y \geq 3\), we have \(x \geq -1\). So \(x \in [-1,1]\). Furthermore \(f(x) = f\!\left(\frac{y-5}{2}\right) = 2\left(\frac{y-5}{2}\right) + 5 = y - 5 + 5 = y\). Thus \(f\) is onto.

    Hence \(f\) is a bijection and so \(|A| = |B|\).

§28 — Countable and Uncountable Sets

Epp 4th ed. pp. 428–441 · 5th ed. (metric) pp. 473–486. Lecture 22 (week 8).

Note that this content will not be assessed in MATH1061.

Problems

  1. Determine whether each of the following sets is countable or uncountable.

    1. \(\{1, 2, \dots, n\}\).

    2. \(\mathbb{Z} \times \mathbb{Z}\).

    3. \(\mathbb{Q} \times \mathbb{Q}\).

    4. \(\mathbb{Z} \times \mathbb{Z} \times \cdots\) (infinitely many copies of \(\mathbb{Z}\)).

    1. Is \(|\mathbb{Z}| = |\mathbb{Q}|\)?

    2. Is \(|\mathbb{Z}| = |\mathbb{Z} \times \mathbb{Z}|\)?

    3. Is \(|\mathbb{R}| = |\mathbb{Q} \times \mathbb{Q}|\)?

    1. Show that the union of a countable number of countable sets is countable.

    2. Show that the set of all finite binary strings is countable.

    3. Use (a) to show that the set of all irrational numbers is uncountable.

Solutions

    1. The set is countable because the function \(f : \{1,2,\dots,n\} \to \mathbb{N}\) defined by \(f(i) = i\) is injective.

    2. The set \(\mathbb{Z} \times \mathbb{Z}\) is countable. Define the function \(f : \mathbb{Z} \times \mathbb{Z} \to \mathbb{N}\) by

    \[f(a,b) = \begin{cases} 2^a 3^b, & a > 0,\ b > 0, \\ 5^{-a} 7^{-b}, & a < 0,\ b < 0, \\ 11^{a} 13^{-b}, & a > 0,\ b < 0, \\ 17^{-a} 19^{b}, & a < 0,\ b > 0, \\ 23^{b}, & a = 0,\ b > 0, \\ 29^{a}, & a > 0,\ b = 0, \\ 31^{-b}, & a = 0,\ b < 0, \\ 37^{-a}, & a < 0,\ b = 0, \\ 1, & (a,b) = (0,0). \end{cases}\]

    We claim \(f\) is injective. Suppose \(f(a,b) = f(c,d)\). By unique prime factorisation, the set of primes dividing \(f(a,b)\) and the set of primes dividing \(f(c,d)\) must be equal. Hence \((a,b)\) and \((c,d)\) lie in the same case. Again, by unique prime factorisation, the corresponding exponents must agree, so \(a = c\) and \(b = d\). Therefore \(f\) is injective.

    1. The set \(\mathbb{Q} \times \mathbb{Q}\) is countable. First define an injection \(e : \mathbb{Z} \to \mathbb{N}\) by

    \[e(z) = \begin{cases} 2z, & z \geq 0, \\ -2z - 1, & z < 0. \end{cases}\]

    Next define an injection \(c : \mathbb{N} \times \mathbb{N} \to \mathbb{N}\) by

    \[c(m,n) = 2^m 3^n.\]

    Write each rational in lowest terms with positive denominator: \(x = \frac{p}{q}\), \(y = \frac{r}{s}\) with \(p, r \in \mathbb{Z}\) and \(q, s \in \mathbb{N}\). Define \(F : \mathbb{Q} \times \mathbb{Q} \to \mathbb{N}\) by

    \[F\left(\frac{p}{q}, \frac{r}{s}\right) = 2^{\,c(e(p),\,q)}\, 3^{\,c(e(r),\,s)}.\]

    If \(F\left(\frac{p}{q}, \frac{r}{s}\right) = F\left(\frac{p'}{q'}, \frac{r'}{s'}\right)\), then by unique prime factorisation we get \(c(e(p),q) = c(e(p'),q')\) and \(c(e(r),s) = c(e(r'),s')\). Since \(c\) is injective, we have \(e(p) = e(p')\), \(q = q'\), \(e(r) = e(r')\), \(s = s'\). Moreover, since \(e\) is injective, we have \(p = p'\) and \(r = r'\). Therefore \(\frac{p}{q} = \frac{p'}{q'}\) and \(\frac{r}{s} = \frac{r'}{s'}\), so \(F\) is injective.

    1. The set \(\mathbb{Z} \times \mathbb{Z} \times \dots\) is not countable. We will prove this by giving an injection \(\mathbb{R} \to \mathbb{Z} \times \mathbb{Z} \times \cdots\).

    For each \(x \in \mathbb{R}\), write \(x = n + r\) with \(n = \lfloor x \rfloor \in \mathbb{Z}\) and \(r \in [0,1)\). Choose the unique decimal expansion of \(r\) that does not end with an infinite tail of 9’s. Write this decimal expansion as

    \[r = 0.d_1 d_2 d_3 \cdots, \qquad d_k \in \{0,1,\dots,9\}.\]

    Then define the function \(\iota : \mathbb{R} \to \mathbb{Z} \times \mathbb{Z} \times \cdots\) by

    \[\iota(x) = (n, d_1, d_2, d_3, \dots) \in \mathbb{Z} \times \mathbb{Z} \times \cdots.\]

    This is injective; if \(\iota(x) = \iota(y)\) then the decimals of \(x\) and \(y\) are equal, so \(x = y\).

    1. Yes. Both \(\mathbb{Z}\) and \(\mathbb{Q}\) are countable and infinite.

    2. Yes. Both \(\mathbb{Z}\) and \(\mathbb{Z} \times \mathbb{Z}\) are countable and infinite.

    3. No. \(\mathbb{Q} \times \mathbb{Q}\) is countable, while \(\mathbb{R}\) is uncountable.

    1. Let \(A_1, A_2, A_3, \dots\) be countable. Then for each \(n\) we can list \(A_n\) as

    \[A_n = \{a_{n,1}, a_{n,2}, a_{n,3}, \dots\}.\]

    Now list the union by going along diagonals:

    \[a_{1,1};\ a_{1,2}, a_{2,1};\ a_{1,3}, a_{2,2}, a_{3,1};\ a_{1,4}, a_{2,3}, a_{3,2}, a_{4,1};\ \dots\]

    Every element of every \(A_n\) appears somewhere in this list, so \(\bigcup_{n=1}^{\infty} A_n\) can be listed in a sequence. Hence \(\bigcup_{n=1}^{\infty} A_n\) is countable.

    1. Let \(B_n\) be the set of all binary strings of length \(n\). Each \(B_n\) is finite (there are \(2^n\) such strings), hence countable. Every finite binary string has some length, so

    \[B = \{\text{all finite binary strings}\} = \bigcup_{n=0}^{\infty} B_n.\]

    By (a), a countable union of countable sets is countable, so \(B\) is countable.

    1. Suppose, for contradiction, that the set of irrationals \(\mathbb{R} \setminus \mathbb{Q}\) were countable. We already know \(\mathbb{Q}\) is countable. Then by part (a),

    \[\mathbb{R} = \mathbb{Q} \cup (\mathbb{R} \setminus \mathbb{Q})\]

    would be a union of two countable sets, hence countable. But \(\mathbb{R}\) is uncountable. This contradiction shows that \(\mathbb{R} \setminus \mathbb{Q}\) is uncountable.

Application — Computability of Functions

Not assessable for MATH1061. Taken from K. Rosen, Discrete Mathematics and its Applications, 7th edition, 1991, pages 170–177.

An important application of the topic of countability to computer science is examining the computability of functions. In computer science, a function is computable if there is a computer program in some programming language that can find the values of the function. If there is no such program, we say the function is uncomputable.

Application problems
    1. Show the set of all computer programs in any given computer language is countable.

    Hint: a computer program written in a programming language can be thought of as a string of symbols from a finite alphabet.

    1. Let \(T\) be the set of all functions from \(\mathbb{Z}^+\) to the set \(\{0,1,2,3,4,5,6,7,8,9\}\). Show that \(T\) is uncountable.

    Hint: start by finding an injection between the interval \((0,1)\) and a subset of \(T\).

    1. Hence show that there are uncomputable functions.
Application solutions
    1. Given a computer language with a finite alphabet \(A\), we let \(P\) be the set of all computer programs in the language. If \(P\) is finite, then it is countable and we are done, so assume \(P\) is infinite.

    Now we use the fact that we can think of each computer program as a series of strings made using the alphabet \(A\). Consider all strings of length \(m\) made from alphabet \(A\), and denote this set by \(A_m\). It is easy to show that there are \(|A|^m\) distinct strings of this type, and since \(A\) is finite, \(|A|^m\) is also finite. Now \(P\) is a subset of \(\bigcup_{i=1}^{\infty} A_m\). Since the union of a countable number of countable sets is also countable (see Question 3(a) from earlier in this section) it follows that \(\bigcup_{i=1}^{\infty} A_m\) is countable and hence \(P\) is countable.

    1. We use the hint and start by defining a function from \((0,1)\) to \(T\) as follows. Recall that every real number in the interval \((0,1)\) can be written in decimal notation as \(0.a_1 a_2 a_3 \dots a_n \dots\) where each \(a_i\) is an integer from 0 to 9. Also recall that \(T\) is the set of all functions from \(\mathbb{Z}^+\) to \(\{0,1,\dots,9\}\). Let \(f : (0,1) \to T\) be the function defined as

    \[f(0.a_1 a_2 a_3 \dots a_n \dots) = g,\]

    where \(g\) is the function in \(T\) such that \(g(n) = a_n\) for all \(n \in \mathbb{Z}^+\).

    So for example, if \(f(0.457013) = g\) then \(g\) is the function where \(g(1) = 4\), \(g(2) = 5\), \(g(3) = 7\), \(g(4) = 0\), \(g(5) = 1\), \(g(6) = 3\) and \(g(n) = 0\) for all \(n \geq 7\).

    Now we will prove that this function is one-to-one. Suppose that \(f(x_1) = f(x_2)\) for \(x_1, x_2 \in (0,1)\). Then for these functions to be equal, they map every integer to the same value. For this to happen, every decimal digit of \(x_1\) must be equal to the same decimal digit of \(x_2\) and so we must have \(x_1 = x_2\).

    Now denote the set of all outputs of \(f\) as \(S\). Hence \(S \subseteq T\). Now if we set the codomain of \(f\) to be \(S\), which is the range of \(f\), we have made this function surjective and hence a bijection. This means that the subset \(S \subseteq T\) is in bijection with the interval \((0,1)\). Since the interval \((0,1)\) is uncountable, we must have that \(S\) is also uncountable and since \(S \subseteq T\), \(T\) must also be uncountable.

    1. Part (a) showed us that the set of all computer programs in a given language is countable. This means that there are a countable number of computable functions. However part (b) showed us that there is an uncountable number of functions in the set \(T\), so we must have that some functions are uncomputable.

§29 — Relations on Sets

Epp 4th ed. pp. 442–449 · 5th ed. (metric) pp. 487–494. Lecture 23 (week 8).

Problems

  1. Define a relation \(P\) on \(\mathbb{Z}\) as follows: for every ordered pair \((m,n) \in \mathbb{Z} \times \mathbb{Z}\),

    \[m \mathrel{P} n \iff m \text{ and } n \text{ have a common prime factor.}\]

    1. Is \(15 \mathrel{P} 25\)?

    2. Is \(22 \mathrel{P} 27\)?

    3. Is \(0 \mathrel{P} 5\)?

    4. Is \(8 \mathrel{P} 8\)?

  2. Let \(X = \{a,b,c\}\). Define a relation \(S\) on the power set \(\mathcal{P}(X)\) as follows: for all sets \(A\) and \(B\) in \(\mathcal{P}(X)\),

    \[A \mathrel{S} B \iff A \text{ has the same number of elements as } B.\]

    1. Is \(\{a,b\} \mathrel{S} \{b,c\}\)?

    2. Is \(\{a\} \mathrel{S} \{a,b\}\)?

    3. Is \(\{c\} \mathrel{S} \{b\}\)?

  3. Let \(A = \{3,4,5\}\) and \(B = \{4,5,6\}\) be sets. Let \(R\) be the ‘divides’ relation. That is, for every ordered pair \((a,b)\) in \(A \times B\), we have \(a \mathrel{R} b\) if and only if \(a \mid b\). Draw an arrow diagram (i.e. the directed graph) of this relation.

Solutions

    1. Yes, since \(15 = 3 \cdot 5\) and \(25 = 5 \cdot 5\) share the prime factor 5.

    2. No, since \(22 = 2 \cdot 11\) and \(27 = 3^3\) share no common prime factor.

    3. Yes, since 0 is divisible by every prime; in particular \(5 \mid 0\) and \(5 \mid 5\), so 0 and 5 have a common prime factor.

    4. Yes, since \(8 = 2^3\) and 2 is a prime factor of 8 (so they share the prime factor 2).

    1. Yes, since \(|\{a,b\}| = 2 = |\{b,c\}|\).

    2. No, since \(|\{a\}| = 1 \neq 2 = |\{a,b\}|\).

    3. Yes, since \(|\{c\}| = 1 = |\{b\}|\).

  3. We have \(A \times B = \{(3,4), (3,5), (3,6), (4,4), (4,5), (4,6), (5,4), (5,5), (5,6)\}\). The pairs \((a,b)\) satisfying \(a \mid b\) are

    \[\{(3,6),\ (4,4),\ (5,5)\} \subseteq A \times B.\]

    So the directed graph has an arrow from 3 (in \(A\)) to 6 (in \(B\)), an arrow from 4 to 4, and an arrow from 5 to 5 — and no other arrows.

§30 — Reflexivity, Symmetry, Transitivity

Epp 4th ed. pp. 449–459 · 5th ed. (metric) pp. 495–505. Lecture 23 (week 8).

Problems

  1. Let \(X = \{1,2,3,4,5\}\) and consider the subset \(R \subseteq X \times X\) given by

    \[R = \{(1,1), (2,2), (3,3), (4,4), (5,5), (1,2), (2,3), (3,4), (4,5), (5,1)\}.\]

    1. Draw an arrow diagram for the binary relation \(R\).

    2. Draw an arrow diagram for the binary relation \(R^{-1}\).

    3. Draw an arrow diagram for the binary relation \(R \cup R^{-1}\).

    4. Is \(R\) symmetric? Is \(R^{-1}\) symmetric? Is \(R \cup R^{-1}\) symmetric?

    5. Is \(R\) transitive? Is \(R^{-1}\) transitive? Is \(R \cup R^{-1}\) transitive?

  2. Suppose \(R\) is a binary relation on the set \(\{1,2,3\}\) such that:

    • \((2,3) \in R\)
    • \((3,2) \in R\)
    • \((1,1) \notin R\)
    • \(R\) is transitive
    • \(R\) is not symmetric
    • \(R\) is not antisymmetric.

    Draw an arrow diagram for the relation \(R\).

Solutions

    1. The arrow diagram for \(R\) has the five vertices \(1, 2, 3, 4, 5\) drawn around a circle, with a loop at every vertex (from the pairs \((1,1), \dots, (5,5)\)) together with the directed 5-cycle

    \[1 \to 2 \to 3 \to 4 \to 5 \to 1.\]

    1. The arrow diagram for \(R^{-1}\) is the same picture with every non-loop arrow reversed: a loop at every vertex, together with the directed 5-cycle

    \[1 \to 5 \to 4 \to 3 \to 2 \to 1,\]

    that is, \(R^{-1} = \{(1,1), (2,2), (3,3), (4,4), (5,5), (2,1), (3,2), (4,3), (5,4), (1,5)\}\).

    1. The arrow diagram for \(R \cup R^{-1}\) has a loop at every vertex and a double-headed arrow between each consecutive pair around the cycle — that is, both \((i, i+1)\) and \((i+1, i)\) for the cycle \(1,2,3,4,5,1\).

    2. \(R\) is not symmetric because \((1,2) \in R\) but \((2,1) \notin R\).

    \(R^{-1}\) is not symmetric because \((2,1) \in R^{-1}\) but \((1,2) \notin R^{-1}\).

    \(R \cup R^{-1}\) is symmetric. If \((a,b) \in R \cup R^{-1}\) then \((a,b) \in R\) or \((a,b) \in R^{-1}\), which means \((b,a) \in R^{-1}\) or \((b,a) \in R\), and thus, in either case, \((b,a) \in R \cup R^{-1}\).

    1. \(R\) is not transitive because \((1,2), (2,3) \in R\) but \((1,3) \notin R\).

    \(R^{-1}\) is not transitive because \((3,2), (2,1) \in R^{-1}\) but \((3,1) \notin R^{-1}\).

    \(R \cup R^{-1}\) is not transitive because \((1,2), (2,3) \in R \cup R^{-1}\) but \((1,3) \notin R \cup R^{-1}\).

  2. One relation on \(\{1,2,3\}\) satisfying all conditions is

    \[R = \{(1,2), (1,3), (2,2), (3,3), (2,3), (3,2)\}.\]

    (It contains \((2,3)\) and \((3,2)\); it is transitive; it is not symmetric since \((1,2) \in R\) but \((2,1) \notin R\); it is not antisymmetric since \(2 \neq 3\) and \((2,3), (3,2) \in R\); and \((1,1) \notin R\).)

    An arrow diagram for this \(R\) has vertices \(1, 2, 3\) with: arrows \(1 \to 2\) and \(1 \to 3\); loops at 2 and at 3; and a double-headed arrow between 2 and 3. There is no loop at vertex 1.

§31 — Equivalence Relations

Epp 4th ed. pp. 459–478 · 5th ed. (metric) pp. 505–523. Lecture 24 (week 9).

Problems

  1. Define the binary relations \(\alpha\), \(\beta\) and \(\gamma\) on \(\mathbb{Z}\) as follows:

    \[\begin{aligned} x \mathrel{\alpha} y &\iff x + y \text{ is an even integer}, \\ x \mathrel{\beta} y &\iff x \text{ is a multiple of } y, \\ x \mathrel{\gamma} y &\iff 3 \text{ is a factor of } x^2 - y^2. \end{aligned}\]

    1. Fill in the table with ‘Yes’/‘No’ to indicate the properties satisfied by each relation.
    \(\alpha\) \(\beta\) \(\gamma\)
    Reflexive
    Symmetric
    Transitive
    Equivalence relation
    1. For each equivalence relation, state its equivalence classes.

  2. Let \(A = \{a,b,c,d,e\}\) be a set and \(\rho\) be a binary relation defined by

    \[\rho = \{(a,a), (a,b), (a,d), (b,b), (c,c), (d,d), (b,a), (d,a), (b,d), (d,b), (c,e), (e,c), (e,e)\}.\]

    For instance, \((a,b) \in \rho\) means \(a \mathrel{\rho} b\).

    1. Draw the directed graph representing \(\rho\).

    2. Is \(\rho\) reflexive? Is \(\rho\) symmetric? Is \(\rho\) transitive?

    3. Is \(\rho\) an equivalence relation on \(A\)? If so, give its equivalence classes.

    4. What is the smallest subset \(\tau\) of \(\rho\) which forms an equivalence relation on \(A\)?

    5. Give the equivalence classes for \(\tau\).

  3. Consider a positive integer \(n\) and any set of strings \(S\). Define a relation \(R_n\) on \(S\) by the condition \(s \mathrel{R_n} t\) if and only if \(s = t\) or both \(s\) and \(t\) have at least \(n\) characters and the first \(n\) characters are the same. Show that \(R_n\) is an equivalence relation.

Solutions

    \(\alpha\) \(\beta\) \(\gamma\)
    Reflexive Yes Yes Yes
    Symmetric Yes No Yes
    Transitive Yes Yes Yes
    Equivalence relation Yes No Yes
    1. The equivalence relations are \(\alpha\) and \(\gamma\).

    For \(\alpha\), notice \(x \mathrel{\alpha} y \iff x + y\) is even \(\iff x\) and \(y\) have the same parity. This means the equivalence classes are \(\mathbb{Z}^{\text{even}}\) and \(\mathbb{Z}^{\text{odd}}\).

    For \(\gamma\), notice \(x \mathrel{\gamma} y \iff 3 \mid (x^2 - y^2) \iff x^2 \equiv y^2 \pmod 3\). But modulo 3 the only possible square residues are

    \[0^2 \equiv 0, \qquad 1^2 \equiv 1, \qquad 2^2 \equiv 1 \pmod 3.\]

    Hence \(x^2 \equiv 0 \pmod 3\) exactly when \(3 \mid x\), and \(x^2 \equiv 1 \pmod 3\) exactly when \(3 \nmid x\). Therefore \(x \mathrel{\gamma} y\) iff either both \(x\) and \(y\) are multiples of 3, or neither is. So the equivalence classes are \(\{n \in \mathbb{Z} : 3 \mid n\}\) and \(\{n \in \mathbb{Z} : 3 \nmid n\}\). Note that another way to write the equivalence classes is \(\{3n \mid n \in \mathbb{Z}\}\) and \(\{3n+1,\ 3n+2 \mid n \in \mathbb{Z}\}\).

    1. A digraph for \(\rho\) (vertex set \(A = \{a,b,c,d,e\}\)) has a loop at every vertex, double-headed arrows between each pair of \(\{a,b,d\}\) — that is between \(a\) and \(b\), between \(a\) and \(d\), and between \(b\) and \(d\) — and a double-headed arrow between \(c\) and \(e\). There are no arrows between \(\{a,b,d\}\) and \(\{c,e\}\).

    2. \(\rho\) is reflexive: we have \((x,x) \in \rho\) for all \(x \in A\) (loops on all vertices).

    \(\rho\) is symmetric: whenever \((x,y) \in \rho\), we have \((y,x) \in \rho\) (no single-direction arrow between any two distinct vertices).

    \(\rho\) is transitive: within \(\{a,b,d\}\) all relations occur, within \(\{c,e\}\) all relations occur, and there are no relations between \(\{a,b,d\}\) and \(\{c,e\}\). That is, whenever \((x,y)\) and \((y,z)\) are in \(\rho\), then \((x,z)\) is also in \(\rho\) (and \(x = z\) is possible).

    1. Yes, \(\rho\) is an equivalence relation. Its equivalence classes are

    \[[a] = [b] = [d] = \{a,b,d\}, \qquad [c] = [e] = \{c,e\}.\]

    1. An equivalence relation \(\tau \subseteq \rho\) must be reflexive, so \((x,x) \in \tau\) for all \(x \in A\). But notice if \(\tau\) is the relation

    \[\tau = \{(a,a), (b,b), (c,c), (d,d), (e,e)\}\]

    then \(\tau\) is an equivalence relation. Therefore this is the smallest relation which is an equivalence relation.

    1. The equivalence classes for \(\tau\) are the singletons:

    \[[a] = \{a\},\ [b] = \{b\},\ [c] = \{c\},\ [d] = \{d\},\ [e] = \{e\}.\]

  3. Let \(S\) be any set of strings and fix \(n \in \mathbb{Z}^+\).

    \(R_n\) is reflexive: for any \(s \in S\), since \(s = s\), by definition \(s \mathrel{R_n} s\).

    \(R_n\) is symmetric: let \(s, t \in S\), and assume \(s \mathrel{R_n} t\). If \(s = t\) then \(t \mathrel{R_n} s\); otherwise both \(s\) and \(t\) must have at least \(n\) characters and, since \(s \mathrel{R_n} t\), they must have the same first \(n\) characters, so \(t \mathrel{R_n} s\).

    \(R_n\) is transitive: let \(s, t, u \in S\) and assume \(s \mathrel{R_n} t\) and \(t \mathrel{R_n} u\). Since \(s \mathrel{R_n} t\) we either have \(s = t\), or \(s\) and \(t\) both have at least \(n\) characters and have the same first \(n\) characters. Similarly, since \(t \mathrel{R_n} u\), either \(t = u\), or \(t\) and \(u\) both have at least \(n\) characters and have the same first \(n\) characters. Thus either \(s = t = u\), or \(s\) and \(u\) both have at least \(n\) characters and they both share the first \(n\) characters with \(t\) (and hence each other). Thus \(s \mathrel{R_n} u\) and so \(R_n\) is transitive.

    Since \(R_n\) is reflexive, symmetric and transitive it is an equivalence relation. \(\square\)

Application — Identifiers

Not assessable for MATH1061. Taken from K. Rosen, Discrete Mathematics and its Applications, 7th edition, 1991, pages 608–618.

In the C programming language, an identifier is the name of a variable, a function or another type of entity.

Each identifier is a nonempty string of characters where each character is a lowercase or uppercase English letter, a digit, or an underscore.

There is no limit on the number of characters used in the identifier; however, for some compilers for some versions of C, there is a limit on the number of characters checked when two names are compared.

For example, Standard C compilers consider two identifiers the same when they agree in their first 31 characters. We see that two identifiers will be considered the same if and only if they are related under the relation \(R_n\) when \(n = 31\) (defined in a previous question).

Application problems

  1. Using \(R_{31}\) on the set of all possible identifiers, determine the equivalence classes of each of the following identifiers:

    1. file_1

    2. Number_of_tropical_storms

    3. Number_of_named_tropical_storms

    4. Number_of_named_tropical_storms_in_the_Atlantic_in_2005

Application solutions

  1. Let \(I\) be the set of all possible identifiers (strings over letters, digits and underscores).

    1. Since file_1 has length \(6 < 31\), its equivalence class is just

    \[[\texttt{file\_1}]_{R_{31}} = \{\texttt{file\_1}\}.\]

    1. Since Number_of_tropical_storms has length \(25 < 31\), its equivalence class is

    \[[\texttt{Number\_of\_tropical\_storms}]_{R_{31}} = \{\texttt{Number\_of\_tropical\_storms}\}.\]

    1. The identifier Number_of_named_tropical_storms has length exactly 31, so two identifiers are \(R_{31}\)-equivalent to it iff they have at least 31 characters and their first 31 characters equal this string. Hence

    \[[\texttt{Number\_of\_named\_tropical\_storms}]_{R_{31}} = \left\{ t \in I : \begin{array}{l} |t| \geq 31 \text{ and the first 31 characters of} \\ t \text{ are } \texttt{Number\_of\_named\_tropical\_storms} \end{array} \right\}.\]

    1. The identifier Number_of_named_tropical_storms_in_the_Atlantic_in_2005 begins with the same first 31 characters Number_of_named_tropical_storms, so it has the same \(R_{31}\)-equivalence class as in (c):

    \[[\texttt{Number\_of\_named\_tropical\_storms\_in\_the\_Atlantic\_in\_2005}]_{R_{31}} = \left\{ t \in I : \begin{array}{l} |t| \geq 31 \text{ and the first 31 characters of} \\ t \text{ are } \texttt{Number\_of\_named\_tropical\_storms} \end{array} \right\}.\]

§32 — Partial Order Relations

Epp 4th ed. pp. 498–515 · 5th ed. (metric) pp. 546–563. Lecture 25 (week 9).

Problems

  1. Let \(\rho\) be a relation on \(\mathbb{Z}\) defined by \(x \mathrel{\rho} y\) if and only if \(x^3 - 1 < y^3\).

    1. Prove \(\rho\) is a partial order relation.

    2. Is \(\rho\) a total order?

  2. Let \(S\) be the set of binary strings of length 6. Define the weight of a binary string as the number of ones it contains. (For example, \(\operatorname{weight}(110110) = 4\).) Define a relation \(\rho\) on \(S\) by

    \[s \mathrel{\rho} t \text{ if and only if } \operatorname{weight}(s) \leq \operatorname{weight}(t).\]

    1. Prove that \(\rho\) is not an equivalence relation.

    2. Prove that \(\rho\) is not a partial order relation.

Solutions

    1. To show that \(\rho\) is a partial order, we need to show that \(\rho\) is reflexive, antisymmetric and transitive.

    Reflexive: let \(x \in \mathbb{Z}\). Then \(x^3 - 1 < x^3\) and so \(x \mathrel{\rho} x\), so \(\rho\) is reflexive.

    Antisymmetric: suppose \(x \mathrel{\rho} y\) where \(x \neq y\). (We must show that then \(y \mathrel{\not\rho} x\).) Now \(x \mathrel{\rho} y\) means that \(x^3 - 1 < y^3\). And since \(x \neq y\) we must have \(x^3 < y^3\) and so \(x^3 \leqslant y^3 - 1\). Hence \(y^3 - 1 \not< x^3\) and so we do not have \(y \mathrel{\rho} x\). Therefore \(\rho\) is antisymmetric.

    Transitive: suppose that \(x \mathrel{\rho} y\) and \(y \mathrel{\rho} z\). Then we know that \(x^3 - 1 < y^3\) and \(y^3 - 1 < z^3\). So \(x^3 - 1 \leqslant y^3 - 1 < z^3\), that is, \(x^3 - 1 < z^3\) so that \(x \mathrel{\rho} z\), which means that \(\rho\) is transitive.

    Hence \(\rho\) is a partial order. \(\square\)

    1. To check whether \(\rho\) is a total order, let \(x\) and \(y\) be any two distinct integers. Then \(x^3 \neq y^3\) and so either \(x^3 < y^3\) or else \(y^3 < x^3\).

    If \(x^3 < y^3\) then \(x^3 - 1 < y^3\) and so \(x \mathrel{\rho} y\).

    If \(y^3 < x^3\) then \(y^3 - 1 < x^3\) and so \(y \mathrel{\rho} x\).

    So for any integers \(x \neq y\), we have either \(x \mathrel{\rho} y\) or else \(y \mathrel{\rho} x\). So any two integers are comparable under \(\rho\). Hence \(\rho\) is a total order.

    1. It is not an equivalence relation because it is not symmetric. For instance, let \(s = 000000\) and \(t = 111111\). Then \(\operatorname{weight}(s) = 0 \leq 6 = \operatorname{weight}(t)\), so \((s,t) \in \rho\), but \((t,s) \notin \rho\).

    2. It is not a partial order because it is not antisymmetric. For instance, let \(s = 100000\) and \(t = 010000\). Then \(\operatorname{weight}(s) = \operatorname{weight}(t) = 1\), so \(s \mathrel{\rho} t\) and \(t \mathrel{\rho} s\), but \(s \neq t\).

§33 — Definitions and Examples of Groups

No relevant textbook pages. Lecture 26 (week 9).

Problems

  1. Show \(\mathbb{Z}_9 - \{0\} = \{1,2,3,\dots,7,8\}\) is not a group under multiplication modulo 9.

  2. Let \(G\) be the set of all odd integers. Define the binary operation \(\star\) on \(G\) by

    \[x \star y = x + y - 1.\]

    Show that \((G, \star)\) is an abelian group.

  3. Show \(\mathbb{Z}_9 - \{0,3,6\} = \{1,2,4,5,7,8\}\) is a group under multiplication modulo 9. Is the group abelian?

  4. Define the group \(H = \mathbb{Z}_5 - \{0\} = \{1,2,3,4\}\) under multiplication modulo 5. Write the Cayley table for \(H\).

  5. Define the group \(G = \mathbb{Z}_2 \times \mathbb{Z}_2\) under component-wise addition modulo 2. That is,

    \[(a,b) + (c,d) = (a + c \bmod 2,\ b + d \bmod 2).\]

    Write the Cayley table for \(G\).

Solutions

  1. It is not a group under multiplication mod 9 because, for example, it is not closed (\(3 \cdot 3 = 0 \notin \mathbb{Z}_9 - \{0\}\)).

  2. Closure: if \(x\) and \(y\) are odd, then \(x + y\) is even, so \(x + y - 1\) is odd. Hence \(x \star y \in G\).

    Associativity: for \(x, y, z \in G\),

    \[(x \star y) \star z = (x + y - 1) + z - 1 = x + y + z - 2 = x + (y + z - 1) - 1 = x \star (y \star z).\]

    Commutativity: for \(x, y \in G\),

    \[x \star y = x + y - 1 = y + x - 1 = y \star x.\]

    Identity: we seek \(e \in G\) such that \(x \star e = x = e \star x\) for all \(x\). Now the requirement \(x \star e = x + e - 1 = x\) implies that \(e = 1\). Note 1 lies in \(G\). Moreover, commutativity implies that \(e \star x = x\) as well. Thus the identity is 1.

    Inverses: given \(x \in G\), we want \(y \in G\) with \(x \star y = 1\). This means \(x + y - 1 = 1\), so \(y = 2 - x\). Since \(x\) is odd, \(2 - x\) is odd, hence \(2 - x \in G\). Thus the inverse of \(x\) is \(x^{-1} = 2 - x\).

    Therefore \((G, \star)\) is an abelian group. \(\square\)

  3. Closure: if \(a, b \in \{1,2,4,5,7,8\}\), then \(3 \nmid a\) and \(3 \nmid b\), hence \(3 \nmid ab\), so \(ab \not\equiv 0, 3, 6 \pmod 9\); thus \(\{1,2,4,5,7,8\}\) is closed under multiplication modulo 9.

    Associativity and commutativity: normal multiplication between integers is associative and commutative, so multiplication modulo 9 in the set \(\{1,2,4,5,7,8\}\) is also associative and commutative. This means the group is abelian.

    Identity: \(1 \in G\) is the identity because \(1 \cdot x \equiv x \pmod 9\).

    Inverse: every element has an inverse. We explicitly calculate the inverses:

    \[\begin{aligned} 1^{-1} &= 1 && \text{because } 1 \cdot 1 \equiv 1 \pmod 9, \\ 2^{-1} &= 5 && \text{because } 2 \cdot 5 = 10 \equiv 1 \pmod 9, \\ 4^{-1} &= 7 && \text{because } 4 \cdot 7 = 28 \equiv 1 \pmod 9, \\ 5^{-1} &= 2 && \text{because } 5 \cdot 2 = 10 \equiv 1 \pmod 9, \\ 7^{-1} &= 4 && \text{because } 7 \cdot 4 = 28 \equiv 1 \pmod 9, \\ 8^{-1} &= 8 && \text{because } 8 \cdot 8 = 64 \equiv 1 \pmod 9. \end{aligned}\]

  4. For \(H = \mathbb{Z}_5 \setminus \{0\} = \{1,2,3,4\}\) under multiplication mod 5, the Cayley table is:

    \(\cdot\) 1 2 3 4
    1 1 2 3 4
    2 2 4 1 3
    3 3 1 4 2
    4 4 3 2 1

  5. For \(G = \mathbb{Z}_2 \times \mathbb{Z}_2 = \{(0,0), (0,1), (1,0), (1,1)\}\) under componentwise addition mod 2, the Cayley table is:

    \(+\) \((0,0)\) \((0,1)\) \((1,0)\) \((1,1)\)
    \(\mathbf{(0,0)}\) \((0,0)\) \((0,1)\) \((1,0)\) \((1,1)\)
    \(\mathbf{(0,1)}\) \((0,1)\) \((0,0)\) \((1,1)\) \((1,0)\)
    \(\mathbf{(1,0)}\) \((1,0)\) \((1,1)\) \((0,0)\) \((0,1)\)
    \(\mathbf{(1,1)}\) \((1,1)\) \((1,0)\) \((0,1)\) \((0,0)\)

§34 — Elementary Properties of Groups

No relevant textbook pages. Lecture 27 (week 10).

Problems

  1. Let \(H = \{3n \mid n \in \mathbb{Z}\}\) be the set of multiples of 3. Prove \(H\) is a subgroup of \((\mathbb{Z}, +)\).

  2. Fix elements \(g\) and \(h\) in a group \(G\) where \(e\) is the group identity. Prove the following statements:

    1. If \(g^{-1} = h\) then \(h^{-1} = g\).

    2. \((g^{-1})^{-1} = g\).

    3. \((gh)^{-1} = h^{-1}g^{-1}\).

  3. Prove \(K = \{(0,0), (0,1), (2,0), (2,1)\}\) is a subgroup of \((\mathbb{Z}_4 \times \mathbb{Z}_2, +)\) where the group operation is componentwise addition:

    \[(a,b) + (c,d) = (a + c \pmod 4,\ b + d \pmod 2).\]

  4. Consider the group \((\mathbb{Z}_{11} - \{0\}, \cdot)\). Show that this group is cyclic.

    Hint: express every element of \(\mathbb{Z}_{11} - \{0\}\) as a power of 2.

Solutions

  1. First we note that \(H \subseteq \mathbb{Z}\), and \(H \neq \emptyset\).

    Closure: if \(x, y \in H\) then \(x = 3a\) and \(y = 3b\) for some \(a, b \in \mathbb{Z}\). Therefore \(x + y = 3a + 3b = 3(a+b) \in H\).

    Identity: the identity of \((\mathbb{Z},+)\) is 0, and since \(0 = 3 \cdot 0\), we have \(0 \in H\).

    Inverse: in \(\mathbb{Z}\), the inverse of \(x\) is \(-x\). In \(H\), if \(x \in H\) then \(x = 3a\) for some \(a \in \mathbb{Z}\). This means \(-x = -3a = 3 \cdot (-a)\), and so \(-x \in H\) also.

    Note that we do not need to check associativity for subgroups, since this property is automatically inherited from the parent group \((\mathbb{Z},+)\). \(\square\)

    1. Suppose \(g^{-1} = h\). Now \(gh = gg^{-1} = e\) and \(hg = g^{-1}g = e\). Thus \(h^{-1} = g\).

    2. By definition, \(gg^{-1} = e = g^{-1}g\), so \((g^{-1})^{-1} = g\).

    3. Notice we have

    \[(gh)(h^{-1}g^{-1}) = ghh^{-1}g^{-1} = e\]

    and

    \[(h^{-1}g^{-1})(gh) = h^{-1}g^{-1}gh = e.\]

    This means \((gh)^{-1} = h^{-1}g^{-1}\).

  3. We draw the Cayley table of \(K = \{(0,0), (0,1), (2,0), (2,1)\}\) under componentwise addition in \(\mathbb{Z}_4 \times \mathbb{Z}_2\):

    \(+\) \((0,0)\) \((0,1)\) \((2,0)\) \((2,1)\)
    \(\mathbf{(0,0)}\) \((0,0)\) \((0,1)\) \((2,0)\) \((2,1)\)
    \(\mathbf{(0,1)}\) \((0,1)\) \((0,0)\) \((2,1)\) \((2,0)\)
    \(\mathbf{(2,0)}\) \((2,0)\) \((2,1)\) \((0,0)\) \((0,1)\)
    \(\mathbf{(2,1)}\) \((2,1)\) \((2,0)\) \((0,1)\) \((0,0)\)

    Notice the operation is closed in \(K\) and every element of \(K\) has an inverse in \(K\). Therefore \(K\) is a subgroup. \(\square\)

  4. Note that 2 is a generator of \((\mathbb{Z}_{11} - \{0\}, \cdot)\), so it is cyclic:

    \[1 = 2^0,\ 2 = 2^1,\ 4 = 2^2,\ 8 = 2^3,\ 5 = 2^4,\ 10 = 2^5,\ 9 = 2^6,\ 7 = 2^7,\ 3 = 2^8,\ 6 = 2^9.\]

§35 — Group Isomorphisms

No relevant textbook pages. No lecture in the published schedule — the Week 10 row jumps from L27 (V34) to L28 (V36).

This content is optional and will not be assessed. The source document carries no practice problems or solutions for this section.

§36 — Definitions and Examples of Fields

No relevant textbook pages. Lecture 28 (week 10).

Problems

    1. Solve \(3x + 4 = 2\) in the field \((\mathbb{Z}_5, +, \cdot)\).

    2. Solve \(8x - 9 = 6\) in the field \((\mathbb{Z}_{11}, +, \cdot)\).

    3. Solve \(4(1-x) = 10\) in the field \((\mathbb{Z}_{13}, +, \cdot)\).

    4. Solve \(16(x-6) = 15\) in the field \((\mathbb{Z}_{19}, +, \cdot)\).

  2. Prove \(\mathbb{Z}_{12}\) with addition modulo 12 and multiplication modulo 12 is not a field.

  3. Define addition and multiplication on \(\mathbb{Q} \times \mathbb{Q}\) by

    \[(a,b) + (c,d) = (a+c,\ b+d)\]

    and

    \[(a,b) \cdot (c,d) = (ac + 3bd,\ ad + bc)\]

    for all \((a,b), (c,d) \in \mathbb{Q} \times \mathbb{Q}\). Prove that \((\mathbb{Q} \times \mathbb{Q}, +, \cdot)\) is a field.

Solutions

    \[\begin{aligned} 3x + 4 &= 2 \\ \to 3x &= -2 \end{aligned}\]

    Now, since \(3 \times 2 = 6 \equiv 1 \pmod 5\), 2 is the multiplicative inverse of 3, and so we multiply both sides by 2 and obtain \(x = -4\). That is, \(x = 1\).

    Check: \(3 \cdot 1 + 4 = 7 \equiv 2 \pmod 5\).

    \[\begin{aligned} 8x - 9 &= 6 \\ \to 8x &= 15 = 4 \end{aligned}\]

    Now, since \(8 \times 7 = 56 \equiv 1 \pmod{11}\), 7 is the multiplicative inverse of 8, and so we multiply both sides by 7 and obtain \(x = 28\). That is, \(x = 6\).

    Check: \(8 \cdot 6 - 9 = 39 \equiv 6 \pmod{11}\).

    1. Since \(4 \times 10 = 40 \equiv 1 \pmod{13}\), 10 is the multiplicative inverse of 4, and so we multiply both sides by 10 and obtain \(1 - x = 100 = 9\). That is, \(x = 1 - 9 = 14 - 9 = 5\).

    Check: \(4(1-5) = -16 \equiv 10 \pmod{13}\).

    1. Since \(16 \times 6 = 96 \equiv 1 \pmod{19}\), 6 is the multiplicative inverse of 16, and so we multiply both sides by 6 and obtain \(x - 6 = 90 = 14\). That is, \(x = 20 = 1\).

    Check: \(16(1-6) = -80 \equiv 15 \pmod{19}\).

  2. \(\mathbb{Z}_{12}\) is not a field because 2 has no multiplicative inverse modulo 12. Indeed \(2 \cdot k \equiv 1 \pmod{12}\) has no solution since \(2k\) is always even.

  3. We show that \((\mathbb{Q} \times \mathbb{Q}, +, \cdot)\) is a field in three steps.

    Step 1 — \((\mathbb{Q} \times \mathbb{Q}, +)\) is an abelian group.

    Closure: for all \((a,b), (c,d) \in \mathbb{Q} \times \mathbb{Q}\), we have \((a,b) + (c,d) = (a+c, b+d) \in \mathbb{Q} \times \mathbb{Q}\).

    Associative: for all \((a,b), (c,d), (e,f) \in \mathbb{Q} \times \mathbb{Q}\) we have \(\big((a,b) + (c,d)\big) + (e,f) = (a+c, b+d) + (e,f) = (a+c+e, b+d+f) = (a,b) + (c+e, d+f) = (a,b) + \big((c,d) + (e,f)\big)\).

    Identity: for all \((a,b) \in \mathbb{Q} \times \mathbb{Q}\) we have \((a,b) + (0,0) = (a+0, b+0) = (a,b) = (0+a, 0+b) = (0,0) + (a,b)\).

    Inverses: for all \((a,b) \in \mathbb{Q} \times \mathbb{Q}\) we have \((a,b) + (-a,-b) = (a-a, b-b) = (0,0)\) and \((-a,-b) + (a,b) = (-a+a, -b+b) = (0,0)\). So \((-a,-b)\) is the inverse of \((a,b)\).

    Commutativity: for all \((a,b), (c,d) \in \mathbb{Q} \times \mathbb{Q}\), we have \((a,b) + (c,d) = (a+c, b+d) = (c+a, d+b) = (c,d) + (a,b)\).

    Thus \((\mathbb{Q} \times \mathbb{Q}, +)\) is an abelian group.

    Step 2 — \(\big((\mathbb{Q} \times \mathbb{Q}) - \{(0,0)\},\ \cdot\big)\) is an abelian group.

    Closure: for all \((a,b), (c,d) \in (\mathbb{Q} \times \mathbb{Q}) - \{(0,0)\}\), we have \((a,b) \cdot (c,d) = (ac+3bd, ad+bc) \in \mathbb{Q} \times \mathbb{Q}\). So we only need to show that \((ac+3bd, ad+bc) \neq (0,0)\). For a contradiction, suppose \((ac+3bd, ad+bc) = (0,0)\). That is, \(ac + 3bd = 0\) and \(ad + bc = 0\). Multiplying the first equation by \(d\) and the second by \(c\) we obtain \(acd + 3bd^2 = 0\) and \(acd + bc^2 = 0\), from which it follows that \(3bd^2 = bc^2\).

    Now, if \(b = 0\), then \(ac = 0\), \(ad = 0\) and \(a \neq 0\), and so \(c = d = 0\); a contradiction. Thus \(b \neq 0\) and we have \(3d^2 = c^2\). This is a contradiction because if \(d = 0\), then \(c \neq 0\) and the equation does not hold, and if \(d \neq 0\), then we have \(3 = \frac{c^2}{d^2}\), which implies \(\sqrt{3} = \frac{c}{d}\) is rational. Thus \((ac+3bd, ad+bc) \neq (0,0)\) and we have closure.

    Associative: for all \((a,b), (c,d), (e,f) \in (\mathbb{Q} \times \mathbb{Q}) - \{(0,0)\}\) we have

    \[\begin{aligned} (a,b) \cdot \big((c,d) \cdot (e,f)\big) &= (a,b) \cdot (ce + 3df,\ cf + de) \\ &= \big(a(ce+3df) + 3b(cf+de),\ a(cf+de) + b(ce+3df)\big) \\ &= (ace + 3adf + 3bcf + 3bde,\ acf + ade + bce + 3bdf) \end{aligned}\]

    and

    \[\begin{aligned} \big((a,b) \cdot (c,d)\big) \cdot (e,f) &= (ac+3bd,\ ad+bc) \cdot (e,f) \\ &= \big((ac+3bd)e + 3(ad+bc)f,\ (ac+3bd)f + (ad+bc)e\big) \\ &= (ace + 3bde + 3adf + 3bcf,\ acf + 3bdf + ade + bce) \\ &= (ace + 3adf + 3bcf + 3bde,\ acf + ade + bce + 3bdf). \end{aligned}\]

    Identity: for all \((a,b) \in (\mathbb{Q} \times \mathbb{Q}) - \{(0,0)\}\) we have \((a,b) \cdot (1,0) = (a,b)\) and \((1,0) \cdot (a,b) = (a,b)\), so \((1,0)\) is a multiplicative identity.

    Commutativity: for all \((a,b), (c,d) \in \mathbb{Q} \times \mathbb{Q}\), we have

    \[(a,b) \cdot (c,d) = (ac+3bd,\ ad+bc) = (ca+3db,\ cb+da) = (c,d) \cdot (a,b).\]

    Inverses: we check that for all \((a,b) \in (\mathbb{Q} \times \mathbb{Q}) - \{(0,0)\}\),

    \[(a,b)^{-1} = \left(\frac{a}{a^2 - 3b^2},\ \frac{-b}{a^2 - 3b^2}\right).\]

    First, note that \(a^2 - 3b^2 \neq 0\), because \(a^2 - 3b^2 = 0\) implies either \((a,b) = (0,0)\) or \(\sqrt{3} = \frac{a}{b}\), both of which are contradictions (the second because \(\sqrt{3}\) is not rational). Thus \((a,b)^{-1} \in (\mathbb{Q} \times \mathbb{Q}) - \{(0,0)\}\), and notice

    \[\begin{aligned} (a,b) \cdot \left(\frac{a}{a^2-3b^2}, \frac{-b}{a^2-3b^2}\right) &= \left(a \cdot \frac{a}{a^2-3b^2} + 3b \cdot \frac{-b}{a^2-3b^2},\ a \cdot \frac{-b}{a^2-3b^2} + b \cdot \frac{a}{a^2-3b^2}\right) \\ &= \left(\frac{a^2 - 3b^2}{a^2 - 3b^2},\ \frac{-ab + ab}{a^2 - 3b^2}\right) \\ &= (1, 0). \end{aligned}\]

    So every nonzero element has an inverse. Thus \(\big((\mathbb{Q} \times \mathbb{Q}) - \{(0,0)\}, \cdot\big)\) is an abelian group.

    Step 3 — the distributive laws. For all \((a,b), (c,d), (e,f) \in \mathbb{Q} \times \mathbb{Q}\) we have

    \[\begin{aligned} (a,b) \cdot \big((c,d) + (e,f)\big) &= (a,b) \cdot (c+e,\ d+f) \\ &= \big(a(c+e) + 3b(d+f),\ a(d+f) + b(c+e)\big) \\ &= \big((ac+3bd) + (ae+3bf),\ (ad+bc) + (af+be)\big) \\ &= (ac+3bd,\ ad+bc) + (ae+3bf,\ af+be) \\ &= (a,b) \cdot (c,d) + (a,b) \cdot (e,f) \end{aligned}\]

    and (since we have already shown that multiplication and addition are commutative)

    \[\begin{aligned} \big((a,b) + (c,d)\big) \cdot (e,f) &= (e,f) \cdot \big((a,b) + (c,d)\big) \\ &= (e,f) \cdot (a,b) + (e,f) \cdot (c,d) && \text{(by the first distributive law)} \\ &= (a,b) \cdot (e,f) + (c,d) \cdot (e,f). \end{aligned}\]

    Thus \((\mathbb{Q} \times \mathbb{Q}, +, \cdot)\) is a field. \(\square\)

§37 — Introduction to Counting

Epp 4th ed. pp. 516–553 · 5th ed. (metric) pp. 564–604. Lecture 29 (week 11).

Problems

  1. The UQ drama club is auditioning for its annual production of Romeo and Juliet. Six people are auditioning for the role of Romeo, and eight other people are auditioning for the role of Juliet. In how many ways can the director cast the lead couple?

  2. There are eleven questions on the final exam, but you decide — based on your successful grade for your in-semester exam — that you will only do five of them. How many combinations of questions are there which you could complete?

    1. How many permutations of \(\{1, 2, \dots, 15\}\) start with an odd number?

    2. How many 5-element subsets of \(\{1, 2, \dots, 15\}\) contain exactly two numbers from \(\{1,2,3,4\}\)?

Solutions

  1. There are 6 choices for Romeo and 8 choices for Juliet, so \(6 \cdot 8 = 48\) ways.

  2. Since the order in which you do the 5 questions does not matter, there are \(\binom{11}{5} = 462\) combinations of questions.

    1. There are 8 odd numbers in the set \(\{1, \dots, 15\}\). Choose the first entry in 8 ways, then permute the remaining 14 numbers, so \(8 \cdot 14!\) permutations.

    2. We choose exactly 2 numbers from the set \(\{1,2,3,4\}\) and exactly 3 numbers from the remaining 11 numbers, so the number of subsets is

    \[\binom{4}{2}\binom{11}{3}.\]

Application — Runtime

Not assessable for MATH1061. Taken from S. Epp, Discrete Mathematics with Applications, 5th edition, pages 787–799.

In many algorithms, the runtime is well approximated by counting how many times the basic instructions are executed.

For programs with nested for-loops and no early exits, the total number of executions of the inner loop body equals the number of choices of the outer-loop index multiplied by the number of choices of the inner-loop index.

Concretely, if \(i\) runs through \(M\) values and, for each fixed \(i\), the variable \(j\) runs through \(N\) values, then the inner loop body is executed exactly \(MN\) times.

Application problems

  1. The following algorithms are written in pseudo-code and contain nested loops. For each algorithm, count the total number of iterations of the inner loop that the algorithm will run.

    for i := 1 to 30
    for j := 1 to 15
        [Statements in body of inner loop. None contain
         branching statements that lead outside the loop.]
    next j
next i
    for i := 5 to 50
    for j := 10 to 20
        [Statements in body of inner loop. None contain
         branching statements that lead outside the loop.]
    next j
next i
    1. (Fix integers \(a, b, c, d \in \mathbb{Z}^+\) with \(a \leq b\) and \(c \leq d\).)
    for i := a to b
    for j := c to d
        [Statements in body of inner loop. None contain
         branching statements that lead outside the loop.]
    next j
next i
Application solutions

    1. \(i = 1, \dots, 30\) gives 30 iterations and \(j = 1, \dots, 15\) gives 15 iterations, so the inner loop runs \(30 \cdot 15 = 450\) times.

    2. \(i = 5, \dots, 50\) gives \(50 - 5 + 1 = 46\) iterations and \(j = 10, \dots, 20\) gives \(20 - 10 + 1 = 11\) iterations, so the inner loop runs \(46 \cdot 11 = 506\) times.

    3. \(i = a, \dots, b\) gives \(b - a + 1\) iterations and \(j = c, \dots, d\) gives \(d - c + 1\) iterations, so the inner loop runs \((b-a+1)(d-c+1)\) times.

§38 — Counting Selections

Epp 4th ed. pp. 516–553 · 5th ed. (metric) pp. 564–604. Lecture 30 (week 11).

Problems

    1. How many distinct arrangements can be made from the letters of the word INDOOROOPILLY?

    2. How many distinct arrangements can be made from the letters of the word INDOOROOPILLY that begin with R and end with P?

    3. How many distinct arrangements can be made from the letters of the word INDOOROOPILLY that begin with I and end with Y?

    4. How many distinct arrangements can be made from the letters of the word INDOOROOPILLY that have all four Os next to each other?

    1. How many distinct arrangements can be made of the letters of the word WOOLLOONGABBA?

    2. How many distinct arrangements are there of the letters of the word WOOLLOONGABBA such that exactly one of the following is true?

    • The two As are not next to each other.
    • The two Bs are not next to each other.
    • The two Ls are not next to each other.

    1. How many one-to-one (injective) functions are there from \(\{a,b,c\}\) to \(\{1,2,3,4,5\}\)?

    2. How many onto (surjective) functions are there from \(\{1,2,3,4,5\}\) to \(\{a,b,c\}\)?

Solutions

    1. There are 13 letters in INDOOROOPILLY, with four O’s, two I’s, two L’s, and the remaining letters appearing once each. Therefore the number of distinct arrangements is

    \[\frac{13!}{4!\,2!\,2!} = 64\,864\,800.\]

    1. Fix R in the first position and P in the last position. There are 11 letters remaining, with four O’s, two I’s, two L’s, and the remaining letters appearing once each. Therefore the number of distinct arrangements is

    \[\frac{11!}{4!\,2!\,2!} = 415\,800.\]

    1. Fix I in the first position and Y in the last position. There are 11 letters remaining, with four O’s, two L’s, and the remaining letters appearing once each. Therefore the number of distinct arrangements is

    \[\frac{11!}{4!\,2!} = 831\,600.\]

    1. Treat the four O’s as one ‘letter’. There are 10 ‘letters’ with two I’s, two L’s, and the remaining ‘letters’ appearing once each. Therefore the number of distinct arrangements is

    \[\frac{10!}{2!\,2!} = 907\,200.\]

    1. There are 13 letters in WOOLLOONGABBA, with four O’s, two L’s, two A’s, two B’s, and the remaining letters appearing once each. Therefore the number of distinct arrangements is

    \[\frac{13!}{4!\,2!\,2!\,2!} = 32\,432\,400.\]

    1. We want exactly one of the pairs AA, BB, LL to be non-adjacent. That is, there are three cases to consider:
    • Case 1: the two A’s are not adjacent but BB and LL are adjacent.
    • Case 2: the two B’s are not adjacent but AA and LL are adjacent.
    • Case 3: the two L’s are not adjacent but AA and BB are adjacent.

    Note that the number of ways Case 1 can occur is exactly the same as the number of ways Case 2 can occur (and similarly for Case 3).

    To count the number of ways Case 1 can occur, we first count arrangements with BB adjacent and LL adjacent. Treat BB as one ‘letter’ and LL as one ‘letter’. Then there are 11 ‘letters’ (W, O, O, LL, O, O, N, G, A, BB, A) to arrange, with four O’s, two A’s, and the remaining ‘letters’ appearing once each. Therefore the number of such arrangements is

    \[\frac{11!}{4!\,2!}.\]

    In order to determine the number of these for which the two A’s are not adjacent, we take this total and subtract those where AA is also adjacent. For this, treat AA, BB, and LL as ‘letters’. Then there are 10 ‘letters’ to arrange, with four O’s and the remaining ‘letters’ appearing once each. Therefore the number of arrangements with AA adjacent and BB adjacent and LL adjacent is

    \[\frac{10!}{4!}.\]

    Hence, the number of ways that Case 1 can occur is

    \[\frac{11!}{4!\,2!} - \frac{10!}{4!}.\]

    Case 2 and Case 3 have exactly the same number of occurrences as Case 1. Also note that the three cases are mutually exclusive (you cannot have both Case 1 and Case 2 occurring at the same time, for example).

    So the number of arrangements of the word WOOLLOONGABBA for which exactly one of the three cases occurs is

    \[3\left(\frac{11!}{4!\,2!} - \frac{10!}{4!}\right).\]

    1. An injective function \(f : \{a,b,c\} \to \{1,2,3,4,5\}\) is determined by choosing distinct images for \(a\), \(b\), \(c\). The number of choices is

    \[5 \cdot 4 \cdot 3 = {}^5P_3 = 60.\]

    1. The number of surjective functions \(f : \{1,2,3,4,5\} \to \{a,b,c\}\) is the total number of functions from \(\{1,2,3,4,5\}\) to \(\{a,b,c\}\) minus the number of functions from \(\{1,2,3,4,5\}\) to \(\{a,b,c\}\) which are not surjective. The former number is \(3^5\). A function which is not surjective must have image of size 1 or 2.
    • In the first case, there are 3 such functions (one for each of the three choices for the image).
    • In the latter case, first choose which 2 elements of \(\{a,b,c\}\) will be in the image. There are \(\binom{3}{2}\) such choices. For this fixed two-element image, the number of functions with this image is

    \[2^5 - 2,\]

    since there are \(2^5\) total functions from a 5-element set to a 2-element set, and we subtract the 2 constant ones (which fall into the first case, having image of size 1). Thus the number of functions from \(\{1,2,3,4,5\}\) to \(\{a,b,c\}\) with an image of size 2 is

    \[\binom{3}{2}(2^5 - 2) = 3 \cdot 30 = 90.\]

    This means there are 93 functions which are not surjective, so the number of surjective functions from \(\{1,2,3,4,5\}\) to \(\{a,b,c\}\) is

    \[3^5 - 93 = 150.\]

§39 — Introduction to Probability

Epp 4th ed. pp. 516–553 · 5th ed. (metric) pp. 564–604. Lecture 31 (week 11).

Problems

  1. To form a committee, five people are to be chosen from a group of ten people.

    1. How many different committees of five can be chosen?

    2. Amongst the ten people, we have Alice, Bob, Eve and Oscar. Alice and Bob are extremely talkative during meetings, and so at most one of these two can be chosen. Eve and Oscar are inseparable, and so a committee must either contain both of them or neither of them. With these constraints, how many different five-person committees can be formed?

    3. What is the probability that none of Alice, Bob, Eve and Oscar are on a committee, given that a random committee of five is chosen subject to the constraints in part (b)?

    1. How many positive two-digit integers are multiples of 3?

    2. What is the probability that a randomly chosen positive two-digit integer is a multiple of 3?

    3. What is the probability that a randomly chosen positive two-digit integer is a multiple of 4?

Solutions

    1. Number of different committees of five is

    \[\binom{10}{5} = \frac{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6}{5 \cdot 4 \cdot 3 \cdot 2} = 252.\]

    1. Let \(A\), \(B\) be Alice and Bob, and let \(E\), \(O\) be Eve and Oscar. There are 6 other people. We split into cases.

    Case 1: the committee contains both \(E\) and \(O\). Then we choose 3 more people from the remaining 8 with the restriction that we cannot choose both \(A\) and \(B\). There are \(\binom{8}{3}\) ways to choose 3 more people from the remaining 8 to be on the committee; of these, there are \(\binom{6}{1}\) ways in which both \(A\) and \(B\) are chosen (such a committee must have \(E, O, A, B\) so there is only one more person to choose from the remaining 6 people). Thus the number of committees for this case is

    \[\binom{8}{3} - \binom{6}{1} = 56 - 6 = 50.\]

    Case 2: the committee contains neither \(E\) nor \(O\). Then we choose 5 people from the remaining 8, again not allowing both \(A\) and \(B\). There are \(\binom{8}{5}\) ways to choose 5 people from the remaining 8 people to be on the committee; of these, there are \(\binom{6}{3}\) ways in which both \(A\) and \(B\) are chosen. Thus the number of committees for this case is

    \[\binom{8}{5} - \binom{6}{3} = 56 - 20 = 36.\]

    Hence the total number is \(50 + 36 = 86\).

    1. To have none of Alice, Bob, Eve, or Oscar on the committee, we must choose all 5 members from the other 6 people, which can be done in \(\binom{6}{5} = 6\) ways. Therefore the probability is

    \[\frac{6}{86} = 0.069767\ldots\]

    1. The positive two-digit multiples of 3 occur as every third number from 12 to 99, so there are

    \[\frac{99 - 12}{3} + 1 = 30\]

    such numbers.

    1. There are 90 positive two-digit integers (9 choices for the first digit and 10 choices for the second digit). Thus, by part (a), the probability that a randomly selected positive two-digit number is a multiple of 3 is

    \[\frac{30}{90} = \frac{1}{3}.\]

    1. The positive two-digit multiples of 4 occur as every fourth number from 12 to 96, so there are

    \[\frac{96 - 12}{4} + 1 = 22\]

    such numbers. Hence the probability that a randomly selected positive two-digit number is a multiple of 4 is

    \[\frac{22}{90} = \frac{11}{45}.\]

§40 — Binomial Coefficients

Epp 4th ed. pp. 565–591 · 5th ed. (metric) pp. 617–641. Lecture 31 (week 11).

Problems

  1. What is the coefficient of \(x^4\) in \((2 + 3x)^7\)?

    1. What is the coefficient of \(x^{12}\) in \((3 - x^3)^6\)?

    2. What is the coefficient of \(x^8\) in \((3 - x^3)^6\)?

  3. What is the coefficient of \(x^2\) in the expansion of \((3-x)^6\)?

  4. What is the coefficient of \(x^5\) in the expansion of \((2x+1)^{20}\)?

  5. What is the coefficient of \(x^5\) in the expansion of \(\left(3x^2 - \dfrac{2}{x}\right)^7\)?

Solutions

  1. By the binomial theorem,

    \[(2 + 3x)^7 = \sum_{k=0}^{7} \binom{7}{k} 2^{7-k} (3x)^k.\]

    The term \(x^4\) appears only when \(k = 4\). This term is \(\binom{7}{4} 2^3 (3x)^4\). Therefore the coefficient of \(x^4\) is

    \[\binom{7}{4} 2^3 3^4 = 35 \times 8 \times 81 = 22\,680.\]

  2. By the binomial theorem,

    \[(3 - x^3)^6 = \sum_{k=0}^{6} \binom{6}{k} 3^{6-k} (-x^3)^k = \sum_{k=0}^{6} \binom{6}{k} 3^{6-k} (-1)^k x^{3k}.\]

    1. The term involving \(x^{12}\) is \(\binom{6}{4} 3^2 (-1)^4 x^{3 \cdot 4}\), so the coefficient of \(x^{12}\) is

    \[15 \times 9 \times (-1)^4 = 135.\]

    1. Every power of \(x\) that appears is a multiple of 3. Since 8 is not a multiple of 3, there is no \(x^8\) term, so the coefficient of \(x^8\) is 0.

  3. By the binomial theorem,

    \[(3-x)^6 = \sum_{i=0}^{6} \binom{6}{i} 3^{6-i} (-x)^i = \sum_{i=0}^{6} \binom{6}{i} 3^{6-i} (-1)^i x^i.\]

    The term \(x^2\) appears only when \(i = 2\), so the coefficient of \(x^2\) is

    \[\binom{6}{2} 3^{6-2} (-1)^2 = \binom{6}{2} 3^4 = 1215.\]

  4. By the binomial theorem,

    \[(2x+1)^{20} = \sum_{k=0}^{20} \binom{20}{k} (2x)^{20-k} (1)^k.\]

    The term \(x^5\) appears only when \(k = 15\), where the term is \(\binom{20}{15}(2x)^5\), so the coefficient of \(x^5\) is

    \[\binom{20}{5} 2^5 = 496\,128.\]

  5. By the binomial theorem,

    \[\left(3x^2 - \frac{2}{x}\right)^7 = \sum_{i=0}^{7} \binom{7}{i} (3x^2)^{7-i}\left(-\frac{2}{x}\right)^i = \sum_{i=0}^{7} \binom{7}{i} 3^{7-i} (-2)^i x^{14-3i}.\]

    The term \(x^5\) appears only when \(14 - 3i = 5\), i.e. \(i = 3\). Therefore the coefficient of \(x^5\) is

    \[\binom{7}{3} 3^4 (-2)^3 = -22\,680.\]

§41 — Inclusion Exclusion

Epp 4th ed. pp. 545–549 · 5th ed. (metric) pp. 595–599. Lecture 32 (week 12).

Problems

  1. Out of 100 people surveyed, 40 people like apples, 50 like bananas and 40 like cherries. 15 like apples and bananas, 16 like bananas and cherries, 17 like cherries and apples and 10 people like all three. How many people do not like any of the three fruits?

  2. How many integers between 1 and 600 inclusive are coprime with 15? That is, how many integers \(n\) satisfy \(1 \leq n \leq 600\) and \(\gcd(n, 15) = 1\)?

  3. Seven friends, named Alice, Bob, Charlie, Delia, Edward, Fred and George, stand in a line for a photograph. How many different arrangements are there of the seven friends standing in a line such that Alice is not next to Charlie and Bob is not next to Delia?

Solutions

  1. Let \(A\) be the set of people who like apples, \(B\) the set of people who like bananas, and \(C\) the set of people who like cherries. We are given

    \[|A| = 40,\ |B| = 50,\ |C| = 40,\quad |A \cap B| = 15,\ |B \cap C| = 16,\ |C \cap A| = 17,\quad |A \cap B \cap C| = 10.\]

    By inclusion–exclusion,

    \[\begin{aligned} |A \cup B \cup C| &= |A| + |B| + |C| - |A \cap B| - |B \cap C| - |C \cap A| + |A \cap B \cap C| \\ &= 40 + 50 + 40 - 15 - 17 - 16 + 10 \\ &= 92. \end{aligned}\]

    Hence the number of people who do not like any of the three fruits is

    \[100 - |A \cup B \cup C| = 100 - 92 = 8.\]

  2. Let \(U = \{1, 2, \dots, 600\}\). Define

    \[A = \{n \in U : 3 \mid n\}, \qquad B = \{n \in U : 5 \mid n\}.\]

    Then \(n\) is not coprime to 15 if and only if \(n \in A \cup B\). We have

    \[|A| = \frac{600}{3} = 200, \qquad |B| = \frac{600}{5} = 120, \qquad |A \cap B| = \frac{600}{15} = 40.\]

    By inclusion–exclusion,

    \[|A \cup B| = |A| + |B| - |A \cap B| = 200 + 120 - 40 = 280.\]

    Therefore the number of \(n\) with \(\gcd(n,15) = 1\) is

    \[|U| - |A \cup B| = 600 - 280 = 320.\]

  3. Let \(U\) be the set of all linear arrangements of the 7 friends, so \(|U| = 7!\). Define

    \[A = \{\text{arrangements in which Alice is next to Charlie}\},\] \[B = \{\text{arrangements in which Bob is next to Delia}\}.\]

    We want to count the number of arrangements in \(U\) which are neither in \(A\) nor in \(B\). The required number is \(|U \setminus (A \cup B)| = |U| - |A \cup B|\).

    To count \(|A|\), treat \(\{\)Alice, Charlie\(\}\) as a single block. There are \(6!\) ways to arrange the 6 “objects”, and for each of these there are 2 ways to order Alice and Charlie within their block (either Alice or Charlie to the left). Hence

    \[|A| = 2 \cdot 6!.\]

    Similarly, \(|B| = 2 \cdot 6!\).

    To count \(|A \cap B|\), treat both pairs \(\{\)Alice, Charlie\(\}\) and \(\{\)Bob, Delia\(\}\) as two single blocks. There are \(5!\) ways to arrange the 5 objects (two blocks plus the other three people), and for each of these there are \(2 \cdot 2\) internal orders of the two blocks. Hence

    \[|A \cap B| = 4 \cdot 5!.\]

    By inclusion–exclusion,

    \[|A \cup B| = |A| + |B| - |A \cap B| = 2 \cdot 6! + 2 \cdot 6! - 4 \cdot 5!.\]

    Therefore

    \[|U \setminus (A \cup B)| = 7! - \left(2 \cdot 6! + 2 \cdot 6! - 4 \cdot 5!\right) = 2640.\]

§42 — The Pigeonhole Principle

Epp 4th ed. pp. 554–565 · 5th ed. (metric) pp. 604–616. Lecture 33 (week 12).

Problems

    1. Explain why, in a group of 50 people, at least five people were born in the same month.

    2. How many people do you need to ensure that at least seven people were born in the same month?

  2. Show that, if we select six distinct integers from the set \(\{1, 2, \dots, 10\}\), two of these must add to give 11.

  3. A computer network consists of six computers. Each computer is directly connected to zero or more of the other computers. Show that there are at least two computers in the network that are directly connected to the same number of other computers.

  4. During a month with 30 days, a baseball team plays at least one game a day, but no more than 45 games. Show that there must be a period of some number of consecutive days during which the team must play exactly 14 games.

  5. Let \(S = \{1, 2, \dots, 100\}\). What is the minimum number of integers you must pick from \(S\) to be sure that at least one pair has a difference which is divisible by 3?

Solutions

    1. There are 12 months (pigeonholes) and 50 people (pigeons). If each month had at most 4 birthdays, then there would be at most \(12 \cdot 4 = 48\) people. Since \(50 > 48\), by the pigeonhole principle some month must contain at least 5 people.

    2. To guarantee at least 7 people are born in the same month, note that if each month had at most 6 birthdays, then there could be at most \(12 \cdot 6 = 72\) people. Therefore with 73 people, by the pigeonhole principle some month has at least 7 people. Moreover, 72 people is not enough (take 6 born in each month), so the minimum is 73.

  2. Partition \(A = \{1, 2, \dots, 10\}\) into the 5 pairs

    \[\{1,10\},\ \{2,9\},\ \{3,8\},\ \{4,7\},\ \{5,6\},\]

    each summing to 11. In order for a selection of distinct integers from the set \(A\) to not have two that add to give 11, at most one integer can be chosen from each of the subsets. Choosing 6 distinct integers from \(A\) forces us to select two numbers from the same pair (there are 5 pairs but 6 choices), by the pigeonhole principle. Those two numbers then add to 11.

  3. For each computer, let its degree be the number of other computers it is directly connected to. With 6 computers, each degree is an integer in \(\{0,1,2,3,4,5\}\). It is impossible to have both a computer of degree 0 and a computer of degree 5 (if one is connected to everyone, then none has degree 0). Hence the degrees must all lie in one of the two 5-element sets

    \[\{0,1,2,3,4\} \quad \text{or} \quad \{1,2,3,4,5\}.\]

    So, in either case, there are only 5 possible degree values available for 6 computers. By the pigeonhole principle, at least two computers have the same degree, i.e. are connected to the same number of other computers.

  4. Let \(g_i\) be the total number of games played in the first \(i\) days, for \(i = 1, 2, \dots, 30\). Then

    \[1 \leq g_1 < g_2 < \cdots < g_{30} \leq 45\]

    because at least one game is played each day. Consider also the 30 numbers

    \[g_1 + 14,\ g_2 + 14,\ \dots,\ g_{30} + 14,\]

    which satisfy \(15 \leq g_1 + 14 < \cdots < g_{30} + 14 \leq 59\). Thus we have 60 integers

    \[g_1, \dots, g_{30},\ g_1 + 14, \dots, g_{30} + 14\]

    all lying in the set \(\{1, 2, \dots, 59\}\) of size 59. By the pigeonhole principle, two of these 60 integers are equal. Since the \(g_i\) are strictly increasing, no equality can occur among the \(g_i\) themselves, and likewise no equality can occur among the \(g_i + 14\) themselves. Therefore we must have

    \[g_j = g_i + 14\]

    for some \(i, j \in \{1, \dots, 30\}\). Then the number of games played from day \(i+1\) through day \(j\) is

    \[g_j - g_i = 14,\]

    so there is a period of consecutive days during which exactly 14 games are played.

  5. First, we note that two integers have a difference that is divisible by 3 if and only if they are congruent modulo 3 (have the same remainder when divided by 3). Now, the set \(S\) can be partitioned into three sets based on their congruence class modulo 3. By the pigeonhole principle, if we select 4 numbers from \(S\), at least two of them must belong to the same congruence class. These two will have difference that is divisible by 3. Note that 4 is the minimum number of selections that guarantees this, since it is possible to select 3 numbers where no two of them have difference divisible by 3 (select one from each congruence class).

§43 — Introduction to Graph Theory

Epp 4th ed. pp. 625–660 · 5th ed. (metric) pp. 677–697. Lecture 33 (week 12).

Problems

  1. For each of the following, state whether or not there exists a simple graph with the stated number of vertices and degrees. Draw the graph, or explain why one does not exist.

    1. Five vertices with degrees 3, 3, 2, 2, 2.

    2. Seven vertices with degrees 4, 2, 3, 1, 1, 1, 1.

    3. Five vertices with degrees 5, 4, 3, 1, 1.

  2. A graph has 10 edges and 6 vertices. Five of the vertices have degree 3. What is the degree of the remaining vertex?

  3. Let \(n\) be a positive integer and let \(K_n\) be the complete graph with \(n\) vertices. How many edges does \(K_n\) have?

  4. A graph has 27 edges and every vertex has degree 2, 3 or 4. If there are 7 vertices that have degree 4 and 4 vertices that have degree 3, then how many vertices of degree 2 are there?

Solutions

    1. Yes, such a simple graph exists. One drawing has vertices \(a, b, c, d, e\) and the six edges

    \[ab,\quad ac,\quad ad,\quad bc,\quad be,\quad de,\]

    giving \(\deg(a) = 3\), \(\deg(b) = 3\), \(\deg(c) = 2\), \(\deg(d) = 2\), \(\deg(e) = 2\) as required.

    1. No such graph exists. If the degrees are 4, 2, 3, 1, 1, 1, 1, then the sum of degrees is

    \[4 + 2 + 3 + 1 + 1 + 1 + 1 = 13,\]

    which is odd. Remember (by the Handshake Theorem) that the sum of the degrees in any graph equals twice the number of edges, and so is always even. So such a graph cannot exist.

    1. No such graph exists. In a simple graph on 5 vertices, the largest possible degree is 4, but the sequence contains a vertex of degree 5, which is impossible. So there is no such simple graph.

  2. Since the graph has 10 edges, the sum of all the degrees is \(2 \cdot 10 = 20\). If five vertices have degree 3, the sum of their degrees is \(5 \cdot 3 = 15\), so the remaining vertex has degree

    \[20 - 15 = 5.\]

  3. In \(K_n\), an edge is determined by choosing 2 distinct vertices, so the number of edges is

    \[\binom{n}{2} = \frac{n(n-1)}{2}.\]

  4. Let \(x\) be the number of vertices of degree 2. Since the graph has 27 edges, the sum of degrees is \(2 \cdot 27 = 54\). Given that there are seven vertices of degree 4 and four vertices of degree 3, and all other vertices have degree 2, we have

    \[2x + 4 \cdot 7 + 3 \cdot 4 = 54.\]

    This means \(x = 7\).

§44 — Walks, Trails and Circuits

Epp 4th ed. pp. 625–660 · 5th ed. (metric) pp. 677–697. Lecture 34 (week 12).

Problems

  1. Let \(G\) and \(H\) be the graphs below.

    \(G\) has vertex set \(\{v_1, v_2, v_3, v_4, v_5\}\) and is drawn with a loop at \(v_2\), a loop at \(v_5\), and two parallel edges between \(v_1\) and \(v_2\). Its degree sequence is

    \[\deg(v_1) = 4,\quad \deg(v_2) = 6,\quad \deg(v_3) = 4,\quad \deg(v_4) = 2,\quad \deg(v_5) = 4,\]

    so \(G\) is connected with 10 edges.

    \(H\) has vertex set \(\{v_1, v_2, v_3, v_4, v_5\}\) and is a simple connected graph drawn as the 4-cycle \(v_1 v_2 v_3 v_4 v_1\) with a single pendant edge \(v_3 v_5\), giving degrees

    \[\deg(v_1) = 2,\quad \deg(v_2) = 2,\quad \deg(v_3) = 3,\quad \deg(v_4) = 2,\quad \deg(v_5) = 1.\]

    For each of the following statements, explain why it is true or false.

    1. \(G\) is a simple graph.

    2. The sum of the degrees of the vertices of \(G\) is 18.

    3. \(G\) contains an Euler circuit.

    4. \(H\) contains an Euler trail.

  2. Does the following graph contain an Euler circuit? (The graph shown is connected and has exactly two vertices of degree 3.)

  3. Let \(G\) be a simple, connected, Eulerian graph with eight vertices and let \(e\) be the number of edges in \(G\). Prove that \(8 \leq e \leq 24\).

Solutions

    1. This statement is false. The graph \(G\) has loops (at \(v_2\) and \(v_5\)) and it also has multiple edges between \(v_1\) and \(v_2\), so \(G\) is not simple.

    2. This statement is false. A loop contributes 2 to the degree, so we have

    \[\deg(v_1) = 4,\quad \deg(v_2) = 6,\quad \deg(v_3) = 4,\quad \deg(v_4) = 2,\quad \deg(v_5) = 4,\]

    and thus the sum of the degrees is \(4 + 6 + 4 + 2 + 4 = 20 \neq 18\).

    1. This statement is true. The graph \(G\) is connected and every vertex has even degree (as computed above), so \(G\) contains an Euler circuit.

    2. This statement is true. In \(H\), the degrees are

    \[\deg(v_1) = 2,\ \deg(v_2) = 2,\ \deg(v_4) = 2,\ \deg(v_3) = 3,\ \deg(v_5) = 1,\]

    so exactly two vertices have odd degree (namely \(v_3\) and \(v_5\)). Therefore \(H\) has an Euler trail (starting at one of \(v_3\), \(v_5\) and ending at the other).

  2. No. The graph is connected, but has two vertices of degree 3, so not all vertices have even degree. Hence there is no Euler circuit.

  3. Since \(G\) is connected and Eulerian, every vertex has even degree, and (because \(G\) is simple and connected with 8 vertices) each vertex has degree at least 2. Hence

    \[2e = \sum_{v} \deg(v) \geq 8 \cdot 2 = 16,\]

    so \(e \geq 8\). Moreover, in a simple graph on 8 vertices we have \(\deg(v) \leq 7\), and because \(G\) is Eulerian all degrees are even, so in fact \(\deg(v) \leq 6\) for every vertex. Therefore

    \[2e = \sum_{v} \deg(v) \leq 8 \cdot 6 = 48,\]

    so \(e \leq 24\). Hence \(8 \leq e \leq 24\). \(\square\)

§45 — Matrix Representations of Graphs

Epp 4th ed. pp. 661–675 · 5th ed. (metric) pp. 698–712. Lecture 35 (week 13).

Problems

  1. Let \(G\) be the graph with vertex set \(\{v_1, v_2, v_3, v_4\}\) and edge set \(\{a,b,c,d,e,f\}\), where the edges are

    edge endpoints
    \(a\) \(v_1 v_2\)
    \(b\) \(v_2 v_3\)
    \(c\) \(v_2 v_3\) (parallel to \(b\))
    \(d\) \(v_3 v_4\)
    \(e\) \(v_2 v_4\)
    \(f\) loop at \(v_4\)

    1. Write down an adjacency matrix for \(G\). Make sure you clearly label the rows and columns of your matrix.

    2. Write down an incidence matrix for \(G\). Make sure you clearly label the rows and columns of your matrix.

    3. Does \(G\) have an Euler circuit? If yes, write it out as a sequence of vertices and edges. If not, determine the fewest number of edges that would need to be added to \(G\) to create a new graph that does have an Euler circuit.

  2. Let \(G\) be the graph with the following adjacency matrix.

    \[\begin{pmatrix} 0 & 1 & 0 & 0 & 3 \\ 1 & 0 & 2 & 0 & 0 \\ 0 & 2 & 0 & 1 & 1 \\ 0 & 0 & 1 & 0 & 1 \\ 3 & 0 & 1 & 1 & 0 \end{pmatrix}\]

    1. Draw the graph \(G\).

    2. Explain whether or not \(G\) has an Euler circuit or an Euler trail.

  3. Let \(G\) be the graph with the following adjacency matrix.

    \[\begin{pmatrix} 0 & 1 & 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 1 \\ 1 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 & 1 & 0 \end{pmatrix}\]

    Explain whether or not \(G\) is Eulerian.

Solutions

    1. The labelled adjacency matrix for \(G\) is
    \(v_1\) \(v_2\) \(v_3\) \(v_4\)
    \(\mathbf{v_1}\) 0 1 0 0
    \(\mathbf{v_2}\) 1 0 2 1
    \(\mathbf{v_3}\) 0 2 0 1
    \(\mathbf{v_4}\) 0 1 1 1

    (where the entry 1 on the diagonal at \(v_4\) records the single loop at \(v_4\)).

    1. The labelled incidence matrix for \(G\) is
    \(a\) \(b\) \(c\) \(d\) \(e\) \(f\)
    \(\mathbf{v_1}\) 1 0 0 0 0 0
    \(\mathbf{v_2}\) 1 1 1 0 1 0
    \(\mathbf{v_3}\) 0 1 1 1 0 0
    \(\mathbf{v_4}\) 0 0 0 1 1 2

    (where the entry 2 in the bottom right indicates the loop that is incident twice with \(v_4\)).

    1. The degrees are

    \[\deg(v_1) = 1,\quad \deg(v_2) = 4,\quad \deg(v_3) = 3,\quad \deg(v_4) = 4,\]

    so \(G\) does not have an Euler circuit (since \(v_1\) and \(v_3\) have odd degree).

    The fewest edges to add to make all degrees even is 1; for example, adding an edge \(g\) between vertices \(v_1\) and \(v_3\) creates a new connected graph in which all of the vertices have even degree.

    An example of an Euler circuit in the new graph is

    \[v_1,\ a,\ v_2,\ b,\ v_3,\ d,\ v_4,\ f,\ v_4,\ e,\ v_2,\ c,\ v_3,\ g,\ v_1.\]

    1. Reading the matrix off (with vertices \(v_1, \dots, v_5\) in matrix order), \(G\) has the edges
    • \(v_1 v_2\) — a single edge,
    • \(v_1 v_5\) — three parallel edges,
    • \(v_2 v_3\) — two parallel edges,
    • \(v_3 v_4\) — a single edge,
    • \(v_3 v_5\) — a single edge,
    • \(v_4 v_5\) — a single edge.
    1. The degrees are

    \[\deg(v_1) = 4,\ \deg(v_2) = 3,\ \deg(v_3) = 4,\ \deg(v_4) = 2,\ \deg(v_5) = 5,\]

    so there is no Euler circuit (since \(v_2\) and \(v_5\) have odd degree), but there is an Euler trail (exactly two odd-degree vertices) from \(v_2\) to \(v_5\).

  3. Let the vertices be \(v_1, v_2, v_3, v_4, v_5, v_6\) in the order of the matrix. Reading the matrix off, \(G\) has edges

    \[v_1 v_2,\quad v_1 v_4,\quad v_2 v_4,\quad v_3 v_5,\quad v_3 v_6,\quad v_5 v_6,\]

    that is, two disjoint triangles: one on \(\{v_1, v_2, v_4\}\) and one on \(\{v_3, v_5, v_6\}\).

    The graph is not connected (e.g. there is no walk from vertex \(v_1\) to vertex \(v_3\)). Hence \(G\) is not Eulerian.

    (Errata note: the source document’s solution to 1(b) was corrected on 11 August 2026 — the note in brackets had previously copied the note from 1(a). The corrected note is the one given above.)

§46 — Trees

Epp 4th ed. pp. 683–701 · 5th ed. (metric) pp. 720–742. Lecture 36 (week 13).

Problems

  1. A tree has 11 vertices, of which exactly four vertices have degree 1 and exactly one vertex has degree 4. Determine the degrees of the remaining six vertices.

  2. Let \(T\) be a tree with \(p\) vertices, where \(p \geq 11\). Suppose \(T\) has precisely six vertices of degree 1 and precisely four vertices of degree 3. Determine the degree of each of the remaining \(p - 10\) vertices of \(T\).

  3. Let \(T\) be a tree on 12 vertices. Suppose \(T\) has exactly three vertices of degree 3 and each of the remaining vertices has degree 1 or 5. Using theorems that were given in lectures, determine the number of vertices of degree 5.

  4. Two trees with the same vertex set \(\{1, 2, \dots, n\}\) are distinct if their edge sets are not equal. Use mathematical induction to prove that for each integer \(n \geq 2\), the number of distinct trees with vertex set \(\{1, 2, \dots, n\}\) is at least \((n-1)!\).

Solutions

  1. A tree with 11 vertices has \(e = 10\) edges, so by the Handshake Theorem, the sum of degrees is \(2e = 20\). We are given that exactly four vertices have degree 1 and exactly one vertex has degree 4. Let \(v_1, v_2, \dots, v_6\) be the remaining vertices whose degrees are to be determined. We have

    \[\begin{aligned} \sum_{i=1}^{6} \deg(v_i) + 4 \cdot 1 + 1 \cdot 4 &= 20 \\ \sum_{i=1}^{6} \deg(v_i) &= 12. \end{aligned}\]

    None of these six vertices \(v_1, v_2, \dots, v_6\) can have degree 1 (since there are exactly four vertices of degree 1, already accounted for). Hence each has degree at least 2. Because six vertices of degree at least 2 have total degree 12, each must have degree exactly 2.

    Thus the remaining six vertices all have degree 2.

  2. \(T\) is a tree with \(p\) vertices and so must have \(p-1\) edges. Now by the Handshake Theorem we know

    \[2(p-1) = \sum_{i=1}^{p} \deg(v_i).\]

    Using the known degrees we can write this as

    \[\begin{aligned} 2(p-1) &= 6 \cdot 1 + 4 \cdot 3 + \sum_{i=1}^{p-10} \deg(v_i) \\ 2(p-1) &= 18 + \sum_{i=1}^{p-10} \deg(v_i) \\ 2p - 2 - 18 &= \sum_{i=1}^{p-10} \deg(v_i) \\ 2(p-10) &= \sum_{i=1}^{p-10} \deg(v_i). \end{aligned}\]

    Rewriting the final line as

    \[\frac{\sum_{i=1}^{p-10} \deg(v_i)}{p-10} = 2, \qquad \text{i.e.} \qquad \frac{\deg(v_1) + \deg(v_2) + \cdots + \deg(v_{p-10})}{p-10} = 2,\]

    we see that the average degree of the remaining \(p-10\) vertices is 2. However, we know that there are no more vertices of degree 1. Thus each of the remaining \(p-10\) vertices must have degree exactly 2.

  3. Let \(x\) be the number of vertices of degree 5 and \(y\) the number of vertices of degree 1. Then, because the tree has 12 vertices, we have

    \[x + y + 3 = 12.\]

    This simplifies to give \(x + y = 9\).

    Now, since \(T\) is a tree on 12 vertices, it must have \(e = 11\) edges. By the Handshake Theorem, the sum of degrees is \(2e = 22\), so

    \[3 \cdot 3 + 5x + 1 \cdot y = 22.\]

    This simplifies to give \(5x + y = 13\).

    Subtracting \(x + y = 9\) from \(5x + y = 13\) gives \(4x = 4\), hence \(x = 1\). Therefore there is exactly one vertex of degree 5.

  4. We prove by induction that the number of distinct trees with vertex set \(\{1, 2, \dots, n\}\) is at least \((n-1)!\).

    Basis: let \(n = 2\). Here \((n-1)! = (2-1)! = 1\). There is exactly one tree on \(\{1,2\}\), namely the single edge \(\{1,2\}\). So the base case is true.

    Inductive hypothesis. Suppose \(k \geq 2\) is an integer and there are at least \((k-1)!\) distinct trees on \(\{1, 2, \dots, k\}\).

    Now consider \(n = k+1\). We aim to show that there are at least \(k!\) distinct trees on vertex set \(\{1, 2, \dots, k, k+1\}\).

    Let \(T\) be a tree on vertex set \(\{1, 2, \dots, k\}\). For any choice of \(m \in \{1, 2, \dots, k\}\), we form a graph \(T_m\) on vertex set \(\{1, 2, \dots, k, k+1\}\) as follows: start with tree \(T\), add a new vertex \(k+1\), and the edge \(\{k+1, m\}\).

    Note that \(T_m\) is connected (since \(k+1\) is joined to \(m\), and \(T\) was connected). Also, \(T_m\) has no circuit because \(T\) is a tree and adding a new leaf does not create a circuit. Thus \(T_m\) is a tree. Moreover, different choices of \(m\) give rise to distinct trees \(T_m\), because their edge sets are different.

    By the inductive hypothesis, there are at least \((k-1)!\) choices for the tree \(T\). Since there are \(k\) choices for the vertex \(m\), there are at least \(k \cdot (k-1)! = k!\) trees \(T_m\) on vertex set \(\{1, 2, \dots, k+1\}\).

    By the principle of mathematical induction, this completes the proof. \(\square\)

See also

Proof Techniques — Direct, Counterexample, Contradiction, Contraposition

The four methods for establishing (or demolishing) a statement of the form \(\forall x \in D, P(x) \to Q(x)\), plus the definitions every proof in this course is expected to unfold.

Definitions to prove from

Proofs are written by demonstrating facts using the definitions — so these have to be at your fingertips. Each is a biconditional: it can be used in either direction.

Term Definition
\(n\) is even \(n = 2k\) for some integer \(k\)
\(n\) is odd \(n = 2k + 1\) for some integer \(k\) (equivalently \(n = 2k - 1\) for some integer \(k\))
\(n\) is prime \(n \in \mathbb{Z}\), \(n > 1\), and \(\forall r, s \in \mathbb{Z}^+\), if \(n = rs\) then (\(r = 1\) and \(s = n\)) or (\(r = n\) and \(s = 1\))
\(n\) is composite \(n \in \mathbb{Z}\), \(n > 1\), and \(\exists r, s \in \mathbb{Z}^+\) such that \(n = rs\) and \(1 < r < n\) and \(1 < s < n\)
\(r\) is rational \(\exists a, b \in \mathbb{Z}\) such that \(r = \frac{a}{b}\) and \(b \neq 0\)
\(d \mid n\) \(d \neq 0\) and \(n = dk\) for some integer \(k\)

Two facts used constantly, and worth naming explicitly when you lean on them:

  • Fact: the sum, difference, and product of integers is an integer.
  • Fact: every integer is either even or odd.

Prime vs. not-prime. For \(n \in \mathbb{Z}^{>1}\), “\(n\) is not prime” means “\(n\) is composite”. For \(n \in \mathbb{Z}\) generally, “\(n\) is not prime” means “\(n\) is composite or \(n \leq 1\)” — the integers split into \(\mathbb{Z}^{\leq 1}\), \(\mathbb{Z}^{\text{prime}}\), and \(\mathbb{Z}^{\text{composite}}\), and \(1\) and everything below it is in none of the latter two. This is why \(n = 1\) is the standard counterexample to “every positive integer is prime or composite”.

Warning: don’t assume something that seems like an “obvious fact” until you have proven it, or can name the lecture or pre-work video it was proved in.

The four methods

All four target \(\forall x \in D, P(x) \to Q(x)\). Recall the truth table for \(p \to q\) — the only row that makes it false is \(p\) true, \(q\) false:

\(p\) \(q\) \(p \to q\)
T T T
T F F
F T T
F F T

Every method below is a different way of attacking that one row.

Direct proof (shows the statement is true)

Let \(x \in D\) and suppose \(P(x)\). \(\dots\) Hence \(Q(x)\). \(\square\)

Why suppose \(P(x)\) is true, when \(P(x) \to Q(x)\) holds anyway if \(P(x)\) is false? Because the only case that could make the statement false is \(P(x)\) true and \(Q(x)\) false. Ruling that out is the whole job — the \(P(x)\)-false rows are true for free.

Disproof by counterexample (shows the statement is false)

Let \(x = \cdots\). Then \(x \in D\) and \(P(x)\) is true. However, \(Q(x)\) is false. So \(x\) is a counterexample and the statement is false.

A counterexample to \(p \to q\) needs \(p\) true and \(q\) false — a value where the hypothesis fails proves nothing.

Proof by contradiction (shows the statement is true)

Suppose the negation of \(\forall x \in D, P(x) \to Q(x)\) is true. That is, suppose \(\exists x \in D\) such that \(P(x)\) is true but \(Q(x)\) is false. \(\dots\) Hence something obviously wrong — \(P(x)\) is false, or \(Q(x)\) is true, or \(1 = 0\). This is a contradiction, and we conclude the original statement is true. \(\square\)

The set-up comes straight from quantified-statements: the negation of \(\forall x \in D, P(x) \to Q(x)\) is \(\exists x \in D\) such that \(P(x) \land \sim Q(x)\).

Useful negations to have ready (see logical-equivalence-laws):

\[\sim(p \lor q) \equiv \sim p \land \sim q \qquad \sim(p \land q) \equiv \sim p \lor \sim q \qquad \sim(p \to q) \equiv p \land \sim q\]

Proof by contraposition (shows the statement is true)

We prove the contrapositive of \(\forall x \in D, P(x) \to Q(x)\), namely \(\forall x \in D, \sim Q(x) \to \sim P(x)\). Let \(x \in D\) and suppose \(\sim Q(x)\). \(\dots\) Hence \(\sim P(x)\). \(\square\)

This is valid because \(p \to q \equiv \sim q \to \sim p\) (see conditional-statements).

Choosing a method

  • Contradiction is the most versatile, but it needs more set-up, so a direct proof is often easier when one is available.
  • Contraposition is the natural choice when \(\sim Q(x)\) gives you something concrete to work with and \(P(x)\) doesn’t — typically when the conclusion is a negative statement (“\(r\) is irrational”, “\(n\) is odd”) whose negation is a definition you can expand.
  • Contradiction vs. contraposition: a contraposition proof is a contradiction proof that never actually needs the assumption \(P(x)\) — you derive \(\sim P(x)\) from \(\sim Q(x)\) alone. If your “contradiction” proof only ever uses the \(\sim Q(x)\) half of the assumption, write it as a contraposition instead.
  • Counterexample vs. contradiction: counterexample is a disproof, contradiction is a proof. They are not variants of each other.

Worked comparisons on the same statements live in the lecture notes: 2026-08-13-proof-by-contradiction proves \(\forall n \in \mathbb{Z}\), if \(n\) is odd then \(3n+2\) is odd both directly and by contradiction, and 2026-08-13-proof-by-contraposition works through statements where one method gets stuck and another doesn’t.

Worked examples

Direct proof

\(\forall n \in \mathbb{Z}\), if \(n\) is odd then \(3n + 2\) is odd.

Let \(n \in \mathbb{Z}\) and suppose \(n\) is odd. Then \(n = 2k + 1\) for some \(k \in \mathbb{Z}\). Now \[3n + 2 = 3(2k+1) + 2 = 6k + 5 = 2(3k + 2) + 1.\] Since \(k \in \mathbb{Z}\), we have \(3k + 2 \in \mathbb{Z}\), and hence \(3n + 2\) is odd. \(\square\)

For any integer \(x\), if \(x + 6 = 4y\) for some integer \(y\), then \(\frac{x}{2}\) is an odd integer.

Let \(x \in \mathbb{Z}\) and suppose \(x + 6 = 4y\) for some \(y \in \mathbb{Z}\). Then \(x = 4y - 6 = 2(2y - 3)\), so \[\frac{x}{2} = \frac{2(2y-3)}{2} = 2y - 3 = 2(y - 1) - 1.\] Since \(y \in \mathbb{Z}\) we have \(2y - 3 \in \mathbb{Z}\), so \(\frac{x}{2} \in \mathbb{Z}\); and \(y - 1 \in \mathbb{Z}\), so by definition \(\frac{x}{2}\) is odd. \(\square\)

The sum of any pair of even integers is even.

Let \(m, n \in \mathbb{Z}\) be even. Then \(m = 2k\) and \(n = 2\ell\) for some \(k, \ell \in \mathbb{Z}\). Hence \(m + n = 2k + 2\ell = 2(k + \ell)\), which is even. \(\square\)

\(\forall n \in \mathbb{Z}^+\), if \(n \geq 4\) then \(2n^2 - 5n + 2\) is composite.

Suppose \(n \in \mathbb{Z}^+\) and \(n \geq 4\). Factor \(2n^2 - 5n + 2 = (2n - 1)(n - 2)\). Since \(n \geq 4\), we have \(2n - 1 \geq 7\) and \(n - 2 \geq 2\), so both factors are integers greater than \(1\). Hence the expression is composite. \(\square\)

Disproof by counterexample

  • For all integers \(m\) and \(n\), if \(2m + n\) is odd then \(m\) and \(n\) are both odd. Counterexample \(m = 4, n = 5\): \(2(4) + 5 = 13\) is odd, but \(m\) is even. (Not \(m=3, n=5\) — there the conclusion holds; not \(m=3,n=6\) or \(m=4,n=6\) — there the hypothesis fails.)
  • \(\forall n \in \mathbb{Z}^{\geq 2}\), \(n\) is composite or \(n + 1\) is composite. Take \(n = 2\): both \(2\) and \(3\) are prime.
  • For each integer \(n \geq 2\), the product of the first \(n\) primes minus \(1\) is prime. Take \(n = 4\): \(2 \cdot 3 \cdot 5 \cdot 7 - 1 = 209 = 11 \cdot 19\), composite.
  • \(\forall x, y \in \mathbb{R}\), if \(y^2 > x^2\) then \(y > x\). Take \(x = 1, y = -2\): \(4 > 1\) but \(-2 \not> 1\).
  • \(\forall x \in \mathbb{R}, \lfloor x^2 \rfloor = \lfloor x \rfloor^2\). Take \(x = \frac{3}{2}\): \(\lfloor \frac{9}{4} \rfloor = 2\) but \(\lfloor \frac{3}{2} \rfloor^2 = 1\). (See modular-arithmetic for the floor function.)

Disproof by proving the negation

When the statement to disprove is existential, a single counterexample isn’t available — you must prove the universal negation.

\(\exists n \in \mathbb{Z}^+\) such that \(n^2 + 5n + 6\) is prime. This is false; the negation is \(\forall n \in \mathbb{Z}^+\), \(n^2 + 5n + 6\) is composite.

Let \(n \in \mathbb{Z}^+\). Now \(n^2 + 5n + 6 = (n+2)(n+3)\). Since \(n \geq 1\) we have \(n + 2 > 1\) and \(n + 3 > 1\), so \(n^2 + 5n + 6\) is composite. \(\square\)

(Because \(n \geq 1\) forces \(n^2 + 5n + 6 \geq 2\), “not prime” here does mean “composite”.)

There is an even integer \(n\) such that \(5n - 4\) is prime. False; prove \(\forall n \in \mathbb{Z}^{\text{even}}\), \(5n - 4\) is not prime.

Suppose \(n\) is even, so \(n = 2k\) for some \(k \in \mathbb{Z}\). Then \(5n - 4 = 10k - 4 = 2(5k - 2)\), which is even. The only even prime is \(2\); if \(2(5k-2) = 2\) then \(k = \frac{3}{5} \notin \mathbb{Z}\). Therefore \(5n - 4\) is not prime. \(\square\)

Proof by contradiction

\(\forall m, n \in \mathbb{Z}\), if \(mn\) is even, then \(m\) is even or \(n\) is even.

Suppose \(\exists m, n \in \mathbb{Z}\) such that \(mn\) is even and both \(m\) and \(n\) are odd. Since \(m\) is odd, \(m = 2k+1\) for some \(k \in \mathbb{Z}\); since \(n\) is odd, \(n = 2\ell + 1\) for some \(\ell \in \mathbb{Z}\) — a new variable, since \(m\) and \(n\) need not be equal. Now \[mn = (2k+1)(2\ell+1) = 4k\ell + 2k + 2\ell + 1 = 2(2k\ell + k + \ell) + 1.\] Since \(k, \ell \in \mathbb{Z}\), \(2k\ell + k + \ell \in \mathbb{Z}\), so \(mn\) is odd. It is impossible for \(mn\) to be both even and odd, a contradiction. Therefore the original statement is true. \(\square\)

\(\forall n \in \mathbb{Z}\), if \(3n^3 - 2\) is odd, then \(n\) is odd.

Suppose the statement is false: \(\exists n \in \mathbb{Z}\) such that \(3n^3 - 2\) is odd and \(n\) is even. Since \(n\) is even, \(n = 2a\) for some \(a \in \mathbb{Z}\). Then \[3n^3 - 2 = 3(2a)^3 - 2 = 24a^3 - 2 = 2(12a^3 - 1).\] Since \(a \in \mathbb{Z}\), \(12a^3 - 1 \in \mathbb{Z}\), so \(3n^3 - 2\) is even — a contradiction. Therefore the original statement is true. \(\square\)

\(\forall m, n \in \mathbb{Z}\), if \(m + n\) is even, then \(m\) and \(n\) are both even or both odd.

Suppose the statement is false. Then \(\exists m, n \in \mathbb{Z}\) such that \(m + n\) is even but exactly one of \(m, n\) is even. Without loss of generality, assume \(m\) is even and \(n\) is odd — the other case (\(m\) odd, \(n\) even) follows because \(m + n = n + m\). Thus \(m = 2a\) for some \(a \in \mathbb{Z}\) and \(n = 2b + 1\) for some \(b \in \mathbb{Z}\). Now \(m + n = 2a + 2b + 1 = 2(a + b) + 1\), and \(a + b \in \mathbb{Z}\), so \(m + n\) is odd. This contradicts the assumption that \(m + n\) is even. Therefore the original statement is true. \(\square\)

For integers \(a\) and \(b\), if \(6a + 3b\) is odd, then \(b\) is odd.

Suppose there exist integers \(a, b\) with \(6a + 3b\) odd and \(b\) even. Then \(b = 2k\) for some \(k \in \mathbb{Z}\), so \(6a + 3b = 6a + 6k = 2(3a + 3k)\), which is even — contradicting that \(6a + 3b\) is odd. \(\square\)

For all positive integers \(x\) and \(y\), \(x^2 - y^2 \neq 1\).

Suppose \(x, y \in \mathbb{Z}^+\) with \(x^2 - y^2 = 1\). Factoring, \(1 = (x-y)(x+y)\). Since \(x, y\) are positive integers we must have \(x - y = 1\) and \(x + y = 1\), so \(2x = 2\) and \(x = 1\); then \(y = 0\), which is not a positive integer — a contradiction. \(\square\)

Proof by contraposition

\(\forall r \in \mathbb{R}\), if \(r^2\) is irrational, then \(r\) is irrational.

We prove the contrapositive: \(\forall r \in \mathbb{R}\), if \(r \in \mathbb{Q}\) then \(r^2 \in \mathbb{Q}\). Suppose \(r \in \mathbb{R}\) with \(r \in \mathbb{Q}\). Then \(r = \frac{a}{b}\) for some \(a, b \in \mathbb{Z}\) with \(b \neq 0\). Now \(r^2 = \frac{a^2}{b^2}\). Since \(a, b \in \mathbb{Z}\) we have \(a^2, b^2 \in \mathbb{Z}\), and \(b^2 \neq 0\) since \(b \neq 0\). So \(r^2 \in \mathbb{Q}\). \(\square\)

\(\forall m, n \in \mathbb{Z}\), if \(mn\) is odd, then \(m\) and \(n\) are both odd.

We prove the contrapositive: \(\forall m, n \in \mathbb{Z}\), if at least one of \(m, n\) is even then \(mn\) is even. Without loss of generality, assume \(m\) is even, so \(m = 2k\) for some \(k \in \mathbb{Z}\). Now \(mn = (2k)n = 2(kn)\), and \(kn \in \mathbb{Z}\), so \(mn\) is even. \(\square\)

For all integers \(a\) and \(b\), if \((ab)^2\) is odd then \(a\) is odd and \(b\) is odd.

Contrapositive: if \(a\) is even or \(b\) is even, then \((ab)^2\) is even. WLOG suppose \(a\) is even, so \(a = 2k\) for some \(k \in \mathbb{Z}\). Then \(ab = 2kb\) and \((ab)^2 = 4k^2b^2 = 2(2k^2b^2)\). Since \(k, b \in \mathbb{Z}\), \(2k^2b^2 \in \mathbb{Z}\), so \((ab)^2\) is even. \(\square\)

For all integers \(x\) and \(y\), if \(x^2(y^2 - 2y)\) is odd then \(x\) and \(y\) are odd.

Contrapositive: if \(x\) is even or \(y\) is even, then \(x^2(y^2 - 2y)\) is even. Here the two cases need separate treatment.

Case 1: \(x\) even. \(x = 2k\) for some \(k \in \mathbb{Z}\), so \(x^2(y^2-2y) = 4k^2(y^2-2y) = 2\big(2k^2(y^2-2y)\big)\), and \(2k^2(y^2-2y) \in \mathbb{Z}\).

Case 2: \(y\) even. \(y = 2\ell\) for some \(\ell \in \mathbb{Z}\), so \(x^2(y^2-2y) = x^2\big(4\ell^2 - 4\ell\big) = 2\big(x^2(2\ell^2 - 2\ell)\big)\), and \(x^2(2\ell^2-2\ell) \in \mathbb{Z}\).

In both cases \(x^2(y^2-2y)\) is even. \(\square\)

For any integer \(n\), \(n^2\) is odd if and only if \(n\) is odd. A biconditional needs both directions, and they want different methods:

(\(\Leftarrow\), direct) Suppose \(n\) is odd, so \(n = 2k+1\). Then \(n^2 = 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1\) is odd.

(\(\Rightarrow\), contraposition) The contrapositive is: if \(n\) is even then \(n^2\) is even. Suppose \(n = 2k\). Then \(n^2 = 4k^2 = 2(2k^2)\) is even. \(\square\)

This lemma — for all integers \(n\), if \(n^2\) is even then \(n\) is even — is used in the proof that \(\sqrt{2}\) is irrational (see rational-and-irrational-numbers).

Common proof errors

The course explicitly tests spotting these:

  • Arguing from examples — a single example never establishes a universal statement. (“\(\forall x \in \mathbb{R}, \exists y \in \mathbb{R}\) s.t. \(xy = 1\). Proof: if \(x = 10\) take \(y = \frac{1}{10}\).” Not a proof — and the statement is false, since \(x = 0\) has no such \(y\).)
  • Using the same variable to mean two different things — “let \(m = 2k+1\) and \(n = 2k\)” only proves the case \(m = n + 1\). Use a fresh variable for each quantity.
  • Assuming what is to be proved — starting from “\(4a + 2b = 2\ell\)” and manipulating both sides assumes the conclusion. Derive the conclusion; don’t begin with it.
  • Jumping to the conclusion — writing down the definition you need to satisfy and then asserting it holds, without exhibiting the witnesses.
  • Forgetting to state assumptions, or forgetting to say where a variable lives (“for some \(r, s \in \mathbb{Z}\)”).
  • Confusion between what is known and what is to be shown.
  • Use of the word any instead of some, and misuse of the word if.

See also