Proof Techniques — Direct, Counterexample, Contradiction, Contraposition

definitions

discrete-math

proof

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.