A question says "prove that \sqrt{2} is irrational," or "show there are infinitely many primes," or "prove that no positive integer solutions exist for x^2 - 2y^2 = 0 with y > 0." Three different problems, one shared recipe. Every proof by contradiction runs the same three steps:

  1. Assume the negation of the conclusion. If the claim is P, assume \lnot P.
  2. Derive a contradiction. Chain lawful deductions until you reach something impossible — a statement and its negation both holding, a violated definition, a number that is both even and odd, a fraction that cannot be in lowest terms.
  3. Declare the original true. Since \lnot P led to an impossibility, \lnot P must be false. Therefore P is true.

That is the entire method. The technique lives in step 2 (deriving the contradiction), but steps 1 and 3 are the scaffolding you reach for the instant you recognise the right situation.

The recognition cue

The reflex fires on any of these stem patterns:

  • "Prove that \sqrt{n} is irrational" for any non-square integer n.
  • "Show that there is no largest prime."
  • "Prove no rational solution exists for..."
  • "Show that \log_2 3 is irrational."
  • "Prove that a specific property P holds for some object" — and every direct argument you try gets stuck.
  • "Disprove that..." — by exhibiting a contradiction when the claim is assumed true.

Each of these screams "direct proof is hard, contradiction is easy." The reason is simple: the negation \lnot P gives you a concrete, usable assumption ("suppose \sqrt{2} = p/q in lowest terms"), while a direct proof of P ("construct the irrationality") has nothing concrete to grab.

The logical spine

The equivalence that powers the method is this:

P \quad\Leftrightarrow\quad \lnot(\lnot P).

Double negation. To prove P, you equivalently disprove \lnot P. And to disprove \lnot P, you assume it and show it is logically impossible — i.e. leads to a contradiction. See Logic and Propositions for the truth-table justification.

Why this works as a proof: mathematics rests on consistency. If a set of assumptions entails a contradiction (a statement Q \land \lnot Q), one of those assumptions must be false. All your other assumptions are lawful theorems and definitions — they cannot be the false one. So the only suspect is the temporary assumption \lnot P. Therefore \lnot P is false, and P is true.

The walk-through — a concrete template

Claim. \sqrt{2} is irrational.

Step 1. Assume the negation: \sqrt{2} is rational. Then \sqrt{2} = p/q for integers p, q with no common factor (lowest terms), q \ne 0.

Step 2. Derive a contradiction.

  • Square both sides: 2 = p^2/q^2, so p^2 = 2q^2.
  • Therefore p^2 is even, so p is even (by the standard lemma). Write p = 2k.
  • Substitute: (2k)^2 = 2q^2 \Rightarrow 4k^2 = 2q^2 \Rightarrow q^2 = 2k^2.
  • Therefore q^2 is even, so q is even.
  • But we assumed p and q have no common factor, and now both are even — both share the factor 2. Contradiction.

Step 3. Declare: the assumption "\sqrt{2} rational" is false. Hence \sqrt{2} is irrational. \blacksquare

The interactive three-stage reflex

The reflex is exactly three moves. Drag the slider (or press Play) to step through the recipe — assume, contradict, declare.

The recognition patterns (where contradiction beats direct)

  • Irrationality claims. Direct construction of an irrational number is hard. Contradiction gives you a rational p/q to squeeze.
  • Non-existence claims. "There is no X with property Y" — assume there is, derive absurdity.
  • Infinitude claims. "There are infinitely many primes" (Euclid's proof): assume finitely many, multiply them, add one, derive a new prime not in the list.
  • Uniqueness claims. "Object X is the unique Y" — suppose two distinct X, X' both satisfy the property, derive X = X'.
  • Extremal claims blocked by well-ordering. "No counterexample can be smallest" — if one existed, a smaller one would exist, contradiction.

If the stem fits any of these templates, reach for contradiction before trying direct.

What counts as a "contradiction"

The endpoint of step 2 must be unambiguously impossible. The standard landing-zones are:

  • Explicit contradiction. A statement Q and its negation \lnot Q both appearing as derived facts.
  • Violated definition. You assumed p/q in lowest terms, but derived \gcd(p, q) > 1.
  • Violated theorem. You derived that an even integer equals an odd integer.
  • Impossible arithmetic. You derived 1 = 0, or n < n, or x^2 < 0 for real x.

Anything that no mathematical world can contain is a legal landing. The contradiction does not need to be flashy — just impossible.

Two worked setups for the exam

Problem 1. Prove: if n^2 is even, then n is even.

This is the contrapositive by contradiction, not the direct. Assume the conclusion is false: n is odd. Then n = 2k+1 for some integer k. Square: n^2 = 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1, which is odd. But we are given n^2 is even. Contradiction. Hence n is even. \blacksquare

Problem 2. Prove: there are infinitely many primes.

Assume finitely many primes: p_1, p_2, \ldots, p_n. Let N = p_1 p_2 \cdots p_n + 1. Then N is either prime itself or has a prime factor. Any prime factor of N leaves a remainder of 1 on division by each p_i, so cannot be any of the p_i. Therefore, there is a prime not in the list — contradicting "finitely many primes exhaust all of them." \blacksquare

The exam reflex

  • See the stem triggers — irrational, no-such, infinitely many, unique, no-smallest.
  • Write step 1 mechanically. "Suppose, for contradiction, that \lnot P..."
  • Aim for a contradiction. The target is often: a number both even and odd, a fraction not in lowest terms, a new object outside a supposedly exhaustive list.
  • Write step 3 mechanically. "This contradicts... therefore P is true."

Steps 1 and 3 are boilerplate. All your energy goes into step 2, and the trigger is pattern recognition on the stem.

Related: Logic and Propositions · Proof by Contradiction · Negation of If A Then B Is A and Not B · Proof by Contrapositive