Proof by Contradiction

To prove a statement, assume it is false and show that this leads to something impossible. The assumption must have been wrong, so the statement is true. The two famous examples are that $\sqrt2$ is irrational and that there are infinitely many primes.

Edexcel YMA01 WMA14Edexcel 9MA0 Paper 2AQA 7357 Paper 1AQA 7357 Paper 2AQA 7357 Paper 3≈ 35 minRevision 5 · updated 2026-10-27

Section 1 of 14

Learning objectives

Write the correct opposite (negation) of a statement as the starting assumption.
Reach and clearly state a contradiction.
Prove that $\sqrt2$ is irrational and that there are infinitely many primes.
Use contradiction for simple results about integers.

Pause and think

Suppose there were a largest even number, $N$ . What can you say about $N+2$ ?

Section 2 of 14

Core idea

Start hereProving by imagining the opposite

A detective says: "Suppose the butler did it. Then he was in the kitchen at 9. But he was seen in town at 9. Impossible, so the butler is innocent." Proof by contradiction assumes the opposite of what you want and shows it leads to nonsense.

The big idea

To prove $P$ : assume "not $P$ ", deduce something impossible, conclude $P$ .

1. √2 is irrational

The classic proof

$\sqrt2=\frac ab,\ \text{lowest terms}$
Assume rational: integers with no common factor.
$2b^2=a^2$
Square, multiply by $b^2$ .
$a^2\text{ even}\Rightarrow a\text{ even}:\ a=2k$
An odd number squared is odd.
$2b^2=4k^2\Rightarrow b^2=2k^2$
Substitute.
$b\text{ even}$
Same argument.

a

b

both even contradicts lowest terms. So

\sqrt2

is irrational ∎.

Worked exampleIf $n^2$ is even, $n$ is even

Prove (used above).

Goal: Proof.
Strategy: Assume $n$ odd.

$n=2k+1$
Assume $n$ odd.
$n^2=4k^2+4k+1=2(2k^2+2k)+1$
Odd.
$\text{contradicts }n^2\text{ even}$
So $n$ is even ∎.

AnswerProved.

✓ Check

7^2=49

odd ✓.

2. Infinitely many primes

Infinitely many primes: no list can be complete

Euclid's argument works for any finite list, so the list of primes can never end.

Euclid

$p_1,p_2,\ldots,p_n$
Assume these are all the primes.
$N=p_1p_2\cdots p_n+1$
Build a new number.
$N\div p_i\text{ leaves remainder }1$
So no $p_i$ divides $N$ .
$N\text{ has a prime factor not in the list}$
Every integer > 1 has a prime factor.

Contradiction: the list was not complete ∎.

Common misconceptionN need not be prime

2\cdot3\cdot5\cdot7\cdot11\cdot13+1=30031=59\times509

. Its factors are just new primes.

Pause and think

Write the assumption for proving "there is no largest even number".

Summary

Assume the opposite.
Reach an impossibility.
Conclude.

Contradiction: Two incompatible statements.
Irrational: Not a fraction of integers.

HistoryThe oldest proofs we still use. Euclid's proof that there are infinitely many primes (Book IX of the Elements, about

300

BC) is still taught almost word for word. The irrationality of

\sqrt2

was known to the Pythagoreans even earlier.

Fun fact"A gambit far finer than chess". The mathematician G. H. Hardy called proof by contradiction "one of a mathematician's finest weapons", far finer than a chess gambit: a chess player may sacrifice a piece, but a mathematician offers the whole game.

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