How does proof by contradiction work?

You assume the opposite of what you want to prove, deduce a statement that is impossible, and conclude that the opposite is false.

When should I use proof by contradiction?

When the statement says that something does not exist or is impossible, because assuming it exists gives you something concrete to work with.

What is the negation of a statement in a proof by contradiction?

It is the exact logical opposite. The negation of 'for all' is 'for some', so one failing case is enough to assume.

TMUA Proof by Contradiction

Proof by contradiction assumes the statement is false and shows that this leads to something impossible. A question may ask which assumption to start with, or which step finishes the proof.

Paper 2 topic. Specification codes: Prf1, Prf2. Last checked 12 November 2026.

The structure

Assume the negation of what you want to prove.
Deduce consequences by valid steps.
Reach a statement that is clearly false, or that contradicts the assumption.
Conclude that the assumption was wrong, so the original statement is true.

Getting the assumption right

The assumption is the exact negation of the claim. The negation of 'no integers satisfy $E$ ' is 'some integers satisfy $E$ '.

Useful tools

Parity (odd or even), divisibility and remainders after division are the usual sources of a contradiction at this level.

Worked examples

Example 1

Prove that there are no integers $m,n$ with $6m+9n=100$ .

Show the worked solution

Assume integers exist. Then $6m+9n=3(2m+3n)$ is a multiple of 3, but 100 is not. This is a contradiction, so no such integers exist.

Answer: No such integers exist

Example 2

Prove that there are no positive integers $a,b$ with $a^2-b^2=2$ .

Show the worked solution

Assume they exist. A square leaves remainder 0 or 1 when divided by 4, so $a^2-b^2$ leaves remainder 0, 1 or 3 (that is $-1$ ). It can never leave remainder 2, a contradiction.

Answer: No such integers exist

Example 3

A proof by contradiction of 'if $a+b+c>15$ then at least one of $a,b,c$ exceeds 5' starts by assuming what?

Show the worked solution

Assume $a+b+c>15$ and that none of $a,b,c$ exceeds 5. Then each is at most 5, so the sum is at most 15, a contradiction.

Answer: That $a,b,c$ are all at most 5 (while $a+b+c>15$ )

Common mistakes

Assuming the wrong negation, for example 'all are bigger than 5' instead of 'none is bigger than 5'.
Finding a false statement but not showing it contradicts the assumption or a known fact.
Using the conclusion you want in an early step.
Showing that an example works and calling it a proof.

Test yourself on this topic with our free practice questions.

Quick answers

Practise it under timed conditions

A free timed sample paper with a worked solution for every question. No sign-up. For two full papers and a weak-spot drill, see Full Access.

Free sample paper

TMUA Proof by Contradiction

The structure

Getting the assumption right

Useful tools

Worked examples

Common mistakes

Quick answers

Practise it under timed conditions

Related topics