Proof by contradiction

The idea

To prove a statement by contradiction, assume it is false and show that this leads to something impossible. Classic examples: $\sqrt{2}$ is irrational, and there are infinitely many primes.

State the opposite of what you want to prove.
Follow it with correct algebra until you reach a contradiction.
Conclude that the assumption was false, so the statement is true.

Watch out Write the negation carefully: the opposite of 'irrational' is 'can be written as $\frac{a}{b}$ in lowest terms'.

See it worked

Worked example

Outline the proof that $\sqrt{2}$ is irrational.

Assume $\sqrt{2} = \frac{a}{b}$ with $a$ and $b$ integers with no common factor.
Then $a^2 = 2b^2$ , so $a$ is even: $a = 2k$ , giving $b^2 = 2k^2$ , so $b$ is even too.
Both even contradicts 'no common factor', so $\sqrt{2}$ is irrational.

Answer The assumption gives $a$ and $b$ both even: a contradiction

Your turn

Question 1Non-calculator3 marks

In the proof that $\sqrt{2}$ is irrational, $a^2 = 2b^2$ and $a = 2k$ . Find $b^2$ in terms of $k$ .

Question 2Non-calculator3 marks

Prove by contradiction that there is no largest even number.

Write your answer, then show the working

Assume there is a largest even number, $N$ .
$N + 2$ is even and $N + 2 > N$ .
This contradicts $N$ being largest, so there is no largest even number.

Answer If $N$ were the largest even number, $N + 2$ would be a larger even number

Question 3Non-calculator3 marks

Prove by contradiction that if $n^2$ is even, then $n$ is even.

Write your answer, then show the working

Assume $n^2$ is even but $n$ is odd: $n = 2k + 1$ .
$n^2 = 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1$ , which is odd.
This contradicts $n^2$ being even, so $n$ is even.

Answer If $n$ were odd, $n^2 = 4k^2 + 4k + 1$ would be odd

In past papers

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