A-level · Proof

Proof by exhaustionExam-style questions

Proving a statement by splitting it into a finite, complete set of cases and checking every one — either a short list of values, or all integers split by parity or remainder. Why the cases must leave no gaps, and why exhaustion only works when the cases really are finite.

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6 questions · A-level level · mark scheme and worked solution on each · no sign-up to try

DifficultyAll Level 2 Level 3 Level 4

Ayesha is using proof by exhaustion to show that, for every integer

n

2n^3 + n

is a multiple of 3

Her proof so far is shown in the box below.

Ayesha's proof
Every integer $n$ can be written in one of the forms $3k$ , $3k + 1$ or $3k + 2$ , where $k$ is an integer.
When $n = 3k + 1$ : $2n^3 + n = 2(27k^3 + 27k^2 + 9k + 1) + 3k + 1 = 54k^3 + 54k^2 + 21k + 3 = 3(18k^3 + 18k^2 + 7k + 1)$ , which is a multiple of 3

Complete Ayesha's proof.

(4 marks)

Level 3 · 4 marksTry it

2
In this question you must show detailed reasoning.

The integer $m$ is not a multiple of 3

Prove by exhaustion, using algebra, that

$2m^2 + 1$

is a multiple of 3

(4 marks)
Level 3 · 4 marksTry it
3
Prove that $n^5 - n$ is divisible by $5$ for every integer $n$ , by exhausting the possible remainders of $n$ on division by $5$ .
Level 3 · 4 marksTry it