GCSE Maths revision · Higher: number and algebra

Algebraic proof

Higher only

The idea

A proof shows something is true for every number, so it uses letters, not examples. For an integer $n$ , an even number is $2n$ , an odd number is $2n + 1$ and consecutive numbers are $n$ and $n + 1$ .

Write the numbers in general form.
Expand and simplify.
Rewrite the result to show the property, for example $2(\dots)$ for even, and say so.

Watch out Checking a few numbers is not a proof.

See it worked

Worked exampleNon-calculator3 marks

$n$ is an integer.

Prove that the sum of any three consecutive even numbers is a multiple of 6.

Write three consecutive even numbers in terms of $n$ . $2n, \quad 2n + 2, \quad 2n + 4$
Add. $2n + (2n + 2) + (2n + 4) = 6n + 6$
Factorise. $6n + 6 = 6(n + 1)$
$n + 1$ is an integer, so the sum is a multiple of 6.

Answer $2n + (2n + 2) + (2n + 4) = 6n + 6 = 6(n + 1)$

Your turn

Question 1Non-calculator4 marks

Kai says, " $n^2 + n + 1$ is a prime number for every positive integer $n$ ."

(a) Show that Kai is wrong.

(b) Prove that $n^2 + n$ is always even for every positive integer $n$ .

Write your answer, then show the working

(a) Try values of $n$ : $n = 1$ gives 3, $n = 2$ gives 7, $n = 3$ gives 13, $n = 4$ gives 21.
$21 = 3 \times 7$ is not prime, so Kai is wrong.
(b) Factorise. $n^2 + n = n(n + 1)$
$n$ and $n + 1$ are consecutive integers, so one of them is even. A product with an even factor is even.

Answer (a) When $n = 4$ , $16 + 4 + 1 = 21 = 3 \times 7$ , which is not prime.; (b) $n^2 + n = n(n + 1)$ , a product of consecutive integers, one of which is even.

Question 2Non-calculator4 marks

A pupil claims that the sum of any three consecutive integers is divisible by 6.

(a) Prove that the sum is always divisible by 3.

(b) Use a counterexample to disprove the pupil’s claim about 6.

Write your answer, then show the working

(a) Represent the consecutive integers around their middle integer. $(n-1)+n+(n+1)$
Collect terms into an integer multiple. $3n$
Writing the integers as n-1, n and n+1 gives sum 3n, a multiple of 3.
(b) Therefore 2+3+4=9, which is not divisible by 6.

Answer (a) Writing the integers as n-1, n and n+1 gives sum 3n, a multiple of 3.; (b) 2+3+4=9, which is not divisible by 6.

Question 3Non-calculator2 marks

Expand and simplify $(2n + 1)^2 - (2n - 1)^2$ .

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