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1 Pure Mathematics

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Question 92

A student was asked to prove, for n∈Z+n \in \mathbb{Z}^+n∈Z+, that

“if n2n^2n2 is a multiple of 5, then nnn must be a multiple of 5”

The start of the student's proof by contradiction is shown in the box below.

Assumption: There exists an integer n∈Z+n \in \mathbb{Z}^+n∈Z+, such that n2n^2n2 is a multiple of 5, and nnn is NOT a multiple of 5.

Case 1: Let n=5k+1n = 5k + 1n=5k+1 for some integer kkk.

n2=(5k+1)2=25k2+10k+1=5(5k2+2k)+1 n^2 = (5k + 1)^2 = 25k^2 + 10k + 1 = 5(5k^2 + 2k) + 1 n2=(5k+1)2=25k2+10k+1=5(5k2+2k)+1

which is not a multiple of 5.

Case 2: Let n=5k+2n = 5k + 2n=5k+2 for some integer kkk.

n2=(5k+2)2=25k2+20k+4=5(5k2+4k)+4 n^2 = (5k + 2)^2 = 25k^2 + 20k + 4 = 5(5k^2 + 4k) + 4 n2=(5k+2)2=25k2+20k+4=5(5k2+4k)+4

which is not a multiple of 5.

a.

Show the calculations and statements required to complete this part of the proof.

[3]
b.

Hence prove, by contradiction, that 5\sqrt{5}5​ is an irrational number.

[4]

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