Question 94 – 1.1 Proof by Contradiction - Questions

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

“if $n^2$ is a multiple of 5, then $n$ 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 $\in \mathbb{Z}^+$ , such that $n^2$ is a multiple of 5, and $n$ is NOT a multiple of 5.

Case 1: Let $n = 5 k + 1$ for some integer $k$ . $n^2 = (5k + 1)^2 = 25k^2 + 10k + 1 = 5(5k^2 + 2k) + 1$ which is not a multiple of 5.

Case 2: Let $n = 5 k + 2$ for some integer $k$ . $n^2 = (5k + 2)^2 = 25k^2 + 20k + 4 = 5(5k^2 + 4k) + 4$ which is not a multiple of 5.

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

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

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

“if $n^2$ is a multiple of 5, then $n$ 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 $\in \mathbb{Z}^+$ , such that $n^2$ is a multiple of 5, and $n$ is NOT a multiple of 5.

Case 1: Let $n = 5 k + 1$ for some integer $k$ . $n^2 = (5k + 1)^2 = 25k^2 + 10k + 1 = 5(5k^2 + 2k) + 1$ which is not a multiple of 5.

Case 2: Let $n = 5 k + 2$ for some integer $k$ . $n^2 = (5k + 2)^2 = 25k^2 + 20k + 4 = 5(5k^2 + 4k) + 4$ which is not a multiple of 5.

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

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

1.1 Proof by Contradiction Questions