Question

Prove that $\sqrt{5}$ is an irrational number.

Hint:

Proof by contradiction is often used to prove irrationality.

Answer 1

Assume, for contradiction, that $\sqrt{5}$ is rational.
Then, $\sqrt{5} = \dfrac{p}{q}$ where $p, q$ are co-prime integers and $q \ne 0$.
Squaring both sides: \[ 5 = \dfrac{p^2}{q^2} \Rightarrow p^2 = 5q^2 \] $\Rightarrow p^2$ is divisible by 5 ⇒ $p$ is divisible by 5 ⇒ let $p = 5k$
Then $p^2 = 25k^2 \Rightarrow 25k^2 = 5q^2 \Rightarrow q^2 = 5k^2$
So $q$ is also divisible by 5.
But this contradicts our assumption that $p$ and $q$ are co-prime.
Hence, $\sqrt{5$ is irrational.}