Question

Prove that $\sqrt 5$ is irrational.

Answer 1

Let $\sqrt 5$ is a rational number.
Therefore, we can find two integers $a, b\ (b ≠ 0)$ $a, b \ (b ≠ 0)$ such that $\sqrt 5=\dfrac{𝑎}{𝑏}$ Let $a$ and $b$ have a common factor other than $1$.

Then we can divide them by the common factor, and assume that $a$ and $b$ are co-prime.

$𝑎=\sqrt 5𝑏$
$⇒𝑎^2=5𝑏^2$

Therefore, $a^2$ is divisible by $5$ and it can be said that $a$ is divisible by $5$.
Let $a = 5k$, where $k$ is an integer
$(5𝑘)^2=5𝑏^2$
$⇒5𝑘^2=𝑏^2$
This means that $b^2$ is divisible by $5$ and hence, b is divisible by $5$.
This implies that a and b have $5$ as a common factor.
And this is a contradiction to the fact that a and b are co-prime.

Hence, $\sqrt 5$ cannot be expressed as $\dfrac{p}{q}$ or it can be said that $\sqrt 5$ is irrational.