Question

How to prove that√5 is irrational number?

Answer 1

Proof by contradiction.

Let's assume that $\sqrt5$ is a rational number. Therefore it can be expressed as $\dfrac{a}{b}$ where $a,b\in\mathbb{Z}$ and $a$ and $b$ are coprime.

$\sqrt5=\dfrac{a}{b}\\\\5=\dfrac{a^2}{b^2}\\\\a^2=5b^2$

Since $a^2$ is divisible by 5, then also $a$ is divisible by 5.

We can write that $a=5c, c\in\mathbb{Z}$.

$(5c)^2=5b^2\\25c^2=5b^2\\b^2=5c^2$

Since $b^2$ is divisible by 5, then also $b$ is divisible by 5.

But if both $a$ and $b$ are divisible by 5, then it contradicts our earlier assumption that those two numbers are coprime. Therefore $\sqrt5$ is irrational number.