Question

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

Answer 1

Lets prove $\sqrt{5}$ is irrational by contradiction.
Lets suppose that $\sqrt{5}$ is rational. It means that we have co-prime integers a and b $(b\ne 0)$ such that $\sqrt{5}=\frac{a}{b}$.
$\Rightarrow b\sqrt{5}=a$
Squaring both sides, we get
$5b^2=a^2$ (1)
It means that 5 is factor of $a^2$
Hence, 5 is also factor of a (2)
If, 5 is factor of a, it means that we can write a=5c for some integer c.
Substituting value of a in (1) , we get
$5b^2=25c^2$
$\Rightarrow b^2=5c^2$
It means that 5 is factor of $b^2$. Hence, 5 is also factor of b (3)
From (2) and (3), we can say that 5 is factor of both a and b. But, a and b are co-prime. Therefore, our assumption was wrong. √5 cannot be rational. Hence, it is irrational.