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

Solution:

Let us assume, to the contrary, that $\sqrt{5}$ is rational. So, we can find coprime integers a and $b(\neq 0)$ such that

$\sqrt{5}=\frac{a}{b}$

$\Rightarrow \sqrt{5} b=a$

Squaring on both sides, we get

$5 b^{2}=a^{2}$

Therefore, 5 divides $\mathrm{a}^{2}$.

Therefore, 5 , divides a

So, we can write a = 5c for some integer c.

Substituting for a, we get

$5 \mathrm{~b}^{2}=25 \mathrm{c}^{2}$

$\Rightarrow \mathrm{b}^{2}=5 \mathrm{c}^{2}$

This means that 5 divides $\mathrm{b}^{2}$, and so 5 divides $\mathrm{b}$.

Therefore, a and b have at least 5 as a common factor.

But this contradicts the fact that a and b have no common factor other than 1.

This contradiction arose because of our incorrect assumption that $\sqrt{5}$ is rational.

So, we conclude that $\sqrt{\mathbf{5}}$ is irrational.