Question

Prove that $\sqrt{7}$ is an irrational number.

Answer 1

Let us assume that $\sqrt{7}$ is rational. Then, there exist co-prime positive integers $a$ and $b$ such that

$\sqrt{7}=\frac{a}{b}$

$⟹a=b\sqrt{7}$

Squaring on both sides, we get

$a^2=7b^2$

Therefore, $a^2$ is divisible by $7$ and hence, $a$ is also divisible by$7$

so, we can write $a=7p,$ for some integer $p.$

Substituting for $a$, we get $49p^2=7b^2⟹b^2=7p^2.$

This means, $b^2$ is also divisible by $7$ and so, $b$ is also divisible by $7$.

Therefore, $a$ and $b$ have at least one common factor, i.e., $7$.

But, this contradicts the fact that $a$ and $b$ are co-prime.

Thus, our supposition is wrong.

Hence, $\sqrt{7}$ is irrational.