Question

Prove that $\sqrt{6}$ is an irrational number

Answer 1

The following proof is a proof by contradiction.

Let us assume that $\sqrt{6}$ is rational number.

Then it can be represented as fraction of two integers.

Let the lowest terms representation be: $\sqrt{6}=\frac{a}{b}$ where $b\ne 0$

Note that this representation is in lowest terms and hence, $a$ and $b$ have no common factors

$a^2=6b^2$

From above $a^2$ is even. If $a^2$ is even, then $a$ should also be even.

$⟹a=2c$

$4c^2=6b^2$

$2c^2=3b^2$

From above $3b^2$ is even. If $3b^2$ is even, then $b^2$ should also be even and again $b$ is even.

But $a$ and $b$ were in lowest form and both cannot be even. Hence, assumption was wrong and hence, $\sqrt{6}$ is an irrational number.