Question

is s the square root 6 rational

Answer 1

Proof by contradiction.

Let assume $\sqrt6$ is rational. Therefore, it can be expressed as a fraction $\dfrac{a}{b}$, where $\text{gcd}(a,b)=1$

$\sqrt6=\dfrac{a}{b}\\\\ 6=\dfrac{a^2}{b^2}\\\\ a^2=6b^2$

This means that $a^2$ must be even, and therefore also $a$ must be even.

So, $a=2k, k\in\mathbb{Z}$

$(2k)^2=6b^2\\\\ 4k^2=6b^2\\\\ 2k^2=3b^2$

This means that $3b^2$ must be even. The only way for this to be even is when $b^2$ is even and therefore also $b$ is even.

But if $a$ and $b$ were even, then earlier assumption that $\text{gcd}(a,b)=1$ would be false. Therefore $\sqrt6$ is irrational.