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
`(a)/(b)`, where
`\text{gcd}(a,b)=1`

`\sqrt6=(a)/(b)\\\\ 6=(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.