Question

Use a proof by contradiction to prove that underroot 3 is irrational.

Answer 1

Let assume that
`\sqrt3` is rational. Therefore we can express it as
`(a)/(b)` where
`a,b\in \mathbb{Z}` and
`\text{gcd}(a,b)=1`.

`(a)/(b)=\sqrt3\\(a^2)/(b^2)=3\\a^2=3b^2`

It means that
`3|a^2` and so also
`3|a`.

Therefore
`a=3k` where
`k\in\mathbb{Z}`.

`(3k)^2=3b^2\\9k^2=3b^2\\b^2=3k^2`

It means that
`3|b^2` and so also
`3|b`.

If both
`a` and
`b` are divisible by 3, then it contradicts our initial assumption that
`\text{gcd}(a,b)=1`. Therefore
`\sqrt3` must be an irrational number.