Irrationality of $\sqrt{d}$

Question

Let $d$ be a square free integer greater than 1. That is, $d$ is not divisible by the square of any prime $p$. Prove that $\sqrt{d}$ is irrational. I know this question is very simple, so I’ve even given my own answer to it.

Answer 1

Personally I prefer doing this way : assume $\sqrt{d}=\frac{p}{q}$ with $p$ and $q$ coprime. Then $q^2d=p^2$. Since $q|q^2d$ it divides $p^2$ and then $p$ since $p$ and $q$ are coprime. That means that $q=1$. This concludes that unless d is a perfect square, $\sqrt{d}$ is irrational.

Answer 2

Here is a simpler argument, which works with the more general hypothesis that $d$ is not a square.

If $a^2=b^2d$, then choose a prime factor $p$ such that $p$ divides $d$ but $p^2$ does not divide $d$. Such prime exists because $d$ is not a square. Now consider the power of $p$ in the factorization of both sides of $a^2=b^2d$. On the left you get an even number but on the right you get an odd number. This is a contradiction.