Proving a simple claim regarding irrational numbers

Question

I am trying to prove the following claim: Let $ n \in \Bbb N$ and let $p$ be a prime number for which $p \nmid n$. Prove that $\sqrt {pn} \notin \Bbb Q$ . I was able to prove it by assuming by contradiction that $\sqrt {pn} \in \Bbb Q$ so there exists $a,b \in \Bbb Z$ such that$ \sqrt { pn}=\frac ab$ . after a few steps I get to the fact that $n=\frac{a^2}{pb^2}$. Obviously this shows that $n \notin \Bbb Z$, which is a contradiction to the fact that $n \in \Bbb N$. The problem is I am having isuues writing the formal proof that $n \notin \Bbb Z$. Any help will be welcomed.

Answer 1

It is not clear to me how $n=\frac{a^2}{p b^2}$ implies that $n$ is not an integer. Say for example that $b=1$ and $a=p$.

Let's go back to the equation $pn=\frac{a^2}{b^2}$. If we assume that $a$ and $b$ are coprime (which we can) then we have that $b=1$. So $pn=a^2$. Now consider the prime decomposition of $pn$ and of $a^2$, which must be equal. What is the power of $p$ on this decomposition? Can you derive a contradiction from this?

It is the same idea that appears on the standard proof that $\sqrt{2}$ is irrational.