I'm trying to prove the statement: Let $x\in\mathbb R$. If $x$ is rational then $\sqrt x$ is irrational. I know that a number $x$ is rational if we can write it as $x=\frac{p}{q}$ for some integers $p, q$, where $q\neq 0$ and we say that $x$ is irrational if it is not rational. Here's what I have so far:
Let $x$ be a real number and let us assume that it is rational. By the definition of rational numbers, there exist integers $p,q$ with $q\neq 0$ such that $x=\frac{p}{q}$. Next, we will take the square root of $x$ which gets us $\sqrt x=\sqrt \frac{p}{q}$.
However, I'm lost on where to go from here. As I was working through the proof, I kind of realized that the statement might be false, so I was trying to think of counterexamples and the one I thought of was let $x=4$, which we can represent as $4=\frac{8}{2}$ and the $\sqrt 4=2$, which is rational. Am I going in the right direction? Any feedback is appreciated.
The square of any rational number, is a rational number whose square root is rational. So what you are trying to prove is false.
However it is true that a rational number $\frac pq$(in lowest terms) has an irrational square root, if and only if at least one of $p$ or $q$ is not a perfect square.