I came across this question, and when it comes to proofs, I'm not very good.
Prove by contradiction that $\sqrt{7}$ is irrational. You may assume that for any integer $x$ and any prime number $p$ we have ($p$ divides $x$)$ \Leftrightarrow $($p$ divides $x^2$).
Assume that we can write $\sqrt{7}$ as a ratio of two numbers, $p$ and $q$. Assuming that this is in simplest form, we have $p/q=\sqrt{7}$. This can be squared to give: $$\begin{align*}\frac{p}q &= \sqrt{7}\\ \left(\frac{p}q\right)^2&=\frac{p^2}{q^2}=7\\p^2&=7q^2 \end{align*}$$ However, this would imply that one of the factors of $p$ would square to get $7$, which is clearly nonsense as only integer values are allowed in the prime factorization. Thus, this is a contradiction, and the proof is false. Thus, $\sqrt{7}$ is irrational.