If $r \in \mathbb{N}$ is not a perfect square, then $\sqrt{r}$ is irrational.
For reference, an integer $n$ is a perfect square if $n=m^2$ for some $m \in \mathbb{Z}$.
Any help proving this would be greatly appreciated.
If $r \in \mathbb{N}$ is not a perfect square, then $\sqrt{r}$ is irrational.
For reference, an integer $n$ is a perfect square if $n=m^2$ for some $m \in \mathbb{Z}$.
Any help proving this would be greatly appreciated.
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Hint: Prove by contradiction. Assume $\sqrt r$ is rational, then $\sqrt r=\frac pq$, for some coprime $p,q$. Then $\frac {p^2}{q^2}=r $, but...