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Suppose $n \in \Bbb N$ has a square root that isn't whole, prove $\sqrt n$ is irrational. The exercise wants me to prove it in the following steps: Suppose $\sqrt n$ is rational, then there exists a minimal natural number $p$ such that $p \sqrt n$ is a natural number. Now, observe the number $q=p \sqrt n -p \lfloor \sqrt n \rfloor$.

I'm not really sure what I can infer from $q$, I tried squaring $q$ but I couldn't infer anything either.

Kantig Shoter
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1 Answers1

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$q<p$, and $q\sqrt{n}\in \mathbb{N}$, but $p$ was the minimal, so you get a contradiction

sti9111
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  • You also need to point out that $q>0$ (otherwise, you have proved $\sqrt n$ is irrational for all $n$, even the ones that are squares), and you need to note that $q$ is a positive integer. – Gerry Myerson Mar 23 '23 at 21:38