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QUESTION

Prove that, for any prime $p,\ \sqrt{p}$ is irrational.



ANSWER

Suppose $\sqrt{p}$ is rational, say $\sqrt{p}=\frac{a}{b}$ where
$a,b\in Z$, and $\frac{a}{b}$ is cancelled to it's lowest terms,
so that gcd$(a,b)=1$. We have $b\sqrt{p}=a$, so, on squaring,
$b^2p=a^2$. Thus $p|a^2$, so by question 2, $p^2|a^2$, say
$a^2=p^2c$. Thus $b^2p=p^2c$, giving $b^2=pc$. Hence $p|b^2$, and
so by question 2, $p|b$. Thus $p|a$ and $p|b$, contrary to
gcd$(a,b)=1$. Thus $\sqrt{p}$ is irrational, as claimed.




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