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Prove by contradiction: For all prime $p$, $\sqrt p$ is irrational.

Hint: Use the following theorem: For every prime $p$ and all integers $a,b$ if $p\mid ab$, then $p\mid a$ or $p\mid b$.

I am currently here: Assume $\sqrt p$ is rational, then, $\sqrt p = \dfrac a b$ where a and b are integers. Then, $p = \dfrac{a^2}{b^2}$

Now I am stuck, I don't understand how to proceed from here nor how to use the theorem given as a hint.

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  • $\begingroup$ Clear of fractions. $\endgroup$ – Lubin Oct 5 '20 at 4:28
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    $\begingroup$ @Atlecx Have you seen the proof that $\sqrt2$ is irrational? The approach is the same, but the divisibility conditions are different. $\endgroup$ – Toby Mak Oct 5 '20 at 8:12
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We add one more point that assuming $\sqrt{p}$ to be rational we have that $a,b$ are coprime.

proceeding from your method $$b^2=\frac{a^2}{p}..(1)$$ or $p|a^2$ or $p|a$ (using the theorem given as Hint). Hence we can assume that $a=pk$ where $k$ is some positive integer.

Can you end it now ?

Hint:Show that using the substituition $a=kp$ in (1) that $p|b$ and hence $a,b$ are not coprime.

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Hint: You can assume it further by making $\sqrt p = \dfrac a b$ where $\dfrac a b$ is the reduced form fraction. Then prove by contradiction

Additional hint: Prove that $\dfrac a b$ is also not in reduced form, which conflicts with the original assumption

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