$n = (\frac{a}{b})^2$, where $a$ and $b$ have no common divisors. This yields
$nb^2 = a^2$
$ra^2b^2 = a^2$ (because $n = ra^2$)
I don't understand why $n$ is equal to $ra^2$.
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1Where is $r$ introduced? – AlexR Mar 06 '15 at 15:40
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For some basic information about writing math at this site see e.g. here, here, here and here. – AlexR Mar 06 '15 at 15:42
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1Duplicate of http://math.stackexchange.com/q/4467/131263. – barak manos Mar 06 '15 at 15:46
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Because $n$ must divide $a^2$ and this is a consequence of $(a,b)=1$ – marwalix Mar 06 '15 at 15:48
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Since n is an integer, so is $({a\over b})^2$. There are two possibilities:
1) ${a\over b}$ is an irrational number which is not possible.
2) ${a\over b}$ is a rational number. Therefore it is an integer because $({a\over b})^2$ is. Since ${a\over b}$ is an integer, there is some $q\in \mathbb Z$ such that $a=b.q$. Hence $n=q^2$.
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