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I was working through a proof that came to a result concerning relatively prime integers (the full proof is fairly long, proving that $\mathbb{Z}_p\times \mathbb{Z}_q\cong \mathbb{Z}_{pq} \Leftrightarrow p,q $ coprime) giving

Let $p$ and $q$ coprime, $n=rp=tq$ gives $$\frac{r}{t}=\frac{q}{p}$$ And since $p,q$ coprime then $r=q$ and $t=p$. Then going on to say $n=pq....$ etc

I just can't get my head around this statement; we can just let $r=2, t=4$ and $q=1, p=2$ and this satisfies all the conditions and doesn't give $r=q$ and $t=p$.

EDIT: SOLVED. See comments below. Thanks

George1811
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  • Generally you need to assume $,r,t,$ coprime to make that deduction. This is the uniqueness of reduced fractions ("unique fractionization"), e.g. see here.. And unless you normalize signs you get only $,|p| = |t|,$ etc. – Bill Dubuque Apr 30 '15 at 14:38
  • Sorry just worked it out, forgot to include that $n$ is the order of the element $(a,b) \in \mathbb{Z}_p \times \mathbb{Z}_q$ and $a^{p}=e_1$ and $b^q=e_2$. Therefore $n$ is effectively the lowest common multiple of $p,q$. Hence as $p,q$ coprime we have $n=pq$ – George1811 Apr 30 '15 at 14:43

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