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Prove that $\sqrt{5}$ is an irrational number. Part of the answer: Let $x^2=5$ and $x=p/q$ where $p$ and $q$ are integer numbers and $\operatorname{hcf}(p,q)=1$. $$\begin{align*} x^2&=5\\\ \left(\frac{p}{q}\right)^2&=5\\\ \frac{p^2}{q^2}&=5 & \cdot q^2 \quad \leftarrow \text{Do I have to write this: }q \neq 0\text{? I mean because it was hcf(p,q)=1}.\\\ p^2&=5q^2 \end{align*}$$ I know how it continues. Thank you for your answer.

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    Could you kindly type it up so that your question is readable? You can look up here (http://meta.math.stackexchange.com/questions/107/faq-for-math-stackexchange/117#117) on how to typeset your post2011-07-11
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    Try to use a more descriptive title.2011-07-11
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    What is going on here? Two downvotes, one for the question, one for a helpful answer and an upvote for an answer that doesn't address the question?2011-07-11
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    @Theo: (I remembered.) Thanks. In view of the clear assertion that he knew how to go on, I thought there was no sense in writing out the rest. (But I will add a couple of style remarks.) Anyway, I prefer the descent version of the same argument.2011-07-11
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    **@all editors:** please note that this question is **not** about the irrationality of $\sqrt{5}$!2011-07-11
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    @John: Yes, $\mathrm{hcf}(p,q)=1$ does not imply $q\neq 0$; note that $\mathrm{hcf}(p,0) = |p|$, so you could have $p=1$, $q=0$ and still have $\mathrm{hcf}(p,q)=1$.2011-07-13

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