# help on proof writing: do I need to explicitly state this trivial assumption?

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 (meta.math.stackexchange.com/questions/107/…) on how to typeset your post –  user17762 Jul 11 '11 at 0:44
Try to use a more descriptive title. –  Tyler Jul 11 '11 at 0:45
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? –  t.b. Jul 11 '11 at 1:11
@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. –  André Nicolas Jul 11 '11 at 1:21
@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 $\mathrm{hcf}(p,q)=1$. –  Arturo Magidin Jul 13 '11 at 19:20
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There is in my opinion no need to remark at this stage that $q \ne 0$, since if we had $q=0$ the earlier expression $$x=\frac{p}{q}$$ would not have made sense.

So we conclude that if $x$ were rational, there would be integers $p$ and $q$ (with $q\ne 0$, but the rest of the proof does not use this, so there is no problem if it is omitted) and $\gcd(p,q)=1$ (this is necessary) such that $$p^2=5q^2.$$ If you now continue in the usual way, the fact that $q \ne 0$ will never need to be used, since what will provide the contradiction is the $\gcd$ condition.

Added: You were in a hurry to get to your question, so perhaps rushed through the first part. It should begin something like this.

Suppose to the contrary that there exist integers $p$ and $q$ such that $\sqrt{5}=\frac{p}{q}$.

Without loss of generality we may assume that the fraction $\frac{p}{q}$ is in lowest terms, that is, that $\gcd(p,q)=1$.

The $x$ stuff is harmless but unnecessary.

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I did not write this: ≠0. This was in my exam and I lost point. I would like to know is it still right if you do not write ≠0 because it was hcf(p,q)=1. –  John Jul 11 '11 at 1:35
@John: In principle you can have $\gcd(p,q)=1$ with $q=0$ (just pick $p=1$). About the loss of a point, my view is that after we have specified that $\gcd(p,q)=1$, whether or not $q=0$ is irrelevant to the argument. When we say that there exist integers $p$ and $q$ such that $\sqrt{5}=p/q$, of course $q\ne 0$. Now if instead we started by saying $q\sqrt{5}=p$, or something like that, then we would have to rule out $q=0$, since this equation does have the solution $p=q=0$. –  André Nicolas Jul 11 '11 at 1:54

If $p^2 = 5q^2$, then $p^2$ is divisible by $5$. Since $5$ is prime that implies $p$ is divisible by $5$; hence $p^2$ is divisible $25$. So $25\cdot(\text{something}) = 5q^2$. Canceling $5$ from both sides, we get $5\cdot(\text{something}) = q^2$. Then, by the same reasoning as above, $q$ is divisible by $5$. Now $p$ and $q$ are both divisible by $5$, so we didn't really have lowest terms.

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I know that. My question was not how it continues. –  John Jul 11 '11 at 1:23
Since the first time you do this is by an assumption, it's clear that $q$ cannot be zero by that. If you are in an introductory proof writing class, I would point it out at that point, because it can't hurt. Otherwise, it's clear enough to omit.