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.
Tell me more
×
Mathematics Stack Exchange is a question and answer site for
people studying math at any level and professionals in related fields. It's 100% free, no registration required.
|
|
|||||||||||||||||
|
|
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. |
|||||||
|
|
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. |
|||||
|
|
In general, you need to assert that numbers are not zero when you divide by them. If you wait until you multiply and cancel, it's "too late," so to speak. 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. It's good that you're thinking about it, though, because asserting that something is not zero before dividing is a proof step that is often forgotten. |
|||
|
|
