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Perhaps I am missing something really obvious, but what I do know is that for some $p,q\in\mathbb{Z}$ such that $r=p/q$ with $q\neq0$, we have that $r^2=p^2/q^2$.

Furthermore, the only way $p^2/q^2$ can be an integer is if $q^2 \mid p^2$; thus, there exists $k\in\mathbb{Z}$ such that $p^2/q^2=k$. If we prove that such $k$ is a perfect square (except when $r=0$, which is a trivial case) then we are finished.

Is there another way to prove this or is it possible to prove that $k$ is in fact a perfect square?

Manuel Osuna
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  • Yes, you can prove that $k$ is a perfect square, but you’ll want to assume that $p$ and $q$ are relatively prime, i.e., that $p/q$ is in lowest terms. – Brian M. Scott May 06 '20 at 22:38
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    Start with $p,q$ coprime, then $p^2=kq^2$, then $p^2 |k$, so $k=p^2k'$, and therefore, $1=k'q^2$, where both $k' $ and $q^2$ are integers. Since both $k'$ and $q^2$ are integers, then the only possibility is $q^2=1$ and $k'=1$, that is $q=1$ or $-1$. – Locally unskillful May 06 '20 at 22:41
  • That is what I was missing! I feel silly, thanks a lot! – Manuel Osuna May 06 '20 at 22:50
  • @Manuel The argument is the comment is either circular or incomplete (you need to justify the unfounded claim that $,p^2\mid k),$ See the dupe for rigorous proofs. – Bill Dubuque May 07 '20 at 00:08
  • @Gone It certainly isn't circular but I do agree it is incomplete, although it is easy to follow what has to be done afterwards. We have that for $p,q\in\mathbb{N}$ coprime then $p^2$ and $q^2$ are also coprime, from that everything else is quite clear. – Manuel Osuna May 07 '20 at 03:01
  • @Manual The problem is that without justifying the inference we can't know if the (crucial!) missing piece is correct or not (It is often incorrect, esp. by beginners). Even your elaboration is incomplete (are you implicitly using FTA or Euclid's Lemma or some equivalent?) There are many errors made here (some even believe the unjustified inferences are "obvious" but don't know how to give a rigorous proof when asked to elaborate). – Bill Dubuque May 07 '20 at 03:51
  • @Gone I somewhat understand what you are trying to say. We need to prove first that any rational can be expressed as a unique quotient of coprimes and my previous elaboration can be proven using FTA. I know a lot of steps are missing for giving a rigorous proof, I am not denying that but all of those results should be already deduced by someone who is tackling such problems. – Manuel Osuna May 07 '20 at 06:23
  • @Manuel Knowing the building blocks (theorems) needed for the proof is not enough. Rather one needs to communicate how to use them to construct a complete, rigorous proof convincing to your peers.The building blocks (basic theorems) are the same for almost all divisibility results. The ingenuity lies in how to assemble them. – Bill Dubuque May 07 '20 at 15:44
  • See also these remarks about intuition vs proof, including a famous remark by Gauss on such. – Bill Dubuque May 07 '20 at 15:48

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