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I am trying to prove the following:

If p and q are distinct primes, then $\sqrt{pq}$ $\notin$ $\Bbb{Q}$.

Here is my proof thus far:

  1. Suppose towards a contradiction that if p and q are distinct primes, that $\sqrt{pq}$ $\in$ $\Bbb{Q}$. If so, there exists some m , n $\in$ $\Bbb{Z}$ such that $\sqrt{pq}$ = $\frac mn$.

  2. Squaring both sides, we see that pq = $\left(\frac{m^2}{n^2}\right)$ .

  3. Multiplying by $\left(\frac nm\right)$, we see that $\left(\frac{m}{n}\right)$ = $\left(\frac{npq}{m}\right)$.

  4. This implies that n|m and m|npq. But, since m, n share no common factors, n $\not\mid$ m.

  5. Hence we have reached a contradiction.

**My concern is step 4 of my proof. Does it follow that since n $\not\mid$ m , $\sqrt{pq}$ $\notin$ $\Bbb{Q}$ ?

Is this "enough" to render a contradiction? It feels fishy to me but I'm not sure. If anybody could help that would be greatly appreciated! I am trying to develop my intuition in regards to this.

Martin Sleziak
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jmoore00
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  • It's fishy all right. Who says $n\vert m$, for instance? – user7530 May 05 '17 at 04:47
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    Your step 4 feels a bit off... why do you directly know that $n\mid m$? I would instead have gone from step 2 by multiplying both sides by $n^2$, you get then $n^2pq=m^2$ and both sides are integers. Here you can then use properties of primes to say that $p\mid m$ and $q\mid m$. You can then continue as the classic greek proof for $\sqrt{2}$'s irrationality does. – JMoravitz May 05 '17 at 04:48
  • Why is it true 4? It is better to simply consider equality $n^2pq=m^2$ assuming $m$ and $n$ have no common factors. Then $p$ devides $m^2$ hence ... – Minz May 05 '17 at 04:51
  • see the answers to https://math.stackexchange.com/questions/1310014/what-is-the-most-rigorous-proof-of-the-irrationality-of-the-square-root-of-3 They contain a proof for the irrationality of the square root of any non-perfect square. – Mark Joshi May 05 '17 at 05:00
  • @JMoravitz : My textbook contains a lemma that says that if m,n,s,t are integers and (m\n) = (s\t) then m | s and n | t. – jmoore00 May 05 '17 at 05:12
  • @MarkJoshi Hmm. I do have a proposition I can use that says that if r is not a perfect square, the square root of r is irrational. could i then say that since p and q are distinct and prime, pq can't be a perfect square, so square root (pq) is irrational? i am worried that my professor will require me to prove that pq can't be a perfect square if p and q are distinct primes. – jmoore00 May 05 '17 at 05:15
  • you could invoke the uniqueness of prime decompositions... the square of an integer will have a prime to the power two which is not the case here. – Mark Joshi May 05 '17 at 05:35
  • Its easy to ptove that pq isn't a perfect square because a product of the relatively prime numbers is a perfect square only if each of the numbers is a perfect square. – kingW3 May 05 '17 at 05:59
  • As the only divisors of $pq$ are $1, p, q, pq$, you can see that none of them can be square root of $pq$ for different $p, q$. – enedil May 05 '17 at 14:00
  • @kingW3 Yes, I know that and we know that but I'm not sure my professor will let me say that I know that. Our book hasn't gone over a precise definition of a perfect square, so I am worried I will not be allowed to use it. – jmoore00 May 05 '17 at 15:27

2 Answers2

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You have $n^2pq=m^2$, but the exponents of $p$ and $q$ are odd.

JMP
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  • So? I'm not quite sure what you are getting at. – jmoore00 May 05 '17 at 05:16
  • we have a contradiction by this stage – JMP May 05 '17 at 05:17
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    @JonMarkPerry But the OP asks about a specific method of proof, and this answer ignores that method. – Bill Dubuque May 05 '17 at 14:32
  • Yes, but his specific method is flawed. This is a step in the right direction. Don't multiply by n/m since there's no guarantee the result is integral. – Jeremy West May 26 '17 at 01:49
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    @Jeremy The OP's method of proof is certainly not flawed. Rather, the flaw lies in your understanding of this method. Again, please ask for elaboration before making false claims based on misunderstandings. – Bill Dubuque May 26 '17 at 16:01
  • @Bill, I 100% understand all the answers posted as well as OP's question and your links. Please don't assume that people who disagree with you are ignorant. There's a comment to the question indicating a lemma that OP is making use of regarding the relationship between equality of fractions and divisibility. Whether it was there when I made this comment or not I don't know. It should be included in the question or else the proof is making some deductions that seem unwarranted, given what one is expected to have already proved when approaching a result like this (which is usually not much). – Jeremy West May 26 '17 at 18:09
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The method you use in step 4 is correct but, we need to assume at the start that wlog $m,n$ are coprime, i.e. $m/n$ is in lowest terms. See here or here for further details

However, this does not yet yield a contradiction. Rather, it yields that $n = 1,$ so if the radical is rational then it is an integer. It remains to finish the proof.

Note $ $ The links are to answers with complete elementary proofs, including John Conway's favorite simple proof. Though there I also append remarks on more advanced ways to conceptualize these elementary proofs, I emphasize that those proofs do not require any knowledge of these more advanced ideas.

Bill Dubuque
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  • I don't think even this saves the proof. Why does n divide m? That assumes m/n = npq/m are integers? Why is that? They were obtained by multiplying integers by n/m which ought to be non-integral. – Jeremy West May 26 '17 at 01:47
  • @Jeremy It is explained in the links I gave. You should ask for clarification before you downvote. – Bill Dubuque May 26 '17 at 04:04
  • I disagree. I can always remove the downvote if convinced otherwise. There's no reason a proof this basic couldn't be self-contained or should depend on significantly more advanced results. I would wager that most people with OP's question are unlikely to be in a position to understand either of your links. Hence, the reason I found your answer less helpful and downvoted. – Jeremy West May 26 '17 at 04:15
  • @Jeremy If you can be more precise about what you have difficulty following then I will be happy to elaborate. The proofs I gave in the links are about as elementary as can be. Though I do append notes on more advanced conceptual ways to view these proofs (e.g. via ideal theory), those tangential remarks can be completely ignored if they are beyond one's level of knowledge. – Bill Dubuque May 26 '17 at 15:17
  • OPs question is a fairly standard exercise early in a course on elementary number theory or intro to mathematical reasoning/proofs. Given his uncertainty about the proof, it would seem that is the level of his question. In that context, results depending on partial fractions, for instance, aren't likely to be relevant. I appreciate your concern for my understanding; I do, in fact, understand your links. I just don't think they are appropriate to the level of the question asked. – Jeremy West May 26 '17 at 18:17
  • @jeremy I'm afraid your are quite confused. There are no "results depending on partial fractions". It appears that you do not understand the proofs in the links. Again, I'm happy to elaborate but I need to know more about your level of knowledge and what you find confusing. – Bill Dubuque May 26 '17 at 18:22
  • Sorry, continued fractions. It's been a while since I looked at the links. I checked them, found them interesting, but both ultimately beyond the scope of OP's question. – Jeremy West May 26 '17 at 18:25
  • I'm not even making claims about what is required by those proofs, to be clear. I'm just saying that they aren't written in a way that is likely to be useful to OPs question. – Jeremy West May 26 '17 at 18:27
  • @jeremy PS This answer explains a bit further the lemma about reduced fractions. The Remark there is exactly what is implicitly used by the OP in step 4. This is a property of (ordinary) fractions known (implicitly) since grade school. As I explain there and elsewhere, it is intimately connected to unique factorization and closely related fundamental arithmetical properties. – Bill Dubuque May 26 '17 at 18:39
  • Let us continue this discussion in chat. – Jeremy West May 26 '17 at 18:41