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In the question How to prove: if $a,b \in \mathbb N$, then $a^{1/b}$ is an integer or an irrational number?

there was an answer (revision 0) by Douglas S Stones which said in complete:

" If $\sqrt{a}=x/y$ where $y$ does not divide $x$, then $a=(\sqrt{a})^2=x^2/y^2$ is not an integer (since $y^2$ does not divide $x^2$), giving a contradiction. "

Two people whose opinions I respect claimed that this proof approach was "totally bogus/circular".

I don't really see how this is circular, or bogus for that matter.

Doesn't the result follow immediately from unique factorization?

So my question is: What is wrong with that answer?

Bill Dubuque
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Aryabhata
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    Of course it follows from unique factorization - that's precisely the point. You have to mention why the inference holds. Many students think that no justification is required - that it's "obvious". – Bill Dubuque Sep 13 '10 at 19:58
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    It begs the question; the assertion is that $x$ doesn't divide $y$ implies that $x^2$ doesn't divide $y^2$; but no justification for this latter implication is offered. One might suppose from the assertion that the writer assumes that no justification is necessary. But there is a domain $R$ and elements $x$ and $y$ of $R$ such that $x^2\mid y^2$ but that $x\not\mid y$. The implication does not follow from the ring axioms (the basic properties of addition and multiplication) but uses special and nontrivial properties of $\mathbb{Z}$. – Robin Chapman Sep 13 '10 at 20:25
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    @Moron: Much of the content of a first number theory course is a consequence of unique factorization. But one cannot omit explicit mention of precisely how those interesting theorems can be proved using unique factorization and related propositions. Moreover, because our hardwired intuition is so rich with knowledge of properties of integer arithmetic, it is especially important to separate informal empirical inferences from logical deductive inference based upon the axiomatic method (ring axioms, induction, etc). Mastering such is crucial in order understand generalizations to other rings. – Bill Dubuque Sep 14 '10 at 00:07
  • @Moron: See my comments below. This "proof" gave no clue whatsoever why the author believed that the equivalent statement is true. In 99% of the cases where I've seen precisely this "proof" presented, the author doesn't even realize that the proof requires invoking uniqueness of factorization or some closely related strong result. Instead the author thinks the equivalent statement is obviously true and requires no further justification. A correct proof cannot leave any such doubts. It must have a unique interpretation to a competent reader. As experts here agreed, this proof failed to do so. – Bill Dubuque Sep 14 '10 at 02:48
  • @Moron: I made no assumptions about your knowledge of other rings. Rather the purpose of my prior remark was merely to stress that the correctness of the proof has nothing to do with these remarks about other rings - they were mentioned for other purposes. – Bill Dubuque Sep 14 '10 at 03:36
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    I find this thread terrifying. – Pierre-Yves Gaillard Sep 14 '10 at 03:57
  • @Pierre: Please do elaborate. I always do appreciate your thoughtful remarks. – Bill Dubuque Sep 14 '10 at 04:37
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    @Bill Dubuque - This thread shows the extent to which we live in a Babel tower! Needless to say that I fully agree with you, Qiaochu Yuan and Robin (I hope I'm not forgetting anybody). But I think Moron and whuber are sincere. 99.9% of the mathematical community would disagree with them, but this doesn't affect their perception of things. In some sense it's good that they say what they think. I often find myself in their situation. There are many mathematical questions on which my opinion is considered as false with even more unanimity than in the case at hand. – Pierre-Yves Gaillard Sep 14 '10 at 06:45
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    @Pierre-Yves: I see little disagreement here, but mainly differences in perspective and emphasis. Moron has made it clear he was seeking a determination of the logical validity of a sequence of statements. The other issues raised by Bill Dubuque and Qiaochu are interesting, important, and valid--how could they not be? The only claim in this entire conversation I would dispute concerns the alleged "circularity" of the proof Moron outlined. A demonstration of circularity requires one to show the consequent was asserted as a supposition. Such a demonstration has not been offered by anyone. – whuber Sep 14 '10 at 15:58
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    @whuber - I haven't really read the whole thread, and I don't think I can explain math better than Bill Dubuque, Qiaochu Yuan and Robin Chapman. So I'll just say that I see the issue as follows. Either you state clearly that you're using the unique factorization property (UFP) and your argument reduces to the usual one, or you give another justification to sustain your claim. I can't help from feeling that you want to use the UFP, but, for reasons I don't understand, you don't want to say you're using it. – Pierre-Yves Gaillard Sep 14 '10 at 17:18
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    @Pierre-Yves: You have an excellent point. A good proof provides not only a logically correct sequence of statements, but also an explicit justification for each one. I actually did invoke the UFP when drafting the response, but then realized it had already been mentioned (vide Qiaochu's earlier response) and deleted it as both redundant and obvious (given the quality of the audience I expected for any question that Moron might pose). It was never my (nor, I believe, anyone's) intention to insinuate that the UFP was unnecessary. – whuber Sep 14 '10 at 20:27
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    @whuber - @Moron - If you want to tell somebody why square root of 2 is irrational, you have to assume that this somebody doesn't know it already! [Thanks to Moron for mentioning the meta thread that I hadn't seen (and thank you to whuber for your very clear comment). I understand your reaction to the way Douglas's answer was edited, but that's another issue. Obviously, you both understand the proof. The question about whuber's accepted answer is NOT "Is it correct?", but "Is it understandable by somebody who doesn't already know this stuff?".] – Pierre-Yves Gaillard Sep 15 '10 at 03:56
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    @whuber - Would you consider kindly adding (for dummies like me) to your answer a sentence saying that you're using the UFP? – Pierre-Yves Gaillard Sep 15 '10 at 08:31
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    @Pierre-Yves: As that's a reasonable request, I have complied, albeit with some trepidation (for I don't really want to reopen this discussion). BTW, there's no need to be self-deprecating: nobody here is a mathematical dummy. Questions of mathematics should never be turned into questions about whether people are smart or not. Nor, for that matter--I'm now taking the opportunity to respond to one of your earlier remarks--should they be resolved by a popular vote :-). – whuber Sep 15 '10 at 15:14
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    possible duplicate of $a^{1/2}$ is either an integer or an irrational number. – Pete L. Clark Sep 15 '10 at 15:18
  • (Sorry: this comment got misplaced.) This question seems to have riled up a lot people (even some usually unrilable ones) without throwing any new mathematical light on the previous question on which it is based. I have voted to close. – Pete L. Clark Sep 15 '10 at 15:35
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    @whuber - Thank you!!! [If you reread the earlier comment of mine you're referring to, you'll see that I completely agree with you.] – Pierre-Yves Gaillard Sep 15 '10 at 15:47
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    To those considering voting to close: please discuss on meta first http://meta.math.stackexchange.com/questions/814 – Bill Dubuque Sep 15 '10 at 17:01
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    @Moron: The argument boils down to the subjective matter of judging which inferences are trivial enough to omit in a proof. Most experts have opined that in this proof, invocation of unique factorization (or equivalent) must be explicitly mentioned. For without doing so there is no way to distinguish a correct proof from the widespread incorrect proofs authored by novices who think that the statement is "obvious" - errors that occurred for centuries before number theory was placed on rigorous foundations by Gauss, Dedekind at al. – Bill Dubuque Sep 15 '10 at 17:18
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    Sadly, I am unable to keep away from this thread! I guess we could close this question as "Subjective and Argumentative". – Aryabhata Sep 15 '10 at 17:27
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    @Moron: Indeed, in the absence of any further information one cannot know if the intended proof was circular, or incomplete, or whatever. The reason I think your question is worth keeping open is that someone could write an interesting answer explaining how to make these decisions when writing good proofs, i.e. what inferences are important enough that they must be explicitly mentioned vs. those so trivial that they are safe to omit without leaving any ambiguity. This is something that is certainly important to learn, and not usually discussed in textbooks. – Bill Dubuque Sep 15 '10 at 17:44
  • Since @Bill created a meta-question, let's keep the discussion of whether to close or not there. – Larry Wang Sep 15 '10 at 18:01
  • @Pierre. Interesting viewpoint. I think a problem at the heart of the matter is that many courses / textbooks on elementary number theory leave much to be desired. They're too specialized to provide a broad enough perspective to someone who doesn't go on to study algebraic number theory. There's too large an abyss between traditional elementary courses and advanced/algebraic courses. I've encountered many PhD's who are clueless about the essence of unique factorization and related topics. I'd have to agree with you, encountering such can be "terrifying" - a real-world irrationality proof! – Bill Dubuque Sep 15 '10 at 18:03
  • I haven't completely followed this discussion, but I must say Robin Chapman's initial point is a very worth one. – Noldorin Sep 15 '10 at 20:53
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    There is more pedantry here than I can stomach - it seems the experts prefer answers which the typical undergraduate would not understand, even to typical undergraduate questions – BlueRaja - Danny Pflughoeft Sep 15 '10 at 21:15
  • To those who think no mention of unique factorization is needed: If you can find even one published textbook at this level that - like the original post - completely omits justification of this inference then I would be quite surprised. – Bill Dubuque Sep 15 '10 at 21:25
  • Closing for 3 votes + Isaac♦ and myself – Larry Wang Sep 15 '10 at 23:36
  • @Bill: btw, I believe there is a proof that gcd(a,b) = 1 implies gcd(a^2,b^2) not equal to b^2, which does not rely on unique factorization. It uses the fact that ax+by is divisible by gcd(a,b), Bezout's identity and the fact that elements of Z don't have inverses wrt to multiplication. (i.e. if I got it right). – Aryabhata Sep 16 '10 at 01:03
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    Of course, e.g. sqrt(N) = B/A, (A,B) = 1 => N = BB/AA, => A|BB => A|B by Euclid's Lemma. QED I mention this frequently on sci.math e.g. http://bit.ly/IrratGCD For a much more general viewpoint see http://bit.ly/FreshGCD But existence of GCDs <=> irreducibles prime <=> uniqueness of factorizations into irreducibles [existence of factorizations into irreducibles is of course trivial in Z]. This is all very well-known. – Bill Dubuque Sep 16 '10 at 01:41
  • @Moron: Here are two nice Monthly surveys on UFDs: http://bit.ly/UFD_Samuel and http://bit.ly/UFD_Cohn – Bill Dubuque Sep 16 '10 at 01:50
  • Note: fix for the above rotted links: IrratGCD and FreshGCD. The fix is to change Google Groups links from "www.google.com/..." to "groups.google.com/..." i.e. replace "www" by "groups", since the former style is no longer supported. – Bill Dubuque Aug 04 '11 at 19:25
  • And here is the fixed url for Pierre Samuel's $1968$ Monthly article Unique factorization – Bill Dubuque May 02 '23 at 22:54

3 Answers3

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The result does follow from unique factorization, but the point being made here is that the stated claim - that if $y$ doesn't divide $x$, it's also true that $y^2$ doesn't divide $x^2$ - is equivalent to the problem statement, so it's circular to use it to prove the problem statement (or in any case it doesn't address the meat of the problem).

Edit: Let me try to make it clearer why I think this proof is not a proof in the sense that no mathematical work has been done. We want to show that if $a$ is a positive integer, $\sqrt{a}$ is either an integer or irrational. What does that mean? That means if $\sqrt{a} = \frac{p}{q}$ where $p, q$ are positive integers, then $q | p$. Equivalently, if $a = \frac{p^2}{q^2}$ where $p, q$ are positive integers, then $q | p$. Equivalently, if $q^2 | p^2$, then $q | p$. Equivalently, if $q$ does not divide $p$, then $q^2$ does not divide $p^2$.

I have done no mathematical work so far. All I have done is unpack definitions. The statement that I have ended up with is 1) exactly as hard to prove as the statement I started with, and 2) true in all of the same rings as the statement I started with. A crucial part of the problem - that we are working in $\mathbb{Z}$ - has not yet been used. To claim that the statement is "obvious" from here is to ignore an essential nontrivial property of $\mathbb{Z}$, namely that it is integrally closed (which follows from unique factorization). There is a reason this property has a name.

Qiaochu Yuan
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    @Moron: The original post did not say anything about unique factorization. It did not even given any hint that the OP thought the equivalent statement needed proof. This is a very common mistake. – Bill Dubuque Sep 13 '10 at 20:13
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    @Moron: the point is that Douglas' answer does not prove A, but only asserts it. In fact A is false in rings larger than the integers, so it is necessary to explicitly point out what properties of the integers are being used here. Anything less cannot be considered a proof of the statement. – Qiaochu Yuan Sep 13 '10 at 20:17
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    @Moron: 1/(c/d) = d/c is something that does need to be spelled out if the problem is to prove that 1/(c/d) = d/c. – Qiaochu Yuan Sep 13 '10 at 20:22
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    @Moron: the two problems are equivalent. I do not see what I am failing to get across here. The answer to "prove A" cannot in any reasonable sense be "well, A, therefore, A." – Qiaochu Yuan Sep 13 '10 at 20:26
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    @Moron: see the answer by whuber below for yet another example of just how widespread such confusion is. Many people think that it does not require proof. The problem is that deep hardwired intution about integers is clouding rational deductive thought. This problem wouldn't arise if the proof was about some abstract structure where humans didn't have primal instincts (pun intended!) – Bill Dubuque Sep 13 '10 at 20:30
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    @Bill: I don't think any of us are under any illusions about what is being stated in this proof (and, IMO, it's rather patronizing to suggest so). It appears that you and Qiaochu may be not making strong enough distinctions between pedagogical and logical concerns. Moron and I recognize his proof as being logically correct and you are also correct in that such a proof requires a statement of equal (or even stronger) content. If there is any "confusion" it lies in not sufficiently making a distinction between content and validity. – whuber Sep 13 '10 at 20:35
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    @Moron: the statement that sqrt(a) is either an integer or irrational can also be stated without constructing the real numbers. To be irrational just means not to exist in Q, which means that there do not exist integers p, q such that a^2 = p/q. Continuing to unpack definitions in this way one gets precisely the second statement. My concern here is not necessarily that A and B are equivalent (though you seem to disagree with me here) but that it is easy to be fooled into thinking that A is nontrivial while B is trivial. They have the same degree of triviality. – Qiaochu Yuan Sep 13 '10 at 20:42
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    @Bill: Whuber isn't trying to prove that statement here! He is only trying to talk about it not being circular. – Aryabhata Sep 13 '10 at 20:42
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    @whuber. You simply restated the original circular proof without giving any hint that you understand why it is circular. – Bill Dubuque Sep 13 '10 at 20:44
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    @Moron: way ahead of you. – Qiaochu Yuan Sep 13 '10 at 20:51
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    @Bill: I'm with Moron here--you have no basis to label his proof "circular." I'm also unaware of any part of mathematics requiring that a logically valid proof "hint" that its writer have any "understanding" of it :-). Validity and understanding are unrelated (although both, of course, are to be encouraged, honored, and recognized). – whuber Sep 13 '10 at 21:03
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    @Moron: No, the equivalence is trivial (except perhaps to a rank novice). The nontrivial part is proving one of these equivalent statements - which typically uses deep fundamental properties of Z - e.g. integral closure or UFD. – Bill Dubuque Sep 13 '10 at 21:04
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    @whuber: please read Qiaochu's comments to your post. – Bill Dubuque Sep 13 '10 at 21:05
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    @whuber: A proof that two statements are equivalent does not constitute a proof of either statement. – Bill Dubuque Sep 13 '10 at 21:07
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    @Moron: An important goal of an elementary number theory course is learning to compose proofs - in particular, learning what details are essential (and must be mentioned) - and what details are trivial (and can be omitted). Any invocation of a high-powered result such as unique factorization must be mentioned explicitly in proofs at this level. Especially since many students erroneously assume that said equivalent statement is obvious and requires no proof. Before Gauss (and even long after) many thought that unique factorization was obvious and required no proof. Intuition is not deduction. – Bill Dubuque Sep 13 '10 at 21:27
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    @Moron, @whuber: I give up. Whatever we are arguing about, it is not mathematics. – Qiaochu Yuan Sep 13 '10 at 22:41
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It seems to me that maybe the best way to describe the situation is this: Douglas Stone's original answer to the original question consisted of rephrasing the question in such a way as to make it accessible to a proof using basic properties on the integers (specifically, unique factorization).

In my opinion, one thing which is being lost in all this discussion is just how important it can be to rephrase a question! Sure, the process of rephrasing contains "no math" as Qiaochu has pointed out. But that doesn't make it useless (and I wouldn't use the word circular here either).

Finding ways to rephrase questions so that they become accessible to the methods available is a basic skill beginning students of mathematics need to learn. For example, much of the material in the early chapters of modern linear algebra books consists of teaching students how to rephrase questions in linear algebra so that they can be solved by row reduction.

I wouldn't accept Douglas Stone's answer as a complete solution to the problem if it were turned in by a student in an elementary number theory class, just as in my linear algebra class, reducing a problem to a question of row reduction isn't a complete solution. But if a student came to me and said he or she was stuck on the problem, the first thing I'd try to do is get them to rephrase the question in precisely the way Douglas Stone did.

Pleasantly, the community has pointed out (both here and at the original question) exactly how to finish the proof after rephrasing it in this useful way.

Dan Ramras
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    It's nice to have a moderate voice join the discussion :-). – whuber Sep 15 '10 at 21:49
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    I think this describes the situation quite well. The rephrasing concept is a very useful one here. In particular, while for those with a lot of mathematical experience (such as Qiaochu) such a step is trivial (and hence their view is that all the work is done in the next step, when one applies something non-trivial like unique factorization to the rephrased problem), to a beginning student, such steps can be genuine blocks, and so it is useful and important to point them out. – Matt E Sep 16 '10 at 03:04
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    The "rephrasing step" is simply squaring the equation to convert it to (equivalent) integral form. Every proof using integer arithmetic has no choice but to eliminate the radical by squaring. I suspect that step is trivial for almost everyone at this level. It's the following steps that are the heart of the matter since they must employ intrinsic factorization properties of Z. – Bill Dubuque Sep 16 '10 at 04:44
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    @Bill: Dear Bill, If "at this level" means people taking part in this discussion, I agree. But I don't think that these steps are trivial to average students (say in an elementary number theory class). Indeed, it's easy for me to imagine that they would find this step more difficult than the step involving appeal to unique factorization (although mathematically, I agree that the latter has depth, while the rephrasing doesn't), just because they probably have some intuition for the latter (and they are not being asked to prove it, just to apply it!), while manipulating definitions ... – Matt E Sep 17 '10 at 06:17
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    @Bill (con'td): ... doesn't come easily to large numbers of students (at least in my experience). – Matt E Sep 17 '10 at 06:20
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It sounds like we only need to be clear about what is being asserted here, because there shouldn't be any dispute. Evidently you want to say that when $a$ is an integer, $\sqrt{a}$ cannot be a non-integral rational number. Assume the contrary in order to achieve a contradiction; that is, suppose there exists a rational number $x/y$ whose square equals $a$ but which itself is non-integral. Because $x/y$ is not an integer, $y$ does not divide $x$. Then $y^2$ does not divide $x^2$, whence $x^2/y^2 = (x/y)^2 = a^2$ is nonintegral, the contradiction. Ergo, this is a valid argument. (Like all logical arguments, it's a complete tautology. But that's not a circularity!)

Edit

In response to a request in the comments below the main question, I am happy to point out that one possible justification for the final step in this demonstration is the unique factorization property of the integers. (UF is not strictly necessary, though: the assertion that for all $x, y$ in a ring $x^2 \mid y^2 \Rightarrow x \mid y$ in and of itself does not guarantee unique factorization.)

whuber
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    This is not a proof of the statement. This is a proof that one of the statements in question implies the other. In fact, as I said, the statements are equivalent, but you haven't proven either. – Qiaochu Yuan Sep 13 '10 at 20:15
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    @Qiaochu: ALL proofs are proofs of equivalence. What kind of idiosyncratic distinction are you making? Of course the logic shows that the truth of the assertion rests on something else, like unique factorization. But that's no basis for objection! – whuber Sep 13 '10 at 20:17
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    @whuber: a pedagogical one. Students are often unaware that the claim that (if y doesn't divide x, then y^2 doesn't divide x^2) needs explicit proof, and if they don't realize this then they'll quickly get into trouble the moment they start talking about rings of integers in number fields larger than Q. – Qiaochu Yuan Sep 13 '10 at 20:19
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    @Qiaochu: I completely agree with you about the pedagogical issue. However, until you mentioned that just now, I saw nothing in your response that suggested this was your motivation. (I did not think to equate "meat" with "pedagogy"!) We all may have been talking at cross purposes: 'Moron' about logic and you (unstated) about making a teaching point. This may have caused some confusion. – whuber Sep 13 '10 at 20:30
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    @whuber: perhaps I can restate my objection as follows. Someone asking about square roots of integers being integer or irrational may well think "I don't know how to prove this," but after seeing that it is equivalent to the claim that if y doesn't divide x, then y^2 doesn't divide x^2, may well think "well, I understand this now." The point is that unless they know how to prove this second statement, they don't know how to prove anything, because the two problems are equivalent in a certain trivial sense: if one unpacks the definitions in the first problem one gets the second. – Qiaochu Yuan Sep 13 '10 at 20:32
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    (cont.) I cannot think of unpacking definitions as being mathematical work in any meaningful sense, so no mathematical work has yet been done. The student who does not understand this does not understand that this problem is not trivial (in the sense that it does not follow from the ring axioms), and also does not understand something important about unpacking definitions. – Qiaochu Yuan Sep 13 '10 at 20:33
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    @Qiaochu: I'm sure you're familiar with Michael Spivak's Calculus on Manifolds. I recollect that his main point in the proof of (the generalized) Stokes theorem is that it is "just" unpacking definitions! But enough of that: let's review how we got here. The original question is purely one concerning logical validity; it got reframed as a pedagogical/philosophical one only within comments made by you and Bill Dubuque. If you guys really want to discuss these issues, wouldn't it be more appropriate to open a specific (community) question about them? – whuber Sep 13 '10 at 21:11
  • @whuber: The statement at hand - although it may seem deceptively trivial to you - is in fact a very powerful property of Z - almost equivalent to unique factorization. You can't simply say it is true without any justification any more than you can say that: p prime => p|a or p|b without justification. Both are deep fundamental properties of Z (viz. integrally closed, and the prime divisor property), whose fundamental nature was revealed only after much introspection by great number theorists (esp. Dedekind, who struggled for a long time before realizing the key role of integral closure). – Bill Dubuque Sep 13 '10 at 21:43
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    @Bill: I agree with you and deeply appreciate the point you are making. It is interesting that Moron's proof explicitly makes the connection between his original question and these issues by showing that one logically implies the other. I don't believe I made any claims of triviality (let's not confuse that with formality, please) and am sorry if somehow I left that impression. All that I have maintained--and this is the last time I will repeat it--is merely that what you say does not bear on the question of the logical correctness of Moron's demonstration. – whuber Sep 13 '10 at 22:42
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    @whuber: I don't question the "logical correctness of Moron's demonstration". As I said in the first comment "of course it follows from unique factorization - that's precisely the point". I object to proofs that don't give any explicit justification - whether by unique factorization, integral closure (rational root test) or some other analogous statement. Without such further explicit remarks there is no way to know if the author knows such justification or, instead, he has no such knowledge but commits the widespread error of assuming the key result is obvious - requiring no proof. – Bill Dubuque Sep 13 '10 at 23:23
  • But this answer says nothing at all about the issue, so why accept it? I'm disappointed that after all this work you've thrown in the towel. – Bill Dubuque Sep 14 '10 at 03:18
  • The answer does no such thing. The "proof" is circular - precisely the same as the original. Three experts have argued this point at length. Do you think they are all wrong? – Bill Dubuque Sep 14 '10 at 03:40
  • @Moron: I'm sad that there appears to be no way to convince you of the essence of the matter. Perhaps you'll be in a better mindset to appreciate these points at some later date. – Bill Dubuque Sep 14 '10 at 04:33
  • @whuber - Thanks!!! [You wrote "in a ring in and of itself" in the last parenthesis. I suspect some typo...] – Pierre-Yves Gaillard Sep 15 '10 at 15:21
  • @Pierre-Yves: Actually, that phrase was intended, although it's difficult to think of any way punctuation could clarify it. I've tried a minor reordering of the words. – whuber Sep 15 '10 at 18:22