# Is this proof of $a^{1/2}$ being either integer or irrational circular/incorrect?

There was an answer by Douglas S Stones:

" 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?

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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
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
I find this thread terrifying. – Pierre-Yves Gaillard Sep 14 '10 at 3:57
@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 6:45
@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

## 3 Answers

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.)

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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
@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
@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
(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
@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

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.

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It's nice to have a moderate voice join the discussion :-). – whuber Sep 15 '10 at 21:49
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 3:04
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 4:44
@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 6:17
@Bill (con'td): ... doesn't come easily to large numbers of students (at least in my experience). – Matt E Sep 17 '10 at 6:20

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.

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If A and B are equivalent, proving A proves B. I don't see the "circularness" (if that's a word) here. I guess you see it that way too, otherwise you wouldn't be talking of the meat there :-) The fact that the proof of A was left out does not mean it is bogus or circular. Just incomplete. – Aryabhata Sep 13 '10 at 20:07
@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
@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
@Bill: Well, the fact that y^2 does not divide x^2 blah is trivial (except perhaps to a rank expert) once you are aware of unique factorization. That is one of the first things taught in any course in basic number theory. I don't really care whether there exist structures which aren't UFD etc. Bringing that up is just irrelevant noise and claiming that it is very relevant is just ridiculous. Anyway, I will stop having this discussion. I got the answer I was seeking. – Aryabhata Sep 13 '10 at 21:10
@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