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