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I was reading various proofs of the irrationality of $\sqrt{2}$ including a geometric proof by Richard Guy involving similar right triangles. Since then, after thinking about it, I wonder why we are ever taught about the irrationality of square roots of any one particular non-square natural at all, as the proof that all non-square naturals have irrational square roots seems obvious. Am I missing something?

First, the proof in words: squares of non-integral rationals are non-integral rationals. Therefore, the square-root of a non-square natural can not be rational.

Suppose $x \in \mathbb{Q} \setminus \mathbb{N}$. Therefore, there exist $ p, q \in \mathbb{N}$ relatively prime with $q >1$ such that $x = p/q$. But $p,q$ relatively prime with $q > 1$ implies that $p^2$ and $q^2$ are relatively prime with $q^2 > 1$. Therefore, $x^2=(p/q)^2 \in \mathbb{Q}\setminus \mathbb{N}$. Therefore, $(\mathbb{Q} \setminus \mathbb{N})^2 \cap (\mathbb{N}\setminus \mathbb{N}^2) = \emptyset$ and consequently, $(\mathbb{Q} \setminus \mathbb{N}) \cap \sqrt{(\mathbb{N}\setminus \mathbb{N}^2)} = \emptyset$. Here, by $\mathbb{A}^p$ we mean the set formed by raising elements of the set $\mathbb{A}$ to the power $p$.

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    $\begingroup$ Are you sure the proof involving similar triangles was by Richard Guy, or was it merely in something he wrote that you read about it? Tom Apostol published such a proof in the Monthly about ten years ago, but I knew the same proof before that, having read it in a book by Otto Toeplitz, who I think attributed it to an ancient Greek. $\endgroup$
    – Michael Hardy
    Jan 20, 2011 at 18:53
  • $\begingroup$ Ok--I think you're right about that. Sorry for the sloppy citation. $\endgroup$
    – Grant Izmirlian
    Jan 20, 2011 at 18:56
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    $\begingroup$ The question it has been closed. It is not of research level. It would however make a fine question on math.stackexchange.com. (To those who answered the question: was it not clear that this was not a research level question? Please try to give a little more consideration to this in the future. Note that you can certainly answer the question on the other site!) $\endgroup$
    – Pete L. Clark
    Jan 20, 2011 at 22:31

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The point is that you use this fact from number theory:

$p$ and $q$ are relatively prime $\Longrightarrow$ $p^2$ and $q^2$ are relatively prime.

While this is not hard, this is not trivial either. You either need unique prime factorization or the Euclidean algorithm.

For $n=2$, you just have to prove that if the square of a number is even, then so is the number itself. This is trivial casebash modulo $2$.

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  • $\begingroup$ Ok then! This clarifies it for me quite nicely. Thanks! $\endgroup$
    – Grant Izmirlian
    Jan 20, 2011 at 18:46
  • $\begingroup$ As I said--it looks like you've resolved the issue completely--but at some level I have trouble distinguishing things I consider to be basic fact from something requiring a less basic result as you've mentioned. But, is not the definition of "relatively prime" given as "have no factors in common". If so then is it not trivial that $p^2$ and $q^2$ have no factors in common? $\endgroup$
    – Grant Izmirlian
    Jan 20, 2011 at 18:53
  • $\begingroup$ It is worth pointing out that this is proved in Euclid's Elements. See Propositions 24 to 27 in Book VII: aleph0.clarku.edu/~djoyce/java/elements/bookVII/bookVII.html $\endgroup$
    – GH from MO
    Jan 20, 2011 at 18:58
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    $\begingroup$ @Grant: It is trivial if you assume unique factorization, but unique factorization usually is proved by similar statements like "if two numbers are relatively prime to a given number, then so is their product". See also my previous comment. $\endgroup$
    – GH from MO
    Jan 20, 2011 at 19:02
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A definition of $a,b$ relatively prime is no common factors but 1 and -1. A useful equivalent is $as+bt=1$ for some integers $s,t$. Using only this and a little algebra one can prove that if $\gcd(a,b)=1$ and $\gcd(a,c)=1$ then also $\gcd(a,bc)=1$. In particular $gcd(a,b^2)=1$ and $\gcd(a^2,b^2)=1$ SO if a/b is a fraction in lowest terms which is not an integer, so is $a^2/b^2$. Hence no proper fraction can be a square root of an integer. (this is a a tool in developing the theory of factorization in integers)

The irrationality of non integer square roots of integers has been discussed in quite some detail on MO recently.

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    $\begingroup$ Showing the equivalence between the definition and the "useful equivalent" is the main work here. When I was 12 I had troubles with accepting that in a proof, but I had no problems with splitting in two cases modulo $2$. Of course, from the viewpoint of somebody who is doing an induction proof as a warmup exercise each morning it's ridiculous. $\endgroup$
    – darij grinberg
    Jan 20, 2011 at 20:27
  • $\begingroup$ Certainly if as+bt=1 then any common divisor divides 1. And for any a,b the smallest possible positive linear combination as+bt is a common divisor (lest the division algorithm yield a smaller combination). This is induction but not very technical. Another proof is: IF $k-1<\sqrt{m}<k$ then the powers $(k-\sqrt{m})^n$ are positive but decrease without bound. Each is of the form $A+B\sqrt{m}$ with $A,B$ integers. If $\sqrt{m}=\frac{p}{q}$ then $A+B\sqrt{m}\ge \frac{1}{q}$. I never liked the even odd proof. $\endgroup$
    – Aaron Meyerowitz
    Jan 21, 2011 at 5:06
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The history of this goes back to Greek mathematics and there is a good historical discussion in Hardy and Wright "An Introduction to the Theory of Numbers" chapter 4 (I have 5th Edition to hand).

I guess pedagogically you might begin with $\sqrt{2}$ as an easy example in order to show, for example, that there is an irrational number before exploring their nature and existence more generally.

And there is also the question of taking care over the properties of numbers on which the proof is based as darij grinberg has posted in an answer which came up as I was typing this one. Hardy and Wright deal with this too, and more generally also with irrational roots of polynomials with integer coefficients.

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