Here is an elementary proof (adapted from Hardy's A Course of Pure Mathematics) that for any integer $k$, $\sqrt{k}$ is either irrational or integral.
- Suppose $\sqrt{k}$ is rational, $\sqrt{k} = \frac{m}{n}$, $m$ and $n$ have no common factor.
- Then $m^2=kn^2$
- Every factor of $m^2$ must divide $kn^2$
- As $m$ and $n$ have no common factor, every factor of $m^2$ must divide $k$
- Hence $k = \lambda m^2,$ $\lambda \in \mathbb{Z}$
- Hence $1 = \lambda n^2 \rightarrow \lambda = n = 1$
- Hence $k=m^2$
So $\sqrt{k}$ is either irrational or integral.
Q.E.D.
My question regards one step in this proof, present here and also in Hardy's proof – step 4. We conclude that, because $m$ and $n$ have no common factors, all of $m^2$'s factors must be factors of $k$ – because none of them could be factors of $n^2$. We've subtly used the 'fact' here that:
The relative primality of $m$ and $n$ implies the relative primality of $m^2$ and $n^2$
And this is where I am concerned, because I can't quite pinpoint why this must be true. Further, this statement we have assumed as 'obvious' is as strong as the whole proof. Indeed, a weak form of the the contrapositive is:
If $m^2 = kn^2, k\in\mathbb{N}$ then $m$ and $n$ have a common factor.
And straight from this, if $k$ is not a perfect square then $\sqrt{k}$ is irrational.
This is my problem – I cannot see why the first statement highlighted above must be true. Of course, it is obvious from the fundamental theorem of arithmetic, but the whole proof is obvious from the fundamental theorem of arithmetic!
How could you prove the first highlighted statement above without FTOA?
Thank you very much 🙂
Best Answer
That's a perceptive observation. In fact the proof works in domains more general than UFDs, e.g. it follows from the monic case of the Rational Root Test, i.e. it works in all integrally-closed domains, which are far from being UFDs. But let's looks closer at the specific proof you give. It employs $\rm\:(m,n)=1\:\Rightarrow\:(m^2,n^2)=1.\:$ This follows from a special case of Euclid's Lemma, viz. $\rm\:(a,b) = 1 = (a,c)\:$ $\Rightarrow$ $\rm\:(a,bc)=1.\:$ Namely,$\rm\: a=m,\,b=n=c\:$ yields $\rm\:(m,n^2)=1.\:$ Finally $\rm\:a=n^2,\,b = m = c\:$ yields $\rm\:(m^2,n^2) = 1.$
This special case of Euclid's Lemma holds true in any GCD domain, in particular in any UFD or any Bezout domain. But it is weaker, i.e. there are domains satisfying this identity which are not GCD domains, i.e. where some elements have no gcd. In fact this is equivalent to Gauss' Lemma, that the product of primitive polynomials is primitive (or the special case for degree $1$ polynomials). Indeed if $\rm\:(m,n)=1\:$ then $\rm\:m\,x\pm n\:$ is primitive, thus so is $\rm\:(m\,x-n)\,(m\,x+n) = m^2\, x - n^2,\:$ hence $\rm\:(m^2,n^2) = 1.\:$ Below is an excerpt of my sci.math post on 2003/5/3 (see here or here) which shows the relationships between various domains closely related to GCD domains. If you wish to understand precisely the class of rings satisfying such properties then see the links in said post, and also google "root-closed" domains.