Hardy goes on by saying that suppose $\frac {p^2}{q^2}=\frac mn,$ where $p$ has no factor in common with $q,$ and $m$ no factor in common with $n.$ Then $n{p^2}=mq^2$. Here is where I get confused. Every factor of $q^2$ must divide $np^2$, and as p and q have no common factor, every factor of $q^2$ must divide n. Hence $n= \lambda q^2$, where $\lambda$ is an integer. But this involves $m=\lambda p^2$. and as m and n have no common factor, $\lambda $ must be unity. Thus $m = p^2, n = q^2$.

I'm just really having trouble understanding the though process here even though it's something probably extremely simple.

  • $\begingroup$ The basic theorem he is using here is that if $a$ is a factor of $bc$ and $a$ and $b$ are relatively prime, then $a$ is a factor of $c$. $\endgroup$ – Thomas Andrews Mar 1 '13 at 18:16
  • $\begingroup$ How can this be the proof when you make no mention of $\sqrt{2}$? $\endgroup$ – ldog Mar 1 '13 at 18:16
  • $\begingroup$ @ldog I think this is meant to be the part of the proof OP didn't understand, not the whole proof. $\endgroup$ – Thomas Andrews Mar 1 '13 at 18:18
  • $\begingroup$ @ldog, because this shows by taking n=1 that there is no rational number whose square is an integer unless the rational number itself is integral. No rational number whose square is 2 is a small subset of that because if $\frac {p^2}{q^2} = 2 $ then $\frac pq = \sqrt 2$ Which means that there is no rational number equivalent to the quantity $\sqrt 2$. $\endgroup$ – AlexHeuman Mar 1 '13 at 18:26
  • 1
    $\begingroup$ @ldog the theorem statement "there is no rational number whose square is 2" also doesn't mention $\sqrt{2}$. It's a theorem about the rational numbers, and the fact that 2 has a square root in some larger number system is not needed. $\endgroup$ – Trevor Wilson Mar 1 '13 at 18:28

Hardy essentially reproves a well-known property about the uniqueness of reduced (lowest-terms) fractions, viz. the Theorem below (sometimes called unique fractionization).

Theorem $\ $ For $\rm\:m,n,x,y\in \Bbb Z,\,$ if $\rm\ gcd(m,n) = 1,\:$ then $\rm\ \dfrac{x}y\, =\, \dfrac{m}n\ \Rightarrow\:\begin{array}{c}x\, =\, k\,m\\ \rm y\, =\, k\,n\end{array}\ \ $ for some $\rm\ k\in \Bbb Z$

Proof $\ $ By Euclid's Lemma, $\rm\ gcd(n,m)=1,\ nx = my\,\Rightarrow\,n\mid y,\:$ so $\rm\ \dfrac{x}m = \dfrac{y}n = k,\:$ for some $\rm\:k\in \Bbb Z.$

Remark $\ $ Hardy's result is the special case where $\rm\:gcd(x,y) = 1\:$ hence $\rm\:k = \pm1,\:$ i.e. two reduced fractions are essentially unique (we can force $\rm\:k = 1\:$ by requiring denominators to be positive).

The theorem can also be proved by Euclidean descent on denominators, and such a proof is often directly "inlined" in irrationality proofs (vs. being called by name). For an example of such, see the irrationality proofs by John Conway and Bill Dubuque in a prior thread here.

Note: The theorem is equivalent to Euclid's Lemma, since if $\rm\:gcd(n,m)=1\:$ and $\rm\:n\mid my,\:$ then $\rm\: nx = my,\:$ for some $\rm\:x\in \Bbb Z,\:$ so $\rm\:m/n =x/y\:$ so $\rm\:n\mid y\:$ by unique fractionization.

Unique fractionization $\!\iff\!$ unique factorization in domains like $\,\Bbb Z\,$ where ever nonunit $\ne 0$ has a factorization into atoms (irreducibles), since the primality of atoms is an immediate consequence of Euclid's Lemma or unique fractionization.

  • $\begingroup$ I don't quite understand yet, but I think with the resources that you provided that I'll be able to figure it out. Thank you. $\endgroup$ – AlexHeuman Mar 1 '13 at 19:25
  • $\begingroup$ @Alex If you let me know precisely which points are not clear, then I will gladly elaborate. $\endgroup$ – Math Gems Mar 1 '13 at 19:27
  • $\begingroup$ I'm very new to this, so the whole thing is quite foreign to me. I'm self-taught, so it might be best for me to look up all the theorems that you've mentioned and go over it until I understand for myself. If after a while I still don't understand I'll ask a more specific question. Thanks for your help, but I don't wish to waste your time. If I could just ask one thing. What is meant by the $\mid$ symbol? $\endgroup$ – AlexHeuman Mar 1 '13 at 19:40
  • $\begingroup$ @Alex You're not wasting my time. I am here to teach. It might help if you could provide a citation to the proof so we can see the entire context. $\endgroup$ – Math Gems Mar 1 '13 at 19:43
  • 1
    $\begingroup$ It's on page 6 of this pdf. I appreciate your help. gutenberg.org/files/38769/38769-pdf.pdf $\endgroup$ – AlexHeuman Mar 1 '13 at 19:45

Hardy is saying that if $$\frac p q$$ is in reduced form, then so is $$\frac{p^2}{q^2}$$

Essentially, if you try to write $\frac{p^2}{q^2} = \frac m n$ then $n$ must be a multiple of $q^2$.

In particular, then, if $\frac{p^2}{q^2}$ is an integer, then $n=1$ and hence $q^2=1$ so $q=\pm 1$ and $\frac p q$ is an integer.

That means $\sqrt 2$ is rational only if $p^2=2$ has a root for some integer $p$.

The nice thing about Hardy's proof is that you can use it to prove more generally that if $D$ is not the square of an integer, then $\sqrt{D}$ is not rational.

  • $\begingroup$ Thank you for this, it is helpful. $\endgroup$ – AlexHeuman Mar 1 '13 at 19:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.