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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.

Hanul Jeon
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AlexHeuman
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  • 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 $c$. – Thomas Andrews Mar 01 '13 at 18:16
  • How can this be the proof when you make no mention of $\sqrt{2}$? – ldog Mar 01 '13 at 18:16
  • @ldog I think this is meant to be the part of the proof OP didn't understand, not the whole proof. – Thomas Andrews Mar 01 '13 at 18:18
  • @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 $\sqrt 2$. – AlexHeuman Mar 01 '13 at 18:26
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    @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. – Trevor Wilson Mar 01 '13 at 18:28
  • @ThomasAndrews that makes things a little clearer for me, but I'm still new to this, so doesn't m play a role in that too? I'm still having trouble understanding. – AlexHeuman Mar 01 '13 at 18:28
  • You may want to check out http://math.stackexchange.com/questions/5/how-can-you-prove-that-the-square-root-of-two-is-irrational. It might be easier to understand some proof before you try to figure out what Hardy was saying in the passage you posted (which I find hard to understand without context.) – Trevor Wilson Mar 01 '13 at 18:31
  • @TrevorWilson I found that proof before, to try and understand Hardy's, and I understood it immediately, but it gave me no insight to this proof. What Thomas Andrews said made some sense to me, but I'm still a little fuzzy on it because of the additional factor of m. The paragraph preceding this says It is "easy" to see that there is no rational number such that its square is 2. In fact we may go further and say that there is no rational number whose square is m/n, where m/n is any positive fraction in its lowest terms, unless m and n are both perfect squares. – AlexHeuman Mar 01 '13 at 18:36
  • I see. I don't have Hardy's book but I hope that someone who does can answer your question. – Trevor Wilson Mar 01 '13 at 18:38
  • Another proof which I understood was this: Suppose, if possible, that p/q is a positive fraction, in its lowest terms, such that $(\frac pq)^2=2$ or $p^2 = 2q^2$. It is easy to see that this involves $(2q-p)^2=2(p-q)^2$; and so (2q-p)/(p-q) is another fraction having the same property. But clearly q<p<2q and so p-q<q. Hence there is another fraction equal to p/q and having a smaller denominator, which contradicts the assumption that p/q is in its lowest terms. – AlexHeuman Mar 01 '13 at 18:39
  • @AlexHeuman: What I was trying to say: why does the theorem statement mention $\sqrt{2}$ when clearly you are trying to prove something general. That is something akin to stating a corollary of a theorem, proving a general theorem, then applying the theorem to show the corollary, without ever stating the general theorem you wanted to prove. Its just bad math writing style in my opinion. – ldog Mar 01 '13 at 19:11
  • @ldog, you're going to have to take that up with G.H. Hardy not me. In his book he explains the rational numbers then states that they are inadequate for suppose the hyp of a right sided triangle whose sides AB=AC=1, then there is no rational number whose square is 2 in order to obtain this quantity. He uses this proof. – AlexHeuman Mar 01 '13 at 19:24

2 Answers2

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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.

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Math Gems
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  • 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. – AlexHeuman Mar 01 '13 at 19:25
  • @Alex If you let me know precisely which points are not clear, then I will gladly elaborate. – Math Gems Mar 01 '13 at 19:27
  • 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? – AlexHeuman Mar 01 '13 at 19:40
  • @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. – Math Gems Mar 01 '13 at 19:43
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    It's on page 6 of this pdf. I appreciate your help. http://www.gutenberg.org/files/38769/38769-pdf.pdf – AlexHeuman Mar 01 '13 at 19:45
  • I'm really getting a better understanding. The key was understanding how Euclid's Lemma fit in. Although that seems obvious now, I didn't have a clue before I saw that theorem. – AlexHeuman Mar 01 '13 at 20:10
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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.

Thomas Andrews
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