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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.

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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 1 '13 at 18:16
    
How can this be the proof when you make no mention of $\sqrt{2}$? –  ldog Mar 1 '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 1 '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 1 '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 1 '13 at 18:28

2 Answers 2

up vote 2 down vote accepted

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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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 1 '13 at 19:25
    
@Alex If you let me know precisely which points are not clear, then I will gladly elaborate. –  Math Gems Mar 1 '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 1 '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 1 '13 at 19:43
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It's on page 6 of this pdf. I appreciate your help. gutenberg.org/files/38769/38769-pdf.pdf –  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.

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Thank you for this, it is helpful. –  AlexHeuman Mar 1 '13 at 19:26

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