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The proof (taken from http://www.themathpage.com/aPreCalc/rational-irrational-numbers.htm#proof):

"To prove that there is no rational number whose square is 2, suppose there were. Then we could express it as a fraction m/n in lowest terms.

That is, suppose $$\frac{m}{n} \cdot \frac{m}{n} = \frac{m \cdot m}{n \cdot n} = 2$$
but that is impossible. Since $\frac{m}{n}$ is in lowest terms, then $m$ and $n$ have no common divisors except 1. Therefore $m \cdot m$ and $n \cdot n$ also have no common divisors - they are relatively prime - and it will be impossible to divide $n \cdot n$ into $m \cdot m$ and get 2."

My problem:

I'm confused about how the fact that $m \cdot m$ and $n \cdot n$ are relatively prime makes it impossible to divide $n \cdot n$ into $m \cdot m$ to get 2.

Since $n$ could be 1, it seems to me that it relies on the assumption that $m$ can't be $\sqrt{2}$, which is what we're trying to prove in the first place, so the argument seems circular to me.

ps: I Wasn't sure about the proof-verification tag; this proof comes from a math Professor so my assumption here is that I'm obviously missing something, I don't mean for the tag to be insulting.

jeremy radcliff
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  • Maybe try writing $m$ and $n$ in terms of their prime factors and think about it that way. – MichaelChirico May 05 '15 at 23:01
  • @MichaelChirico, sorry I don't understand, could you be more specific? Are you saying that $n$ can't be 1? In that case, how would that work if you switch 2 with 4? – jeremy radcliff May 05 '15 at 23:10
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    Nice spotting of the gap! We do need to show that $n\ne 1$, that is, that $\sqrt{2}$ is not an integer. This is clear, since $1^2\lt 2$ and if $m\ge 2$ then $m^2\gt 2$. That's what goes 'wrong" if we try, for example, to prove that $\sqrt{25}$ is not an integer, using a proof along the same lines. There is an additional gap, showing that indeed if $m$ and $n$ are relatively prime so are $m^2$ and $n^2$. For that we need unique factorization, or Euclid's Lemma (if a prime divides $ab$ then it divides at least one of $a$ and $b$). – André Nicolas May 05 '15 at 23:14
  • @AndréNicolas, thank you, it was driving me crazy. In the absence of the step showing that $n \neq 1$, is it accurate to say that the argument is circular since it implicitly assumes what we're trying to prove? I'm wondering if I'm thinking about the terminology "circular argument" in the right way. – jeremy radcliff May 05 '15 at 23:26
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    No it is not circular, it just is missing the verification that $\sqrt{2}$ is not an integer, which is a different assertion than the fact it is not rational, and quite a bit easier to prove. – André Nicolas May 05 '15 at 23:31
  • @AndréNicolas, ah ok, the distinction makes sense. Thank you. – jeremy radcliff May 05 '15 at 23:33
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    @jeremy The "proof" is also missing the justification that $,m,n,$ coprime $,\Rightarrow,$ $m^2,n^2$ coprime, which is the key inference. This is a consequence of the fundamental theorem of arithmetic (existence and uniqueness of prime factorizations), or related properties such as Euclid's Lemma, Bezout's gcd identity, gcd distributive law, etc. – Bill Dubuque May 06 '15 at 01:45
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    Euclid's Lemma says that if $,\gcd (\color{#c00}{n,m})=1,$ then $\ \color{#c00}{n\mid m},k\ ,\Rightarrow\ n\mid k.\ $ Therefore, applying this Lemma in our case above we deduce that: $\ 2n^2 = m^2,$ $\Rightarrow$ $,\color{#c00}{n\mid m},m,\Rightarrow, n\mid m,\ $ so $,m/n,$ is an integer which squares to $,2,,$ contradiction. That's one rigorous way to complete the proof. See this thread for many other proofs. – Bill Dubuque May 06 '15 at 01:58
  • @BillDubuque, thank you for the explanation. It took me a while to walk through the Lemma and see exactly its implications, but what seemed to be very subtle details in the proof are now a lot clearer. – jeremy radcliff May 06 '15 at 15:41
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    @jeremy Great. If anything remains unclear please feel welcome to ask questions. – Bill Dubuque May 06 '15 at 15:46

1 Answers1

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If you assume your number is rational, this means you can write it as a fraction in it's lowest terms, i.e. it cannot be factorised any more. This is, if your $\sqrt{2}=\frac{m}{n} $, where $m,n \in \mathbb{Z} $ cannot be simplified. In other words, if you write $m$ and $n$ as a product of primes you will not have any common primes in expansions of $m$ and $n$. So when you take $m \cdot m=m^{2}$ and $n \cdot n=n^{2}$ and write them as product of primes, they will not have any common primes, and hence cannot be factorised any more. So the only time this can be $2$ is when $n=1$. This however is a contradiction with your earlier assumption $\frac{m}{n}=\sqrt{2}$ as this would imply $m=\sqrt{2} \Rightarrow \sqrt{2} \in \mathbb{Z}$, and here is your contradiction.

Hope it makes sense...

stanislaw.bi
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