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Rudin has gone through complex scenarios to prove this theorem, but isn't the following correct?

By contradiction, let $\exists p\in \mathbb{Q}; p^2=2$. Take $p=a/b$ where $a,b\in \mathbb{Z}$ and $(a,b)=1$. So, $a^2/b^2=2\Rightarrow a^2=2b^2\Rightarrow a=\sqrt{2}b$, yielding $\sqrt{2}b\in \mathbb{R}\backslash\mathbb{Q}\Rightarrow a\in\mathbb{R}\backslash\mathbb{Q}$, contradicting $a\in \mathbb{Z}$. $\square$

Mill
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    This is not correct; if $p\in \mathbb{Q}$ then you have derived $a=\sqrt{2}b=pb\in \mathbb{Q}$, which is not a contradiction. – vadim123 Jul 01 '14 at 15:31
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    After you derive $a^2 = 2b^2$, can you apply unique prime factorization? Also, who is Rudin and what complicated stuff did he do on this question? – Zubin Mukerjee Jul 01 '14 at 15:34
  • @ZubinMukerjee, Walter Rudin, he wrote a few famous analysis textbooks. Specifically, he showed that sets of rationals where $p^2>2$ and $q^2<2$ have no greatest lower bound and no least upper bound (respectively) that belong to either set. – Silynn Jul 01 '14 at 15:36
  • @vadim123 I think you comment is a fine answer! I think you should post it. – amWhy Jul 01 '14 at 15:36
  • @vadim123 Absolutely! I need more concentration now. – Mill Jul 01 '14 at 15:37
  • See http://math.stackexchange.com/questions/5/how-can-you-prove-that-the-square-root-of-two-is-irrational (I'm not saying this is a duplicate, because this newer question is mainly about critiquing a proposed proof). – David K Jul 01 '14 at 16:05

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Here's the correct way to prove this, anyway:

Let $p\in \mathbb{Q}$, so $\exists a,b\in \mathbb{Z}; p=a/b$ where a and b are not both even. Thus, $a^2=2b^2$, yielding $a^2$ is even, hence a is even, so a is divisible by 4. It follows that the $2b^2$ is divisible by 4. So $b^2$ is even implying b is even. That contradicts the assumption that not both a and b are even. $\square$

Mill
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You started assuming that $\sqrt{2}$ is rational, as $b$ is rational and then $\sqrt{2}b$ is rational!

hrkrshnn
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