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Possible Duplicate:
Irrationality proofs not by contradiction

I've been puzzled for some days now, and I can't come up with an answer. I'm trying to come with a direct proof that $\sqrt{2}$ is irrational. Can somebody check of this is a valid proof ?


Let $p,q$ positive integers. Then $p=2^{p_1}\cdot3^{p_2}\cdot5^{p_3}...$ and $q=2^{q_1}\cdot3^{q_2}\cdot5^{q_3}\cdot...$ where $(p_n)$ and $(q_n)$ are positive integers. Then $p_1-q_1$ is an integer. So $2(p_n-q_n)\ne1$.

Therefore the prime factorization of $\frac{p^2}{q^2}$: $$\frac{p^2}{q^2}=2^{2(p_1-q_1)}\cdot3^{2(p_2-q_2)}\cdot5^{2(p_3-q_3)}\cdot...$$

Because $p_1-q_1$ is an integer, $2(p_1-q_1)\ne1$. Therefore $\frac{p^2}{q^2}\ne2$. Which implies $\frac{p}{q}\neq\sqrt 2$.

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    Duplicate of http://math.stackexchange.com/q/20567/2642012-12-30
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    Also covered in some detail [here](http://mathoverflow.net/questions/32011/direct-proof-of-irrationality) on MO2012-12-30
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    This is a generalization for you: Show that if $n$ is an integer and $r$ is a positive rational,then $n^r$ is rational if and only if $n^r$ is an integer. Proof: Let $r = \frac {p}{q}$. Then $n^r$ is the root of the monic polynomial $x^q - n^p$, which only has integer roots (by the rational root theorem / integer root theorem). Thus $n^r$ is an integer. The converse is obvious.2012-12-30
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    @Zev: I know you're a bit rusty, but there was nothing wrong with just closing this! :-)2012-12-30
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    The answers to the older question are a bit more abstract than what is asked here. **Yes**, the standard proof that $\sqrt2$ is irrational is straightforward to rephrase such that it stars with "Let $p,q\in\mathbb Z$ be arbitrary ..." and ends with "... in each of these cases $(\frac pq)^2\ne 2$ and thus $\frac pq\ne\sqrt 2$".2012-12-30
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    Removing the banner mentioning what question the people who think this is a duplicate think this is a duplicate of, will not change the fact that it is marked as a duplicate and cannot be answered.2014-11-11
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    I have put back the banner, yet retained the other changes. It can now be voted on for reopening. You can also post an "answer" to [this question](http://meta.math.stackexchange.com/questions/6424/requests-for-reopen-votes) describing why you would like this question reopened.2014-11-11

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