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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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    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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