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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 $\frac{p}{q}\neq\sqrt 2$.


marked as duplicate by TMM, Asaf Karagila, Qiaochu Yuan Dec 30 '12 at 0:07

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    $\begingroup$ Duplicate of math.stackexchange.com/q/20567/264 $\endgroup$ – Zev Chonoles Dec 30 '12 at 0:00
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    $\begingroup$ Also covered in some detail here on MO $\endgroup$ – Old John Dec 30 '12 at 0:01
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    $\begingroup$ 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 $\frac pq\ne\sqrt 2$". $\endgroup$ – Henning Makholm Dec 30 '12 at 0:18
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    $\begingroup$ 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. $\endgroup$ – robjohn Nov 11 '14 at 13:18
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    $\begingroup$ 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 describing why you would like this question reopened. $\endgroup$ – robjohn Nov 11 '14 at 13:26