# Direct proof that $\sqrt{2}$ is irrational? [duplicate]

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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 YuanDec 30 '12 at 0:07

• 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 $\frac pq\ne\sqrt 2$". – Henning Makholm Dec 30 '12 at 0:18