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I have to prove that $\sqrt 3$ is irrational. let us assume that $\sqrt 3$ is rational. This means for some distinct integers $p$ and $q$ having no common factor other than 1,

$$\frac{p}{q} = \sqrt3$$

$$\Rightarrow \frac{p^2}{q^2} = 3$$

$$\Rightarrow p^2 = 3 q^2$$

This means that 3 divides $p^2$. This means that 3 divides $p$ (because every factor must appear twice for the square to exist). So we have, $p = 3 r$ for some integer $r$. Extending the argument to $q$, we discover that they have a common factor of 3, which is a contradiction.

Is this proof correct?

brinkle martyn
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    @AaronMaroja This looks like a proof verification, and not exactly the same proof as that question... – miradulo Jul 16 '15 at 18:18
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    I would add the step $9r^2 = 3q^2 \implies q^2 = 3r^2$ for clarity. – Marconius Jul 16 '15 at 18:18
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    @AaronMaroja I don't think that is a duplicate, at least of that question. The thing to prove is the same, but both questions are proof-verifications, and the proofs are quite different. – ajotatxe Jul 16 '15 at 18:19
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    @DonkeyKong It's just that, there are so many different questions and duplicates over this topic. – Aaron Maroja Jul 16 '15 at 18:20
  • @AaronMaroja That's fair enough :) I'll try and hunt down a better duplicate for you. – miradulo Jul 16 '15 at 18:21
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    @brinkle The proof is correct, and rather classical. – ajotatxe Jul 16 '15 at 18:21
  • In the main, it's fine. It's probably slightly easier to consider the parities of the number of (not necessarily distinct) prime factors $p^2$ and of $q^2$, and then note $p^2 = 3q^2$. – Ken Jul 16 '15 at 18:22
  • You may as well note, without the need to assume $\gcd(p,q)=1$, that $p^2=3q^2$ is impossible because LHS has an even while RHS has an odd power of $3$. – user26486 Jul 16 '15 at 19:05

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Another proof, without arithmetic theorems:

Suppose $x=\sqrt 3$ is rational. Let $q$ be the smallest positive integer such that $qx$ is an integer, and $q'= q(x-1)$. Note it is a natural number since $qx$ is; furthermore $$q'x =qx^2-qx=3q-qx$$ is a natural number.

However, since $1<3<4$, we know $1<x<2$, hence $0<x-1<1$, so that $$0<q'<q$$ which contradicts the minimality of $q$.

John Bentin
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Bernard
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