If we set $\sqrt{2} = \frac{a}{b}$ where $a,b$ are positive and coprime, we get $2 = \frac{a^2}{b^2}$. However, as $a^2,b^2$ must also be coprime, this fraction can only be an integer if $b^2 = 1$, so $b=1$. Then $a^2 = 2$, which is clearly false as $a^2 > 3$ for $a > 2$ and $a$ is not 0 or 1.
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How do you get the relation a>2? – Mar 01 '22 at 03:54
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@RamanujanXV Because a is assumed to be an integer and cannot be 0 or 1 – Golden_Ratio Mar 01 '22 at 04:17
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Yep, I'm just taking cases on the positive integers. – Fnark Man Mar 01 '22 at 04:18
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@Golden_Ratio I understand that, but why not a$\geq$2? – Mar 01 '22 at 04:39
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You might find this question of mine interesting: https://math.stackexchange.com/questions/1311228/what-is-the-most-unusual-proof-you-know-that-sqrt2-is-irrational – marty cohen Mar 01 '22 at 04:46
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@RamanujanXV Well tbh I think the last bit in OP's proof can be tidied up. OP can just say it is clear $b\neq 1$ since $2$ is not a perfect square. – Golden_Ratio Mar 01 '22 at 04:48
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@Golden_Ratio I agree with you. – Mar 01 '22 at 04:51
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1The crucial claim that $(a,b)=1\Rightarrow (a^2,b^2)=1$ requires proof. This well-known proof is already posted here many times, e.g. see the linked dupes. Please search for answers before posting questions. – Bill Dubuque Mar 01 '22 at 09:20
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Looks okay to me. You can simply state that $a^2,b^2$ being coprime implies they cannot cancel out to an integer since $b\neq 1$, which is proof 4 here.
Golden_Ratio
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This gives a very easy proof for any number which isn't a perfect square right? Essentially if the square root is not an integer, it can't be rational. – Fnark Man Mar 01 '22 at 04:20
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1Please strive not to add more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Mar 01 '22 at 09:20