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I am given the following hint: Express $ q $ as a quotient of integers $ m/n $ where $ m,n $ are mutually prime, and show that $ m^2 = 2n^2 $ leads to a contradiction.

Proof solution: $2$ divides the right-hand side of this equation, hence $2\vert m^2$ ($2$ divides $m^2$). Since $2$ is prime, we must have $2\vert m$. So in fact $4\vert m^2$, and from the same equation again, $2\vert n^2$ so $2\vert n$. But now $2\vert m$ and $2\vert n$ which contradicts that $m$ and $n$ are mutually prime.

What I don't understand is:

  • Why does contradicting $ m^2 = 2n^2 $ where $ m,n $ are mutually prime disprove no rational $q$ such that $q^2 = 2$? Why is the mutual prime condition necessary?
  • "Since $2$ is prime, we must have $2\vert m$." Why does $2$ being prime entail that $2$ must divide $m$?
  • Could the statement have been proved without the mutual prime condition?
Bill Dubuque
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nico
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  • First part: if $q$ is rational then it is always expressable in the form $m/n$ for some co-prime $m$ and $n$. You will show this isn't possible. For the second part: Euclid's lemma which is a fundamental property of primes. For the third, you could use approaches other than contradiction -- but why would you want that, if it is so easy this way? – Sahaj Feb 21 '24 at 11:09
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    The Wikipedia page on $\sqrt 2$ answers all your questions and offers different proofs for its irrationality as well. – Sahaj Feb 21 '24 at 11:12
  • @Sahaj Ok, thank you for the clarification! I wasn't aware of the first 2 facts, then of course the third question doesn't really make sense... – nico Feb 21 '24 at 11:13

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