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satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1site is not working for me idea is start by contradiction \[p=(\frac{a}{b})^n\] so \[b^np=a^n\] then note that all the primes in \(a^n\) how powers that are as multiple of \(n\) whereas on the left the power of \(p\) is not

henpen
 2 years ago
Best ResponseYou've already chosen the best response.0Sorry for rebumping, the connection was so bad that it failed to notify me of your answer.

henpen
 2 years ago
Best ResponseYou've already chosen the best response.0Much more concise than other answers on sqrt2 proof thanks!
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