Does anyone know of a simple direct proof that if p is prime, then √p is irrational?
I have always seen this proved by contradiction and have been trying unsuccessfully to prove it constructively. I searched this site and could not find the question answered without using contradiction.
Answer
Maybe an elementary proof that I gave in a more general context, but I can't find it on the site, so I'll adapt it to this case.
Set n=⌊√p⌋. Suppose √p is rational and let m be the smallest positive integer such that m√p is an integer. Consider m′=m(√p−n); it is an integer, and
m′√p=m(√p−n)√p=mp−nm√p
is an integer too.
However, since 0≤√p−n<1, we have $0\le m'
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