Sunday, 22 December 2013

real analysis - Direct proof that if p is prime, then sqrtp is irrational.




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 mp is an integer. Consider m=m(pn); it is an integer, and
mp=m(pn)p=mpnmp


is an integer too.



However, since 0pn<1, we have $0\le m' smallest positive integer such that mp is an integer, it implies m=0, which means p=n, hence p=n2, which contradicts p being prime.



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