How to prove that when $m$ is not a full power of any natural number, then $m^{(1/2) }$ is irrational
I have tried this but got nothing;
$\sqrt{m}=\frac{p}{q}$ were $p,q$ are natural and $\gcd(p,q)=1$;
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Mithlesh Upadhyay
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Sead Mejzini
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4You might find this interesting: http://math.stackexchange.com/questions/4467/sqrt-a-is-either-an-integer-or-an-irrational-number – Olivier Oloa Aug 26 '16 at 12:48
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From $\sqrt m=\frac ab$ with $\gcd(a,b)=1$, we find $a^2=mb^2$. Let $p$ be a prime dividing $b$. Then $p\mid mb^2=a^2$, hence also $p\mid a$. This contradicts $\gcd(a,b)=1$. Hence no such prime exists. We conclude $q=1$.
Hagen von Eitzen
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This proof if exactly a special case of the proof of the Rational Root Test applied to $,x^2 - m.,$ If one already knows this test then is is simpler to invoke it by name (vs. repeat the proof in this special case). – Bill Dubuque Aug 26 '16 at 13:40
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Should we assume on the beginning that b must be different from 1 – Sead Mejzini Aug 26 '16 at 18:57