I want to prove that the equation $$x^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\dots+a_0=0$$ has no solutions in the form of $p/q$ when $p$ and $q$ and coprime and $q>1$. With this polynomial, $a_n=1$ and $a_i$ is an integer. I have already proven this; I want to see how other people prove it.
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1You mean $q\gt 1$. Same proof as irrationality of $\sqrt{2}$. – André Nicolas Jul 29 '14 at 04:23
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2Multiply everything by $q^n$, you get something like $p^n + q(\text{integer}) = 0 \quad\implies\quad q | p$, a contradiction! – achille hui Jul 29 '14 at 04:25
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Showing that $\mathbb Z$ is integrally closed is a special case of the rational root theorem. – Jonas Meyer Jul 29 '14 at 05:05
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Here's a generalization: http://math.stackexchange.com/q/150554/ – Jonas Meyer Jul 29 '14 at 05:06
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@achillehui Wow! your proof is way shorter than mine. – Joao Jul 29 '14 at 07:04
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@Joao, actually, I've a typo in my comment. should be $q|p^n$ instead of $q|p$ but the basic idea is the same. – achille hui Jul 29 '14 at 07:13
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@Joao It's the standard proof $,(a,b) = 1,\ 0 = b^n f(a/b) = a^n + b(\cdots) = 0,\Rightarrow,b\mid a^n,\Rightarrow, b = 1,$ by Euclid's Lemma (or unique factorization, or Bezout, etc). For a more general proof that works universally for both gcds and invertible ideals see here. – Bill Dubuque Jul 29 '14 at 14:50