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Prove that if $c$ is a positive rational and $k$ is a positive integer, then $c^{1/k}$ is either an integer or irrational, using the rational roots theorem.

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  • $\begingroup$ it is enough to prove for natural $c$ $\endgroup$
    – Norbert
    Oct 15, 2013 at 19:59
  • $\begingroup$ I suppose Matina Manos you mean $c$ is a positive integer! $\endgroup$
    – P..
    Dec 24, 2013 at 11:12

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The way you have posed the problem is: Given $c>0$ over rationals, then the $k$-th root of $c$ is irrational (or integer). That's simply not true! For example, if $c=4/9$ and $k=2$, then $\sqrt{4/9}=2/3$ is neither irrational nor an integer.

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Well, by definition $c^{1/k}$ is a root of the polynomial $x^k - c$...

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  • $\begingroup$ I do not know what that has to do with my question $\endgroup$ Oct 16, 2013 at 0:50

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