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I don't understand this proof for the case where $m^{1/n} \in \mathbb{Q \cap Z}^C$.

Then $\exists \; a, b \in \mathbb{Z} \qquad \ni \qquad\gcd(a,b)=1 \; \text{ and } \; m^{1/n} = \dfrac{a}{b}$.

If $\gcd(a,b)=1$, then $\gcd(a^n,b^n)=1$.

Thus, $m = \dfrac{a^n}{b^n} \in \mathbb{N}$.

But how do I deduce the contradiction: $\dfrac{a^n}{b^n} \not\in \mathbb{N}$?

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    The only way $\frac{a^n}{b^n}$ can be an integer is if $b^n|a^n$.... But if $\gcd(a^n,b^n) = 1$ that is impossible. – fleablood Jun 05 '18 at 06:46

3 Answers3

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$\gcd(a^n, b^n) = 1$ means $a^n$ and $b^n$ have no common factors other than 1 (or $-1$).

$\frac {a^n}{b^n}$ means $b^n$ divides evenly into $a^n$ and $b^n$ itself is a factor of $a^n$. So $a^n$ and $b^n$ have $b^n$ as a factor in common.

The only possible way these are both true, that $a^n$ and $b^n$ have no common factor othere than $1$, and $a^n$ and $b^n$ have $b^n$ as a common factor is if $b^n = \pm 1$.

So $b = 1$ and well, .... now what?

fleablood
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If $\frac {a^n}{b^n} = m$ for some $m \in N$, then $a^n = m\times b^n$ and thus you have $b^n|a^n$. So, $gcd(a^n,b^n)= b^n$, a contradiction.

Shak
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Assume that $|b| \neq 1$. Further assume that $a^n = kb^n, k \in \mathbb{N}$. Using Bezout's lemma: $ra^n + sb^n = 1$ for some $r,s \in \mathbb{Z}\implies rkb^n+sb^n= 1\implies b^n = \pm 1\implies |b| = 1$, contradiction to $|b| \neq 1$. Thus the assertion follows.

DeepSea
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