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How would you prove $\sqrt[5]{672}$ is irrational? I was trying proof by contradiction starting by saying:

Suppose $\sqrt[5]{672}$ is rational ...

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Let $a$ be a rational number that is not an integer, it can be written as $\frac{a}{b}$ where $b$ does not divide $a$. exponentiating yields $\frac{a^5}{b^5}$ where $b^5$ does not divide $a^5$. Conclusion: $\sqrt[n]{a}$ si either an integer or irrational for any natural numbers $n,a$.

So we just have to check whether there is an integer $n$ so that $n^5=672$:

$1^5=1,2^5=64, 3^5=273, 4^5=1024$. So $\sqrt[5]{672}$ is irrational.

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    $\begingroup$ While it is true that $\,b\nmid a\,\Rightarrow\,b^5\nmid a^5,\,$ in problems like this is is essential to justify that claim. $\endgroup$ – Bill Dubuque Mar 16 '15 at 18:10
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$$672^{1/5}=\frac pq,$$($p$ and $q$ relative primes) then $$p^5=672q^5,$$ which is possible only if $p$ is a multiple of $7$, which in turn implies that $q$ is a multiple of $7$.

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The polynomial $X^5-672$ is irreducible over $\mathbb Q$ (Eisenstein for $p=3$), thus has no rational zero.

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Find the prime factorization of 672.

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