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Given $\ $ $ a, b, c \in \mathbb{N} $, $\ $ $ a,b,c > 1 $, $\ $ $ b > c $, $\ $ $b$ and $c$ coprime, $a$ and $c$ coprime, is $ a^{\frac{b}{c}} $ irrational?

Examples: $2^{\frac{4}{3}}$, $13^{\frac{25}{12}}$.

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  • $\begingroup$ $2^{\frac{2}{1}}$ $\endgroup$ – Lee Jun 1 '20 at 11:38
  • $\begingroup$ Sorry, a b c > 1 $\endgroup$ – Andrea Frasca Jun 1 '20 at 11:39
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    $\begingroup$ $8^{4/3} = 16 \;$ $\endgroup$ – amWhy Jun 1 '20 at 11:40
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    $\begingroup$ What is true is it's either irrational or an integer. We can prove this with the monic special case of the rational root theorem. $\endgroup$ – J.G. Jun 1 '20 at 11:41
  • $\begingroup$ take for example any $a=q^2$, $c=2$, then $a^{\frac{b}{c}}=q^b$ $\endgroup$ – Lee Jun 1 '20 at 11:41
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For $x\in \Bbb N$ and for prime $p$ let $F_p(x)$ be the largest $n\in \{0\}\cup \Bbb N$ such that $p^n$ is a divisor of $x.$

$(\bullet)$ . If $a,c\in \Bbb N$ and if $c$ is a divisor of $F_p(a)$ for $every$ prime $p$ that divides $a$ then $a^{1/c}\in \Bbb N....\;$ I.e. then $a=(a')^c$ for some $a'\in \Bbb N.$

If $a,b,c,d,e\in \Bbb N$ with $\gcd (b,c)=1$ and if $a^{b/c}=d/e$ then $a^be^c=d^c.$

So for any prime $p$ that divides $a$ we have $bF_p(a)+cF_p(e)=F_p(a^be^c)=F_p(d^c)=cF_p(d).$

So $c$ is a divisor of $bF_p(a)$ because $bF_p(a)=c(F_p(d)-F_p(e)\,).$

Now $c|bF_p(a)\implies c|F_p(a)$ because $\gcd (b,c)=1.$ So by $(\bullet)$ we have $a=(a')^c$ for some $a'\in \Bbb N.$ Hence, $d/e=a^{b/c}=((a')^c)^{b/c}=(a')^c\in \Bbb N.$

So if $a,b,c\in \Bbb N$ and $\gcd(b,c)=1$ then either $a^{b/c}$ is an integer or $a^{b/c}$ is irrational.

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  • $\begingroup$ Thanks a lot. But, can it be proved that at least one between a^(b/c) and (a+1)^(b/c) is irrational? $\endgroup$ – Andrea Frasca Jun 1 '20 at 12:50
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    $\begingroup$ Yes, if $\gcd(b,c)=1$ and $c>1.$ Because $a$ and $a+1$ cannot both be $c$-th powers of members of $\Bbb N....$ $\endgroup$ – DanielWainfleet Jun 1 '20 at 13:04
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No, it is not true always. For example, $a=4, b=5, c=2$, then $a^\frac{b}{c} = 32$ which is rational. However, if you cannot find any $x\in\mathbb{N}$ such that $x^c=a$, then $a^\frac{b}{c}$ is always irrational. In my example, $x$ was $2$.

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    $\begingroup$ Thanks! What if a and c are coprime? $\endgroup$ – Andrea Frasca Jun 1 '20 at 11:40
  • $\begingroup$ Ok, then take $a=9, b=5, c=2$. I was about to add something to the answer to get closer to what you may have meant but your comment made me to reply first. $\endgroup$ – Beyond Infinity Jun 1 '20 at 11:42

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