# Is the following constructed number always irrational?

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

• $2^{\frac{2}{1}}$ – Lee Jun 1 '20 at 11:38
• Sorry, a b c > 1 – Andrea Frasca Jun 1 '20 at 11:39
• $8^{4/3} = 16 \;$ – amWhy Jun 1 '20 at 11:40
• 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. – J.G. Jun 1 '20 at 11:41
• take for example any $a=q^2$, $c=2$, then $a^{\frac{b}{c}}=q^b$ – Lee Jun 1 '20 at 11:41

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.

• Thanks a lot. But, can it be proved that at least one between a^(b/c) and (a+1)^(b/c) is irrational? – Andrea Frasca Jun 1 '20 at 12:50
• Yes, if $\gcd(b,c)=1$ and $c>1.$ Because $a$ and $a+1$ cannot both be $c$-th powers of members of $\Bbb N....$ – DanielWainfleet Jun 1 '20 at 13:04

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 $$2$$.

• Thanks! What if a and c are coprime? – Andrea Frasca Jun 1 '20 at 11:40
• 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. – Beyond Infinity Jun 1 '20 at 11:42