Can any root, such as a square root or a cube root, be rational?

Question

I've heard of this and most roots are irrational such as $\sqrt{8}$ and $\sqrt[3]{25}$. So, can any of these roots be rational? I think so as I'm typing this. I think these are rational: $\sqrt[4]{256}$ and $\sqrt{144}$. I know that a rational number can be written as two fractions. Do the last two simplify to get a number that can be written as fractions? I think so. I think the third one simplifies to $4$ and the fourth one simplifies to $12$, while I think the first one simplifies to about 2.83 rounded to the nearest hundredth. Am I on the serious right track? I really hope this is a good train track that I'm standing on right here as I'm typing this.

Answer 1

Suppose $a$ and $b$ are integers with no bigger common divisors than $1$, and $m,n>1$ are integers. Suppose $$ \sqrt[n]{m} = \frac a b. $$ Then $$ b^n m = a^n. $$ Consider prime factorizations of $b$ and $a$: $$ \left(q_1^{e^1}\cdots q_k^{e_k}\right)^n m = p_1^{f_1}\cdots p_\ell^{f_\ell}. $$ Since none of $p_1,\ldots,p_\ell$ is a divisor of $\left(q_1^{e^1}\cdots q_k^{e_k}\right)^n$, we must have $p_1^{f_1}\cdots p_\ell^{f_\ell}$ dividing $m$, but that makes the left side bigger than the right side unless $b=1$. From that we conclude that $a/b$ is an integer. So it's only when the $n$th root is an integer that the $n$th root is rational.

Perhaps the hard part of the proof is what I haven't yet mentioned: prove that an integer cannot have more than one prime factorization. That was tacitly used above. (Maybe one can get by with a somewhat weaker result.)