I'm not quite sure I understand this one. Show that the specified real number is rational: $7^{2/3}$

Question

This is the first problem in this discrete math assignment, and I'm a little bit confused because I thought that the square root, cube root, nth root of a non-square, non-cube, etc. were not rational numbers. Is that untrue, and if so I don't even know how to start working on this one. Just as a reference, this problem is from the book Discrete Mathematics and Applications by Kevin Ferland p. 175 #2. !

Answer 1

Proof that the number $7^\frac{2}{3}$ is irrational:

Suppose otherwise, i.e. that $7^\frac{2}{3} = \frac{a}{b}$ where $a, b$ are integers such that $\gcd(a, b) = 1$. Then we may write

$$49b^3 = a^3$$

Which implies $7 | a \iff a = 7k, k \in \mathbb{Z} \implies 49b^3 = 343k^3 \iff b^3 = 7k^3 \implies 7 | b$. That is, both $a$ and $b$ are divisible by $7$, which contradicts our supposition $\gcd(a,b) =1$ $\square$.