How do we prove that $(\frac{2}{3})^{1/5}$ is irrational?

Question

How do we prove that $(\frac{2}{3})^{1/5}$ is irrational? What I'm doing is this:

I'm trying to prove by contradiction, so let's say $(\frac{2}{3})^{1/5}$ is rational. This means I can write it as quotient of two integers $\frac{x}{y}$.

$$\left(\frac{2}{3}\right)^{1/5} = \frac{x}{y} $$

$$\left(\frac{2}{3}\right) = \frac{x^5}{y^5}$$

I'm stuck here.

Answer 1

As u have written $\frac{2}{3} = \frac{x^5}{y^5}$, Now $2 y^5 = 3 x^5$. Do a prime factorization of $x,y$ and also $3 | y$ and hence $2 \times 3^{j5} \times b = 3 \times 3^{k5} \times c$ where $3$ doesnt divide $b,c$ with $j > 0$. Since prime factorization is unique, we have $5j = 5k+1$. A contradiction.

Answer 2

I don't know how far you are into learning proof techniques, but we could try a proof by contradiction here.

What we want to do is assume $\frac{2}{3}^{1/5}$ is rational, so $\frac{2}{3}^{1/5} = \frac{a}{b}$, where a and b only have a common divisor of 1 (this is the term we will contradict).

Why do we assume $\frac{a}{b}$ doesn't have a common divisor? Well, if they did then $\frac{a}{b}$ would equal a whole number!!

so lets rewrite that to this:

$b\sqrt[5]{2} = a\sqrt[5]{3}$, then multiply both sides by $2^5$ and $3^5$,

$(3^5)(b^5)(2) = (2^5)(a^5)(3)$, which can be rewritten as $2k(b^5) = 2w(a^5)$, $k, w \in \mathbb{Z}$ .

This means b and a are both even values, meaning a and b share a common divisor other than 1, therefore this is a contradiciton of our original statement!

(I am a new proof student, so please correct my mistakes if this is incorrect. Thank you!)