Is the cubed root of x irrational if and only if x is irrational?

Question

Is the cubed root of x irrational if and only if x is irrational? Hoping for simple answers. Thank you very much.

Answer 1

No. Most numbers' cube roots are irrational, including but not limited to 2, 3, 4, 5...

Answer 2

We have that whenever $x^3$ is irrational, $x$ is also irrational, but you don't have the converse:

For example, $\sqrt[3]{2}$ is irrational, but $2$ is rational.

Answer 3

This breaks down into two statements:

• $\sqrt[3]{x}\in\mathbb{Q}\Rightarrow x\in\mathbb{Q}$

• $x\in\mathbb{Q}\Rightarrow \sqrt[3]{x}\in\mathbb{Q}$

The first statement is true, but the second isn't. To see why the forward implication is true, suppose that $\sqrt[3]{x}$ is rational, then it can be written as some $\frac{p}{q}$ with $p,q\in\mathbb{Z}$ and $q\neq 0$. Then $x=(\sqrt[3]{x})^3=(\frac{p}{q})^3=\frac{p^3}{q^3}$ is also written as a ratio of two integers (with $q^3\neq 0$) and is therefore rational.

For the reverse implication, you can prove that $\sqrt[3]{2}$ is an irrational number using the same technique as proving that $\sqrt{2}$ is irrational. (See for example this question or this wiki-page), giving an example of when $x\in\mathbb{Q}$ but $\sqrt[3]{x}\not\in\mathbb{Q}$, showing that the reverse implication is false.

Answer 4

Theorem: $\sqrt[n]{a}$ for $a\in\Bbb Z$ and $n\in\Bbb Z_{\ge 2}$, if real, is either an integer or irrational.

Proof: If it's irrational, we're done. Otherwise $\sqrt[n]{a}=\frac{k}{l}$ for some coprime $k,l\in\Bbb Z$.

Then $al^n=k^n$. Then $l=1$, since if not, then $p\mid l\,\Rightarrow\, p\mid k^n$, by Euclid's lemma $p\mid k$, so $\gcd(k,l)\ge p$, contradicts $\gcd(k,l)=1$. QED

So $\sqrt[3]{a}$ for an integer $a$, if not an integer, is always irrational. So $\sqrt[3]{2},\sqrt[3]{3}$, etc. are irrational.