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.

Did you try looking for some examples/counterexamples?
Edit: Is there any reason this is tagged "discrete-mathematics"? — LeviathanTheEsper, Jul 07 '15 at 17:58
Instead of editing to "Donotanswer $\times\infty$", once your question has been answered simply accept an answer from those given below to designate the question as "answered" so that your question can be helpful to other users in the future. Editing out your question does not help future users. — JMoravitz, Jul 07 '15 at 18:08

score 4 · Answer 1 · edited Nov 06 '21 at 23:28

4

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.

edited Nov 06 '21 at 23:28

Neel Sandell

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answered Jul 07 '15 at 17:53

Tryss

14,310

$x^3$ irrational does not imply $x$ is rational, you have $\sqrt{2}^3=2\sqrt{2}$ and both irrational. – LeviathanTheEsper Jul 07 '15 at 17:57
Sorry, it was a misstype, it was obviously x^3 irrationnal imply x irrationnal – Tryss Jul 07 '15 at 17:59
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@Tryss, spelling nitpick: "rational" and "irrational" with one n each. – hmakholm left over Monica Jul 07 '15 at 18:00
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@HenningMakholm : I'm french, and in french the similar word has two n, so when I'm not carefull (like here), you can bet that I'll make this mistake :'( – Tryss Jul 07 '15 at 18:03
Imagine myself, that the word in my language is "Irracional". Some years ago I used to write it like that in english. – LeviathanTheEsper Jul 07 '15 at 18:05

score 2 · Answer 2 · answered Jul 07 '15 at 17:53

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

answered Jul 07 '15 at 17:53

RK01

803

score 1 · Answer 3 · edited Apr 13 '17 at 12:21

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.

score 1 · Answer 4 · answered Jul 07 '15 at 18:02

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.

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

4 Answers4