4

Prove that $\sqrt[3]{5-\sqrt{2}}$ is not a rational number

My attempt:

Consider the polynomial $ (x^3-5)^2 - 2 = x^6 -10x^3 + 23 = 0 $. By the rational root theorem, we can conclude that $ \pm 1$ and $ \pm 23 $ are the only possible rational solutions*. Since none of those are solutions and $ \sqrt[3]{5 - \sqrt{2}} $ is, it follows that $ \sqrt[3]{5-\sqrt{2}} $ is not a rational number.

Is this proof correct? What are some other ways of proving this?

*The case of $\pm 1$ is trivial. That of $\pm 23$ is not; I could factor out 23s to make it easy to do by hand. $23^6 - 10*23^3 + 23 = 23(23^2(23^3-10)+1)$. Since $23^3>10$, it follows that the function does equal $0$ for $23$. $-23$ would be the same except that we would replace $-10$ with $+10$.

Bart Michels
  • 26,355
William Muenzinger
  • 716
  • 4
    You want $(x^3-5)^2-2$ (according to the title). Also, how do you know that $\sqrt[3]{5-\sqrt{2}}$ isn't $2$, or any of these others? – vadim123 Jan 06 '14 at 15:36
  • The title was a typo, but we'll go with it because that's what people responded to. Edit: Ok i've modified my proof to fit the titular problem. – William Muenzinger Jan 06 '14 at 17:10
  • The case of $\pm 1$ is trivial. That of $\pm 23$ is not; I could factor out 23s to make it easy to do by hand. $23^6 - 10*23^3 + 23 = 23(23^2(23^3-10)+1)$. Since $23^3>10$, it follows that the function does equal 0 for 23. -23 would be the same except that we would replace -10 with +10. – William Muenzinger Jan 06 '14 at 17:24

3 Answers3

18

Let $\displaystyle\sqrt[3]{5-\sqrt2}=a$ where $a$ is rational

Cubing either sides, $\displaystyle5-\sqrt2=a^3\iff 5-a^3=\sqrt2$ which is irrational unlike the Left Hand Side

Community
  • 1
lab bhattacharjee
  • 274,582
8

Hint $\ x^3\not\in \Bbb Q\Rightarrow x\not\in\Bbb Q,\,$ since $\,x\in \Bbb Q\Rightarrow x^3\in\Bbb Q$

Bill Dubuque
  • 272,048
0

A simple, algebraic consideration: Let $\alpha=\sqrt[3]{(5-\sqrt2)}.$ Then $\mathbb Q(\alpha)$ contains $\sqrt2,$ so $[\mathbb Q(\alpha):\mathbb Q]\gt1,$ i.e. $\alpha$ is irrational.
In essence this is nothing than the answer by Bill, but expressed more abstracly.

awllower
  • 16,536