0
$\begingroup$

Prove that $\sqrt[3]{2} + \sqrt{2}$ is irrational

This question is from Spivak’s book 18 of the third edition, but the solution is not intuitive. It suggests you to work out the first six powers of this number. Are there any more elegant approaches to this proof?

$\endgroup$
  • $\begingroup$ See homework-help, or mathhelpforum. $\endgroup$ – Dietrich Burde Sep 26 '19 at 9:26
  • $\begingroup$ See Prove $2^{1/3} + 2^{1/2}$ is irrational. $\endgroup$ – Dietrich Burde Sep 26 '19 at 9:27
  • $\begingroup$ @DietrichBurde Can you explain how these are similar questions? One only has cubed roots, my question has both square roots and cubed roots. In particular, the trick shown by kingW3 does not work in my case. $\endgroup$ – Snowball Sep 26 '19 at 20:33
  • $\begingroup$ I'm unable to post the answer, but assuming $\sqrt[3]{2} + \sqrt{2} = \frac{p}{q}$, where $p,q \in \mathbb{Z}^+$, and then isolating $\sqrt[3]{2}q$ on one side, and then taking the cube of both sides, should provide an obvious contradiction with terms involving $\sqrt{2}$ being irrational, and others being rational $\endgroup$ – Snowball Sep 26 '19 at 21:58
1
$\begingroup$

Let $y=2^{1/6}$ and $x=2^{1/3} + 2^{1/2}=y^2+y^3$. Suppose $x \in \mathbb Q$.

Since $y^3=x-y^2$, any power of $y$ can be expressed as a quadratic in $y$ over $\mathbb Q(x)=\mathbb Q$. In particular, from $y^6=2$ we obtain a quadratic equation for $y$ and so we have the contradiction that $[\mathbb Q(y):\mathbb Q]=2$.

| cite | improve this answer | |
$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.