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

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

My attempt: Suppose $$\sqrt{2} +\sqrt[3]{2}$$ is rational then $$\exists$$ $$x\in \mathbb{Q}$$ such that $$\sqrt{2} +\sqrt[3]{2}=x$$ Rewriting the above equation as $$x-\sqrt{2}=\sqrt[3]{2}$$ cubing the above equation gives us $$x^3-3x^2\sqrt{2}+6x-2\sqrt{2}=2$$ This implies that $$\sqrt{2}=\frac{x^3+6x-2}{3x^2+2}\in\mathbb{Q}$$ but this is absurd . Therefore ,$$\sqrt{2} +\sqrt[3]{2}$$ is irrational. Does this look good ? Can this be done any other way.

This question is from chapter 2 of Spivak's Calculus. In there he's given a hint which says "start by working with the 6th power " of this expression.. I don't see how that helps . Can you show me how it can be done using the hint the Author has provided ? Thank you

• Maybe they want you to find a certain 6-degree polynomial of which $x$ is a root, and then to show it has no rational roots? – Ivan Neretin Oct 9 '19 at 8:54
• @IvanNeretin You mean, like $x^6-6x^4-4x^3+12x^2-24x-4=0$? But this has already been done here at this site. – Dietrich Burde Oct 9 '19 at 9:04
• @DietrichBurde That makes this question a duplicate. – Ivan Neretin Oct 9 '19 at 9:08
• Well, yes, but on the other hand it is about proof verification in part $1$, and in part $2$ it asks "Can this be done any other way". So I am not sure if it is an exact duplicate. – Dietrich Burde Oct 9 '19 at 9:10

The OP shows great ingenuity with their (now fixed up) answer.

You can also go with Spivak's hint:

From Spivak's Exercise 18.a we know that if $$u$$ satisfies

$$\tag 1 u^n + a_{n-1}u^{n-1} + \dots + a_0 = 0$$

for integers $$a_{n-1}, \dots, a_0$$, then $$u$$ is either irrational or an integer.

Let

$$\quad u = 2^{\frac{2}{6}} + 2^{\frac{3}{6}}$$

It can be show that $$u$$ isn't an integer (see next section).

Moreover, it can be shown (by solving a system of linear equations) that

$$\tag 2 u^6 -6u^4-4u^3+12u^2 -24u - 4 = 0$$

Since

$$\quad \sqrt{2} \lt \frac{3}{2}$$

$$\quad \sqrt[3]{2} \lt \frac{4}{3}$$

we can write

$$\quad \sqrt{2} + \sqrt[3]{2} \lt \frac{3}{2} + \frac{4}{3} \lt 3$$

Since

$$\quad \sqrt{2} \gt \frac{5}{4}$$

$$\quad \sqrt[3]{2} \gt \frac{5}{4}$$

we can write

$$\quad \sqrt{2} + \sqrt[3]{2} \gt \frac{5}{4} + \frac{5}{4} \gt 2$$

So $$\sqrt{2} + \sqrt[3]{2}$$ can't be an integer.

You've got the right idea, but there are two (small) problems:

• the equation $$x^3-3x^2\sqrt{2}+6x+2\sqrt{2}=2$$ is wrong, the last sign on the left should be $$-$$, as in $$(-1)^3=-1$$.

• you should explain why it is possible to divide by $$3x^2-2$$. In fact, once you fix the first point I mentioned, it's going to be $$3x^2+2$$ and it's going to be much easier.

• @Arnauld Mortier , Thanks for pointing out . – user655800 Oct 9 '19 at 8:56

We can use field theory to conclude that the degree $$[\Bbb Q(\sqrt{2}+\sqrt[3]{2}):\Bbb Q]=6>1.$$

The minimal polynomial of $$\sqrt{2}+\sqrt[3]{2}$$ is $$x^6-6x^4-4x^3+12x^2-24x-4=0,$$ see the first link.

References:

Finding the minimal polynomial of $\sqrt 2 + \sqrt[3] 2$ over $\mathbb Q$.

Finding a basis for the field extension $\mathbb{Q}(\sqrt{2}+\sqrt[3]{4})$

The first reference shows in a very elementary way that $$\sqrt{2}\in \Bbb Q(\sqrt{2}+\sqrt[3]{2})$$. Hence $$\sqrt{2}+\sqrt[3]{2}$$ is irrational.

• Haven't learned abstract algebra yet... – user655800 Oct 9 '19 at 8:57
• thanks for that first link. – user655800 Oct 9 '19 at 9:00
• I see. You have asked "Can this be done any other way", so I thought that this is another way, even if you haven't seen it - but this could be a good motivation. – Dietrich Burde Oct 9 '19 at 9:03