Show that $\sqrt[3]{2} + \sqrt[3]{4}$ is irrational

Question

Our professor asked us this to prove that

$$ \sqrt[3]{2} + \sqrt[3]{4} \notin \Bbb Q. $$

I know how to prove each one separately that it is irrational, but when it comes to summing two irrational numbers its not certain what the result will be.

How do I solve it in a more certain way?

Answer 1

Let us rewrite $$\sqrt[3]2+\sqrt[3]4=\color{blue}{1+\sqrt[3]2+\sqrt[3]4}-1=\color{blue}{\frac{\left(\sqrt[3]2\right)^3-1}{\sqrt[3]2-1}}-1=\frac{1}{\sqrt[3]2-1}-1.$$ If the left side were rational, what could we say about $\sqrt[3]2$ ?

Answer 2

Hint: Find a polynomial with integer coefficients for which $\alpha = \sqrt[3]{2} + \sqrt[3]{4}$ is a root. Then you can use the Rational Roots Theorem.

Answer 3

Hint. If $\sqrt[3]{2}+\sqrt[3]{4}$ rational, so is $1+\sqrt[3]{2}+\sqrt[3]{4}$, and so is $$ \frac{1}{1+\sqrt[3]{2}+\sqrt[3]{4}}=\frac{(\sqrt[3]{2})^3-1}{(\sqrt[3]{2})^2+\sqrt[3]{2}+1}=\sqrt[3]{2}-1 $$ and hence so is $\sqrt[3]{2}$.

Answer 4

Suppose the sum is a rational number $r$, then $\sqrt[3]{4} = -\sqrt[3]{2} + r$, and cubing both sides: $4 = r^3 - 3r^2a + 3ra^2 - 2$, with $a = \sqrt[3]{2}$. Thus: $6 - r^3 = -3r^2a + 3r(r-a) \to a = \dfrac{6-r^3-3r^2}{-3r^2-3r} \in \mathbb{Q}$. But it can be proved that $a = \sqrt[3]{2} \notin \mathbb{Q}$, a contradiction, and we are done.

Answer 5

Write $\sqrt[3]{2} + \sqrt[3]{4} = \alpha + \alpha^2$, with $\alpha=\sqrt[3]{2}$.

Suppose $\alpha + \alpha^2 = q \in \Bbb Q$. Then $\alpha^2 + \alpha^3 = q \alpha$. Since $\alpha^3=2$, we get

$\quad\alpha + \alpha^2 = q$

$\quad\alpha^2 + 2= q \alpha$

Subtracting these, we get $\alpha-2=q-q \alpha$, and so $\alpha=\frac{q+2}{q+1}\in \Bbb Q$, a contradiction since $\alpha$ is irrational.

This argument is essentially dividing the polynomial $x^3-2$ by $x^2+x-q$, getting remainder $(q+1)x-(q+2)$.