# Is $a\sqrt[3]{2} + b\sqrt[3]{4}$ irrational?

I need to prove that $$a\sqrt[3]{2} + b\sqrt[3]{4}$$ is irrational, while $a$,$b$ are non zero rationals.

I know that $\sqrt[3]{2} + \sqrt[3]{4}$ is irrational and I also know how to prove it, but I can't think of any reasonable implication that would state: $a\sqrt[3]{2} + b\sqrt[3]{4}$ is rational $\implies$ $\sqrt[3]{2} + \sqrt[3]{4}$ is rational, so I could show it is a contradiction.

• It is not always irrational; consider $a = b = 0$. – Megadeth Nov 23 '15 at 13:05
• Are their conditions imposed on the values of $a,b$ i.e. are $a,b$ rational? – SchrodingersCat Nov 23 '15 at 13:06
• sorry...i corrected it – Filip Sulik Nov 23 '15 at 13:07
• yeas a,b are rational – Filip Sulik Nov 23 '15 at 13:09
• – Martin R Nov 23 '15 at 13:12

If $b\ne0$, then $a\sqrt[3]{2} + b\sqrt[3]{4}$ is rational iff $\sqrt[3]{2}$ is a root of a quadratic polynomial over $\mathbb Q$.

$\sqrt[3]{2}$ is a root of $X^3-2$, which is irreducible over $\mathbb Q$.

So, every polynomial having $\sqrt[3]{2}$ as root must be a multiple of $X^3-2$. In particular, $\sqrt[3]{2}$ is not a root of a quadratic polynomial over $\mathbb Q$.

Therefore, if $a\sqrt[3]{2} + b\sqrt[3]{4}$ is rational, then $b=0$. But $\sqrt[3]{2}$ is irrational, and so $a=0$.

All this can be summarized as:

$1, \sqrt[3]{2}, \sqrt[3]{4}$ are linearly independent over $\mathbb Q$ and so the only way that $a\sqrt[3]{2} + b\sqrt[3]{4}$ is rational is when $a=b=0$.

• That is the thing...I need to prove that they really are linearly independent...thats why am trying to prove this. – Filip Sulik Nov 23 '15 at 13:20
• Could you please explain it to me little further ? – Filip Sulik Nov 23 '15 at 13:25
• thanks a lot....this helped – Filip Sulik Nov 23 '15 at 13:35

Assume $a\sqrt[3]2+b\sqrt[3]4$ is rational. This gives that $$(a\sqrt[3]2+b\sqrt[3]4)^2=2b^2\sqrt[3]2+a^2\sqrt[3]4+4ab$$is also rational. Subtracting $4ab$ doesn't change rationality. This means that $$a(2b^2\sqrt[3]2+a^2\sqrt[3]4)-2b^2(a\sqrt[3]2+b\sqrt[3]4)=(a^3-2b^3)\sqrt[3]4$$ is rational. But $\sqrt[3]4$ is irrational, so that can only be true if $a^3=2b^3$, which, under the assumption that $b\neq0$, implies that $\frac ab=\sqrt[3]2$ is rational. This is clearly a contradiction, and we are done.

Set $\alpha=\sqrt[3]{2}$ and note that $\{1, \alpha, \alpha^{2} \}$ is linearly independent over $\mathbb{Q}$.You will contradict this if you proceed by contradiction.

Suppose $a\sqrt[3]{2} + b\sqrt[3]{4}$ is rational, ($=y$, say)

So $a\sqrt[3]{2} + b\sqrt[3]{4} = y$ and
$a\sqrt[3]{2} = y - b\sqrt[3]{4}$

Cubing both sides, we have $$(a\sqrt[3]{2})^3 = (y - b\sqrt[3]{4})^3$$ $$2a^3=y^3-4b^3 + 3y\cdot 4^{\frac{2}{3}} - 3y^2\cdot 4^{\frac{1}{3}}$$

Equating rational and irrational parts, we have $$3y\cdot 4^{\frac{2}{3}} - 3y^2\cdot 4^{\frac{1}{3}}=0$$ or, $$3y \cdot 4^{\frac{1}{3}}\left[4^{\frac{1}{3}} - y\right] = 0$$

Now our first assumption that $y$ is rational holds iff $y=0$

But then $\frac{a}{b}=-\frac{\sqrt[3]{4}}{\sqrt[3]{2}}$ implies $a,b$ irrational.