# Prove that $\sqrt[3]{5} + \sqrt{2}$ is irrational

I tried with both squaring and cubing the statement, it got messy, here's my latest attempt:

Assume for the sake of contradiction: $$\sqrt[3]{5} + \sqrt{2}$$ is rational

$$\sqrt[3]{5} + \sqrt{2}$$ = $$\frac{a}{b}$$ $$a,b$$ are odd integers $$> 0$$ and $$b\neq 0$$

$${(\sqrt[3]{5} + \sqrt{2})}^3$$ = $$\frac{a^3}{b^3}$$

by multiplying by $$b^3$$:

$${(\sqrt[3]{5} + \sqrt{2})}^3 \times b^3$$ = $${a^3}$$

so: $$a^3$$ is divisible by $${(\sqrt[3]{5} + \sqrt{2})}^3$$ which means $$a$$ is divisible by $${(\sqrt[3]{5} + \sqrt{2})}$$

doing the same thing with $$b$$ i found :

$$\frac{a^3}{{(\sqrt[3]{5} + \sqrt{2})}^3}$$ = $${b^3}$$

so: $$b^3$$ is divisible by $${(\sqrt[3]{5} + \sqrt{2})}^3$$ which means $$b$$ is divisible by $${(\sqrt[3]{5} + \sqrt{2})}$$ (wrong)

$${(\sqrt[3]{5} + \sqrt{2})}$$ is a common divisor for both $$a$$ & $$b$$ which is a contradiction, thus $$\sqrt[3]{5} + \sqrt{2}$$ is irrational. (wrong)

• What does "divisible by $(\sqrt[3] 5 +\sqrt 2)$" mean? These aren't integers. – lulu Oct 28 '18 at 21:36
• I am not too sure how you got $(\sqrt[3] 5+\sqrt 2)^3\cdot a^3=b^3$ – Mohammad Zuhair Khan Oct 28 '18 at 21:38
• If $a, b > 0$ then there is no need to say that $b \ne 0$ – Rolazaro Azeveires Oct 30 '18 at 1:23
• There is no need to require both a and b to be odd. You only need them to be coprime. That is, e.g. $2/3$ is a valid rational number, in irreducible form, although 2 is even. – Rolazaro Azeveires Oct 30 '18 at 1:27

You can't do divisibility in irational and rational numbers. When you are operating with divisibility you have to have an integers. It is a relation defined on integer numbers.

Suppose it is rational, then exist rational number $$q$$ such that $$\sqrt[3]{5} + \sqrt{2}= q$$ so $$5 = (q-\sqrt{2})^3 = q^3-3q^2\sqrt{2}+6q-2\sqrt{2}$$

So we have $$\sqrt{2}(\underbrace{3q^2+2}_{\in\mathbb{Q}}) = \underbrace{q^3+6q-5}_{\in\mathbb{Q}}$$

so $$\sqrt{2}= \underbrace{q^3+6q-5\over 3q^2+2}_{\in\mathbb{Q}}$$

• There is a proof-verification tag. – Ennar Oct 28 '18 at 21:41
Assume that $$\sqrt[3]{5} + \sqrt{2}=r$$ where r is a rational number.
We have $$\sqrt[3]{5} =r-\sqrt{2}$$
Raise to the third power to get $$5=r^3-3r^2 \sqrt 2 +6r - 2\sqrt 2$$ Solving for $$\sqrt 2$$ we get $$\sqrt 2 = \frac {5-r^3-6r}{-3r^2-2}$$
The $$RHS$$ is a rational number which is impossible.