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

Question

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)

Answer 1

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.

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

A contradiction.

Answer 2

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.