1b) $\displaystyle \sqrt[3]{2}$ is an irrational number.

My proof(by contradiction):

$\displaystyle \sqrt[3]{2}$ is a positive number such that its cube is 2.

Assume $\displaystyle \sqrt[3]{2}$ is rational.

$\displaystyle \exists$ integers P, Q such that $\displaystyle \sqrt[3]{2}$ = $\displaystyle \frac{P}{Q}$, fully simplified.

$\displaystyle (\sqrt[3]{2})^3 = (\frac{P}{Q})^3 \rightarrow 2 = \frac{P^3}{Q^3} \rightarrow 2Q^3 = P^3$

(Corollary: If $\displaystyle n^3$ is even, then n is even. (proven in 1a))

An integer n is even if there exists an integer k such that n = 2k.

$\displaystyle 2Q^3 = (2k)^3 = 8k^3 \rightarrow Q^3 = 4k^3 = 2(2k^3)$

$\displaystyle 2k^3$ is an integer, therefore Q is even, along with P.

Since P and Q are even, they share a common factor. Since we defined that $\displaystyle \frac{P}{Q} $ was fully simplified, we have a contradiction.