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

Let's argue indirectly:

Assume that $\sqrt[3]{2}$ is rational. Then we can find integers $m$ and $n$ where $$\sqrt[3]{2} = \frac{m}{n} \quad \textrm{(which is in lowest terms)}$$ Cubing both sides, we find $$\begin{align}2 = \frac{m^3}{n^3} &\Rightarrow 2n^3 = m^3\\ &\Rightarrow 2 \mid m^3\\ &\Rightarrow 2 \mid m\\ &\Rightarrow m = 2k_1 \quad \textrm{for some integer } k_1\end{align}$$ Now, substituting our new expression for $m$ into a previous equation, we find $$\begin{align} 2n^3 = m^3 &\Rightarrow 2n^3 = (2k_1)^3\\ &\Rightarrow 2n^3 = 8{k_1}^3\\ &\Rightarrow n^3 = 2 \cdot 2{k_1}^3\\ &\Rightarrow 2 \mid n^3\\ &\Rightarrow 2 \mid n\end{align}$$ But this contradicts the assumption that we could find such a fraction $m/n$ in lowest terms (2 divides both $m$ and $n$).

Hence, our assumption must be rejected, and the opposite must be true: $\sqrt[3]{2}$ is irrational.