Prove that the cube root of 2, the number whose cube is 2, is irrational…

Step 1. Suppose the opposite. Suppose the cube root of $2$ is $\dfrac{a}{b}$, in lowest terms. Cubing, $a^3 = 2b^3$.

Step 2. 2 divides a. $a^3$ is even, so the prime $2$ divides $a$. Write $a = 2c$.

Step 3. Put it back. $8c^3 = 2b^3$, so $b^3 = 4c^3$.

Step 4. 2 divides b. $b^3$ is even, so $2$ divides $b$.

Step 5. The contradiction. $2$ divides both $a$ and $b$, which have no common factor.