6 is not a prime, but √6 is still irrational. Prove it by contradiction, using only the prime 2.

Step 1. Suppose the opposite. Suppose $\sqrt{6} = \dfrac{a}{b}$ in lowest terms. Then $a^2 = 6b^2$.

Step 2. 2 divides a. $a^2 = 2 \times 3b^2$ is even, and $2$ is a prime, so $2$ divides $a$. Write $a = 2c$.

Step 3. Put it back. $4c^2 = 6b^2$, so $2c^2 = 3b^2$.

Step 4. 2 divides b. The left side is even, so $3b^2$ is even. The prime $2$ does not divide $3$, so it divides $b^2$, and then it divides $b$.

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