Prove that √7 is irrational. Give a reason for every step.

Step 1. Suppose the opposite. Suppose $\sqrt{7} = \dfrac{a}{b}$, with whole numbers $a$ and $b$ that have no common factor.

Step 2. Square. $7 = \dfrac{a^2}{b^2}$, so $a^2 = 7b^2$.

Step 3. 7 divides a. $7$ divides $a^2$, and $7$ is a prime, so $7$ divides $a$. Write $a = 7c$.

Step 4. Put it back. $49c^2 = 7b^2$, so $b^2 = 7c^2$. Then $7$ divides $b^2$, so $7$ divides $b$.

Step 5. The contradiction. $7$ divides both $a$ and $b$, but they have no common factor.