Worked in full: √3 is irrational

Assume √3 = a/b in lowest terms. Then 3b² = a², so 3 | a² and (Theorem 1.2) 3 | a. Write a = 3c: 3b² = 9c² ⇒ b² = 3c², so 3 | b too. Both share 3 — contradiction. So √3 is irrational.