Answer:
Suppose that √3 is rational. Then
√3 = a/b, where a and b are relatively prime integers with b ≠ 0. It follows that
3 = a²/b², so 3b² = a². Since 3 | a², it follows that 3 | a. Let a = 3c, so 3b² = 9c². We then have b² = 3c². Hence, 3 | b², so
3 | b. However since a and b are relatively prime, 3 cannot divide both a and b. This contradiction proves that √3 is an irrational number.