Prove by contradiction that sqrt(3) is irrational. (5 marks)

To the contrary assume that sqrt(3) is a rational number. Thus we can write sqrt(3) = a/b where a and b are coprime integers and b is non-zero. (1 mark for this or equivalent statement - may have that this fraction is irreducible)
By rearrangement we get that b * sqrt(3) = a, and therefore that 3b² = a². (1 mark)
Therefore 3 divides a² and, as 3 is prime, 3 divides a, so we can write a = 3k where k is an integer. (1 mark)
By substitution we then get that 3b²=(3k)²=9k² so b² = 3k². (1 mark)
Therefore 3 divides b² and, as 3 is prime, 3 divides b. However, 3 also divides a so a and b are not coprime (alternatively may have that a/b is not irreducible). Hence we have a contradiction, so sqrt(3) can not be rational and must be irrational. (1 mark for this or equivalent statement - must use either irreducible or coprime consistently throughout)