(b) Given, f = {(ab, a + b): a, b ∈ Z } Let a = 1, b = 0 ∴ab = 0 and a + b = 1 Again, let a = 0, b = 0 ∴ab = 0 and a + b = 1 Hence, we see that two different values of a and b we get same value, hence it is way one. It is clear that for different values of a and b, we get all values of Z in a + b. Hence, it is onto. Hence, it is surjective.