HCU MCA 2013 Paper

(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.