Consider the function f: N → N, where N is the set of natural numbers, defined by f(n) = n²+n+1. Show that the function f is one-one but not onto.

Ans: To prove that f is one to one, it is needed to prove that for any two integers n and m, if f(n) = f(m) after that n = m.

f(n) = f(m) ⇔ n² + n + 1 = m² + m + 1

⇔ n² + n = m² + m

⇔ n(n + 1) = m(m + 1)

⇔ n = m.

As product of consecutive natural numbers begining from m and n are equal iff m = n. Next f is not onto as for any n (odd or even) n² + n + 1 is odd. This entails that there are even elements in N that are not image of any element in N.