Let \( f : \mathbb{N} \to \mathbb{N} \) be defined by \( f(x) = x^2 + x + 1 \). Then \( f \) is:

Since \( f(x_1) = f(x_2) \implies x_1^2 + x_1 = x_2^2 + x_2 \implies (x_1 - x_2)(x_1 + x_2 + 1) = 0 \implies x_1 = x_2 \) for naturals, it is one-one. It is not onto since values like 2, 4, 5 have no pre-image in \( \mathbb{N} \).