The function is mapped from natural numbers to integer numbers

Conclusion: the function is neither injective nor surjective (hence not bijective).

Injectivity check:

• Compute f(1) = 1^2 - 3·1 + 3 = 1 and f(2) = 2^2 - 3·2 + 3 = 1.

• Since f(1) = f(2) but 1 ≠ 2, the function is not injective.

Surjectivity check:

• To be surjective onto the integers, every integer must be an output. Consider the integer 0 and solve n^2 - 3n + 3 = 0.

• The discriminant is Δ = (−3)^2 − 4·1·3 = 9 − 12 = −3 < 0, so there is no real (and hence no natural) solution producing 0 as an output.

• Therefore 0 is not in the image, so the function is not surjective onto Z.

Since the function fails both injectivity and surjectivity, it is not bijective.