Question

2. Check the injectivity and surjectivity of the following functions:
(i) f:N → N given by f(x) = x²
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Answer 1

f : N → N given by f (x) = x²

We can see that for x, y ϵ N,

f(x) = f(y)

⇒ x² = y²

⇒ x = y

⇒ f is injective.

Now, let 2 ϵ N. But, we can see that there does not exists any x in N such that

f(x) = x² = 2, therefore, f is not surjective.

Therefore, function f is injective but not surjective.

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