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Question
Check the injectivity and surjectivity of the following function:
f : N → N given by f(x) = x2
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Solution
f : N → N given by f(x) = x2
Injectivity:
Suppose f(x1) = f(x2)
⇒ `x_1^2 = x_2^2`
⇒ x1 = x2 ...(because x1, x2 ∈ N)
∴ f is one-one (injective).
Surjectivity:
There are many elements in the codomain N which have no pre-image in the domain N.
For example, 3 ∈ N is an element of the codomain, but for f(x) = x2 there is no x ∈ N for which f(x) = 3.
∴ f is not onto (surjective).
Hence, f is injective but not surjective.
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