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Question
Check the injectivity and surjectivity of the following function:
f: N → N given by f(x) = x2
Solution
f: Z → Z given by f(x) = x2
Z = {O, ±1, ±2, ±3,...}
(a) f : Z → Z
Let -1, 1 ∈ Z, f (-1) = f(1)
⇒ 1 = 1
But -1 ≠ 1 ∴f is not one-on-one, i.e., f is not injective.
(b) There are many such elements that belong to co-domain but have no pre-image in its domain z.
e.g., 2 ∈ Z (co-domain). But `2^(1//2) != Z` (domain)
∴ Element 2 has no pre-image in its domain Z.
f is not onto i.e., f is not surjective.
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