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Question
Check the injectivity and surjectivity of the following function:
f: Z → Z given by f(x) = x2
Solution
f: Z → Z is given by,
f(x) = x2
It is seen that f(−1) = f(1) = 1, but −1 ≠ 1.
∴ f is not injective.
Now,−2 ∈ Z. But there does not exist any element x ∈ Z such that f(x) = x2 = −2.
∴ f is not surjective.
Hence, function f is neither injective nor surjective.
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