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Question
Check the injectivity and surjectivity of the following function.
f : R → R given by f(x) = x2
Solution
f : R → R given by f(x) = x2
f(x1) = f(x2), x1, x2 ∈ R
∴ x12 = x22
∴ x12 – x22 = 0
∴ (x1 + x2) (x1 – x2) = 0
∴ x1 = x2 or x1 = – x2
∴ f is not injective.
Therefore all negative integers of codomain are not images under f.
∴ f is not surjective.
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