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Question
Check the injectivity and surjectivity of the following function.
f : R → R given by f(x) = x3
Solution
f : R → R given by f(x) = x3
Let x13 = x23
∴ x13 – x23 = 0
∴ `(x_1 - x_2) underbrace((x_1^2 + x_1 x_2 + x_2^2))_(> 0 "for all" "x"_1, "x"_2 "as it's discriminant" < 0)` = 0
∴ x1 = x2
∴ f is injective.
Let y = x3
∴ x = `y^(1/3)`
∴ For every y ∈ R, there is some x ∈ R
∴ f is surjective.
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