Check the injectivity and surjectivity of the following functions:
$(i) f : Z \to Z$ given by $f (x) = x^{3}$

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Solution

$f : Z \to Z$ is given by, $f (x) = x^{3}$
It is seen that for x, $y \in Z, f (x) = f (y) \Rightarrow x^{3} = y^{3} \Rightarrow x = y$
Therefore, f is injective.
Now, $2 \in Z$ . But, there does not exist any element x in domain Z such that $f (x) = x^{3} = 2$ . Therefore, f is not surjective.
Hence, function f is injective but not surjective.

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