Check the injectivity and surjectivity of the following functions:
$(i) f : Z \to Z$ given by $f (x) = x^{2}$

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Solution

$f : R \to R$ is given by $f (x) = x^{2}$ . It is seen that f(-1)=f(1)=1 but $- 1 \neq 1,$
Therefore, f is not injective.
Now, $- 2 \in R$ . But there does not exist any element $x \in Z$ such that $f (x) = x^{2} = - 2$
Therefore, f is not surjective.
Hence, function f is neither injective nor surjective.

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