Q. 2 B3.6( 49 Votes )

# Check the injecti

Answer :

It is given that f : Z → Z given by f (x) = x^{2}

We can see that f(-1) = f(1) = 1, but -1 ≠ 1

⇒ f is not injective.

Now, let -2 ϵ Z. But, we can see that there does not exists any x in Z such that

f(x) = x^{2} = -2

⇒ f is not surjective.

Therefore, function f is neither injective nor surjective.

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