Question

Let $Z$ be the set of integers and $f:Z\to Z$ be defined as $f\left(x\right)=x^2,x\in Z$, then function is

• A. bijection
• B. injection
• C. surjection
• D. none of these

Answer 1

Answer: (D)

none of these

As $x=1$ and $x=-1$ both having same image that is 1 hence $f\left(x\right)=x^2$ is not injective.

Also as given domain is $Z$ and there is no preimage for $x=-1$ in $Z$ hence given function $f\left(x\right)=x^2$ is not surjective.

Finally as given function is neither injective nor surjective hence obviously it is not bijective.