Question

On the set of integers $Z$, define $f:Z\to Z$ as

$f\left(n\right)=\left\{\begin{array}{l}\frac{n}{2},n\text{is even}\\ 0,n\text{is odd}\end{array}\right.$

Then $f$ is

• A. injective but not surjective
• B. neither injective nor surjective
• C. surjective but not injective
• D. bijective

Answer 1

The correct option is C

surjective but not injective

Explanation of correct option:

Finding $f$:

Given

$f\left(n\right)=\left\{\begin{array}{l}\frac{n}{2},n\text{is even}\\ 0,n\text{is odd}\end{array}\right.$

Thus every odd value of $n$ yields zero i.e. $f\left(1\right)=f\left(3\right)=\cdots =f(2n+1)=0$, for all $n\in Z$.

So, it is a many-one function.

Hence, the function cannot be injective.

Let $n\in Z$ be any nonzero integer. Then $\exists 2n\in Z$ such that $f\left(2n\right)=n$.

For $0\in Z$, $\exists (2n+1)\in Z$ such that $f(2n+1)=0$, for all $n\in Z$.

So, the function is surjective.

Hence, option (C) is the correct answer.