Question

Let $f:N\to N$ be defined by $f\left(n\right)=\{\begin{array}{cc}\frac{n+1}{2},& \text{is n is odd}\\ \frac{n}{2},& \text{is n is even}\end{array}$ for all $n\in N$. State whether the function $f$ is bijective. Justify your answer.

Answer 1

Solve for one-one.
$\{\begin{array}{cc}\frac{n+1}{2},& \text{is n is odd}\\ \frac{n}{2},& \text{is n is even}\end{array}$, for all $n\in N$
Now for one-one.
$f\left(1\right)=\frac{n+1}{2}=\frac{1+1}{2}=1$ (since $1$ is odd)
$f\left(2\right)=\frac{2}{2}=1$ (since $2$ is even)
Since, $f\left(1\right)=f\left(2\right)$
but $1\ne 2$
both $f\left(1\right)\&f\left(2\right)$ have same image $1$
$f$ is not one-one.

Solve for onto.
$f\left(n\right)=\{\begin{array}{cc}\frac{n+1}{2},& \text{is n is odd}\\ \frac{n}{2},& \text{is n is even}\end{array}$ for all $n\in N$
When $n$ is odd,
$n=2k-1$ for $k\in N$
$f(2k-1)=\frac{2k-1+1}{2}=k$
When $n$ is even
$n=2k$ for some $k\in N$ such that
$f\left(2\right)=\frac{2k}{2}=k$
Thus, for every $n\in N,f\left(n\right)\in N$
Therefore, $f$ is onto.
Hence, $f$ is onto but not bijective function.