Q.Let $f: \mathbf{N} \rightarrow \mathbf{N}$ be defined by $f(n) = \begin{cases} \frac{n+1}{2}, \text{ if } n \text{ is odd} \\ \frac{n}{2}, \text{ if } n \text{ is even} \end{cases}$ for all $n \in \mathbf{N}$. State whether the function $f$ is bijective. Justify your answer.

Concept: One-One (Injective) Function — each element of the domain maps to a distinct element in the codomain.

Step 1 — Check injectivity:

Take two different inputs, say $n=1$ (odd) and $n=2$ (even).

$f(1) = \frac{1+1}{2} = 1$, and $f(2) = \frac{2}{2} = 1$.

So $f(1) = f(2)$ but $1 \neq 2$. Hence $f$ is not one-one.

Step 2 — Check surjectivity:

For any $y \in \mathbf{N}$, we can find a preimage:

If $y$ is odd, take $n = 2y-1$ (odd) → $f(2y-1) = y$.

If $y$ is even, take $n = 2y$ (even) → $f(2y) = y$. …

First, see what $f$ does. For an odd number $n = 2k-1$:

For an even number $n = 2k$:

So each consecutive pair $\{2k-1,\ 2k\}$ maps to the same value $k$; in short, $f(n) = \left\lceil \tfrac{n}{2} \right\rceil$ (e.g. $f(1)=1,\ f(2)=1,\ f(3)=2,\ f(4)=2,\dots$).

1. Is $f$ one-one (injective)?

A function is one-one if distinct inputs give distinct outputs. Here

Thus $1 \ne 2$ but $f(1) = f(2)$.

$f$ is not one-one.

2. Is $f$ onto (surjective)?

We must check that every $k \in \mathbf{N}$ is an output. For any $k \in \mathbf{N}$, take the odd number $n = 2k-1 \in \mathbf{N}$:

So every natural number $k$ is attained. …

Method: Deciding if a piecewise function is bijective

For a function defined by cases, test injectivity and surjectivity through the cases — a clash across two cases often settles one-one at once.

Steps

Step 1: One-one — try small inputs drawn from different cases. If an odd-input and an even-input give the same value, injectivity fails immediately (here $f(1)=f(2)=1$). …

Common Mistakes

Mistake 1: Assuming the function is one-one because each case formula looks invertible.

Why it's wrong: the two cases overlap in output — an odd input and an even input can give the same value, e.g. $f(1)=f(2)=1$. Correct approach: test inputs from different cases before deciding injectivity.

Mistake 2: Reading "not bijective" as "not onto". …