Q.Show that the function $f: \mathbf{R} \to \{x \in \mathbf{R} : -1 < x < 1\}$ defined by $f(x) = \frac{x}{1+|x|}$, $x \in \mathbf{R}$ is one one and onto function.

Concept: Bijection Proof — show injectivity (one-one) and surjectivity (onto) separately.

Step 1: One-one.

Take $x_1, x_2 \in \mathbf{R}$ with $f(x_1) = f(x_2)$.

Then $\frac{x_1}{1+|x_1|} = \frac{x_2}{1+|x_2|}$.

If $x_1$ and $x_2$ have the same sign (or are zero), cross-multiplying gives $x_1 = x_2$. If they have opposite signs, the left side is non-negative and the right side non-positive (or vice versa), forcing both to be zero, so $x_1 = x_2 = 0$. Hence $f$ is injective.

Step 2: Onto.

Let $y \in (-1, 1)$. We need $x$ such that $f(x) = y$.

If $y \ge 0$, take $x = \frac{y}{1-y} \ge 0$; then $|x| = x$ and $f(x) = y$.

If $y < 0$, take $x = \frac{y}{1+y}$ (negative); then $|x| = -x$ and $f(x) = y$.

Thus every $y$ in $(-1,1)$ has a preimage, so $f$ is surjective.

Why a Bijection Proof Works

To show a function is both one-one (injective) and onto (surjective), we need to prove two things:

1. One-one: Different inputs give different outputs. For a real function, showing it is strictly increasing (or strictly decreasing) is often the cleanest route — because if $a < b$ implies $f(a) < f(b)$, then $f(a) = f(b)$ can only happen when $a = b$.

2. Onto: Every element in the codomain $(-1,1)$ is actually hit by some input. This means we need to show the range of $f$ is exactly $(-1,1)$.

The function $f(x) = \frac{x}{1+|x|}$ is cleverly designed: the $|x|$ in the denominator "clamps" the output between $-1$ and $1$ without ever reaching them. For positive $x$, it becomes $\frac{x}{1+x}$; for negative $x$, it becomes $\frac{x}{1-x}$ (since $|x| = -x$ when $x<0$). Both are simple rational functions that are easy to analyze.

Step-by-Step Proof

1. Write the function piecewise.

For $x \ge 0$, $|x| = x$, so

For $x < 0$, $|x| = -x$, so

Notice that for $x<0$, the denominator $1-x > 1$, so $f(x)$ is negative (since numerator is negative, denominator positive).

2. Show $f$ is one-one (injective).

We'll prove $f$ is strictly increasing on all of $\mathbf{R}$.

Case 1: $x \ge 0$.

Consider $f(x) = \frac{x}{1+x}$. For $0 \le a < b$, we have

So $f$ is strictly increasing on $[0, \infty)$.

Case 2: $x < 0$.

For $a < b < 0$, write $a = -p$, $b = -q$ with $p > q > 0$ (since $a<b<0$ means $-a>-b>0$). Then

Since $p > q$, and $t \mapsto \frac{t}{1+t}$ is strictly increasing for $t>0$ (shown in Case 1), we have $\frac{p}{1+p} > \frac{q}{1+q}$, so $-\frac{p}{1+p} < -\frac{q}{1+q}$, meaning $f(a) < f(b)$. So $f$ is also strictly increasing on $(-\infty, 0)$.

At the junction $x=0$:

For any $x<0$, $f(x) < 0 = f(0)$. For any $x>0$, $f(x) > 0 = f(0)$. So the function is strictly increasing across $0$ as well.

Thus $f$ is strictly increasing on all of $\mathbf{R}$, which implies it is one-one.

3. Show $f$ is onto (surjective).

We need to prove that for any $y \in (-1,1)$, there exists an $x \in \mathbf{R}$ such that $f(x) = y$.

Case 1: $y \ge 0$.

We look for $x \ge 0$ such that $\frac{x}{1+x} = y$. Solve:

Since $0 \le y < 1$, $1-y > 0$, so $x \ge 0$ is valid. Check: $f\left(\frac{y}{1-y}\right) = \frac{y/(1-y)}{1 + y/(1-y)} = \frac{y/(1-y)}{(1-y+y)/(1-y)} = y$. So every $y \in [0,1)$ is hit.

Case 2: $y < 0$.

We look for $x < 0$ such that $\frac{x}{1-x} = y$ (since $|x| = -x$ for $x<0$). Solve:

Since $-1 < y < 0$, $1+y > 0$, so $x = \frac{y}{1+y}$ is negative (numerator negative, denominator positive). Check: $f\left(\frac{y}{1+y}\right) = \frac{y/(1+y)}{1 - y/(1+y)} = \frac{y/(1+y)}{(1+y - y)/(1+y)} = y$. So every $y \in (-1,0)$ is hit.

Together, every $y \in (-1,1)$ is attained: for $y \ge 0$ we use $x = \frac{y}{1-y}$, and for $y < 0$ we use $x = \frac{y}{1+y}$.

4. Conclude.

Since $f$ is both one-one and onto, it is a bijection from $\mathbf{R}$ to $(-1,1)$.

Method: Proving a bijection for an absolute-value function via cases

For $f(x)=\frac{x}{1+|x|}$, split at $x=0$ to remove the modulus, then prove one-one (monotonic) and onto (solve for a preimage) piece by piece.

Steps

Step 1: Write the function piecewise

For $x\ge 0$, $f(x)=\frac{x}{1+x}$; for $x<0$, $f(x)=\frac{x}{1-x}$. Handle each piece separately.

Step 2: One-one via strict monotonicity

Show each piece is strictly increasing (compare $f(b)-f(a)$ for $a<b$), and check the pieces join without a downward jump at $x=0$. A strictly increasing function is one-one.

Step 3: Onto by inverting on each range

For a target $y$ in the codomain, solve $f(x)=y$ per case: $y\ge 0$ gives $x=\frac{y}{1-y}$, $y<0$ gives $x=\frac{y}{1+y}$; confirm the sign of $x$ matches the case. Every $y\in(-1,1)$ then has a preimage.

Common Mistakes

Mistake 1: Trying to handle $\frac{x}{1+|x|}$ without splitting the modulus into cases.

Why it's wrong: $|x|$ behaves differently for $x\ge 0$ and $x<0$, so a single manipulation mishandles the sign. Correct approach: write $\frac{x}{1+x}$ for $x\ge 0$ and $\frac{x}{1-x}$ for $x<0$, then work each case.

Mistake 2: Assuming a piecewise function is increasing just because each piece is.

Why it's wrong: the pieces could jump downward at the join. Correct approach: check $x=0$ — here $f(x)<0$ for $x<0$ and $f(x)>0$ for $x>0$ around $f(0)=0$, so it is strictly increasing overall.