Miscellaneous · Q1: $f(x)=\frac{x}{1+|x|}$ is a bijection onto $(-1,1)$

Basics you need

Solution

1. Every output lands inside $(-1,1)$

   The denominator satisfies $1+|x| \ge 1 > 0$⁠, so it is never zero and $f(x)$ is defined for every real $x$⁠. Now compare the size of the top and the bottom:

   $$|f(x)| = \frac{|x|}{1+|x|} < 1, \qquad \text{because } |x| < 1+|x|.$$

   So $-1 < f(x) < 1$ for every $x\in\R$⁠. The function really does map $\R$ into the stated codomain $(-1,1)$⁠.

   The question names $(-1,1)$ as the codomain. Before asking "is every point of $(-1,1)$ reached?", we must confirm that no output escapes it. A fraction whose numerator is smaller in size than its positive denominator always has size less than 1. Note the inequality is strict: $f(x)$ can get as close to $\pm1$ as we like but never equals $\pm1$⁠, which matches the open interval.
   For a fraction $\frac{p}{q}$ with $q>0$⁠, $\left|\frac{p}{q}\right| = \frac{|p|}{q}$⁠, and this is less than 1 exactly when $|p| < q$⁠.

2. The sign rule: $f(x)$ has the same sign as $x$

   Since the denominator $1+|x|$ is positive, the sign of $f(x)$ is the sign of its numerator $x$⁠:

   $$\begin{gathered} x > 0 \Rightarrow f(x) > 0, \\ x = 0 \Rightarrow f(x) = 0, \\ x < 0 \Rightarrow f(x) < 0. \end{gathered}$$

   So the two formulas for $|x|$ give two formulas for $f$⁠:

   $$f(x) = \begin{cases} \dfrac{x}{1+x}, & x \ge 0, \\[2mm] \dfrac{x}{1-x}, & x < 0. \end{cases}$$

   The first formula gives outputs in $[0,1)$⁠, the second outputs in $(-1,0)$⁠.

   This step is what makes the case analysis safe. If we know $f(x_1)=f(x_2)$⁠, the two outputs have the same sign, so $x_1$ and $x_2$ have the same sign too, and we may use the same formula for both. Without this observation we would also have to handle mixed cases, such as $x_1 \ge 0 > x_2$⁠.
   Dividing by a positive number never changes the sign. Absolute value: $|x|=x$ for $x\ge0$ and $|x|=-x$ for $x<0$ (Basic 1).

3. One-one: equal outputs force equal inputs

   Suppose $f(x_1) = f(x_2) = c$⁠. By Step 2 there are three possibilities for the common value $c$⁠.

   Case $c = 0$⁠. Only $x = 0$ gives output 0 (the numerator must be 0), so $x_1 = x_2 = 0$⁠.

   Case $c > 0$⁠. Then $x_1 > 0$ and $x_2 > 0$⁠, so both use $f(x) = \frac{x}{1+x}$⁠:

   $$\begin{aligned} & \frac{x_1}{1+x_1} = \frac{x_2}{1+x_2} \\ \Longrightarrow\ & x_1(1+x_2) = x_2(1+x_1) \\ \Longrightarrow\ & x_1 + x_1x_2 = x_2 + x_1x_2 \\ \Longrightarrow\ & x_1 = x_2. \end{aligned}$$

   Case $c < 0$⁠. Then $x_1 < 0$ and $x_2 < 0$⁠, so both use $f(x) = \frac{x}{1-x}$⁠:

   $$\begin{aligned} & \frac{x_1}{1-x_1} = \frac{x_2}{1-x_2} \\ \Longrightarrow\ & x_1(1-x_2) = x_2(1-x_1) \\ \Longrightarrow\ & x_1 - x_1x_2 = x_2 - x_1x_2 \\ \Longrightarrow\ & x_1 = x_2. \end{aligned}$$

   In every case $x_1 = x_2$⁠, so $f$ is one-one.

   We cross-multiply because it clears the fractions. That is legal since $1+x_1$⁠, $1+x_2$ (in the positive case) and $1-x_1$⁠, $1-x_2$ (in the negative case) are all positive, never zero. After expanding, the product term $x_1x_2$ appears on both sides and cancels, leaving $x_1 = x_2$⁠. That cancellation is the whole reason this function is one-one. The three cases cover every possible value of $c$⁠, so the proof is complete.
   Proof by cases (Basic 4): the cases $c=0$⁠, $c>0$ and $c<0$ exhaust all real numbers. One-one means $f(x_1)=f(x_2) \Rightarrow x_1=x_2$ (Basic 3).

4. Onto: solve $f(x) = y$ for every $y$ in $(-1,1)$

   Take any $y$ with $-1 < y < 1$⁠. We look for a real $x$ with $f(x) = y$⁠. The sign rule tells us which case to search in.

   If $0 \le y < 1$⁠: look for $x \ge 0$⁠, so solve $\dfrac{x}{1+x} = y$⁠:

   $$\begin{aligned} & x = y(1+x) = y + yx \\ \Longrightarrow\ & x - yx = y \\ \Longrightarrow\ & x(1-y) = y \\ \Longrightarrow\ & x = \frac{y}{1-y}. \end{aligned}$$

   Because $y<1$⁠, $1-y>0$⁠, so the division is allowed. Because $y\ge0$⁠, $x = \frac{y}{1-y} \ge 0$⁠, so this $x$ really is in the case we assumed. Check: $1 + x = \frac{1-y+y}{1-y} = \frac{1}{1-y}$⁠, so

   $$f(x) = \frac{x}{1+x} = \frac{y/(1-y)}{1/(1-y)} = y. \ ✓$$

   If $-1 < y < 0$⁠: look for $x < 0$⁠, so solve $\dfrac{x}{1-x} = y$⁠:

   $$\begin{aligned} & x = y(1-x) = y - yx \\ \Longrightarrow\ & x + yx = y \\ \Longrightarrow\ & x(1+y) = y \\ \Longrightarrow\ & x = \frac{y}{1+y}. \end{aligned}$$

   Because $y > -1$⁠, $1+y>0$⁠, so the division is allowed, and because $y<0$⁠, $x = \frac{y}{1+y} < 0$⁠, so $x$ is in the assumed case. Check: $1 - x = \frac{1+y-y}{1+y} = \frac{1}{1+y}$⁠, so

   $$f(x) = \frac{x}{1-x} = \frac{y/(1+y)}{1/(1+y)} = y. \ ✓$$

   Every $y\in(-1,1)$ has a preimage, so $f$ is onto $(-1,1)$.

   This is where the open interval matters. The formula $x = \frac{y}{1-y}$ breaks down exactly at $y = 1$ (division by zero), and $\frac{y}{1+y}$ breaks down at $y=-1$⁠. These are precisely the two values excluded from the codomain. If the question had used $\R$ as the codomain, onto would be false: no $x$ gives $f(x)=1$⁠. The check at the end of each case is not optional. Solving an equation finds the only candidate; the check confirms that it lies in the right case and really works.
   Onto means: for each $y$ in the codomain, find an $x$ in the domain with $f(x)=y$ (Basic 3). Multiplying both sides by a non-zero quantity keeps an equation equivalent (Basic 5).

5. Conclude, and read off the inverse

   $f$ is one-one (Step 3) and onto $(-1,1)$ (Step 4), so it is a bijection from $\R$ to $(-1,1)$⁠. The two formulas of Step 4 combine into one, because $1-y$ for $y\ge0$ and $1+y$ for $y<0$ are both $1-|y|$⁠:

   $$f^{-1}(y) = \frac{y}{1-|y|}, \qquad -1<y<1.$$
   A bijection can be undone, and Step 4 has already done the undoing: it found the unique input for each output. Writing the result as one formula shows the pleasing symmetry. $f$ squeezes the whole real line into $(-1,1)$ by dividing by $1+|x|$⁠, and $f^{-1}$ stretches it back out by dividing by $1-|y|$⁠.
   A function has an inverse exactly when it is one-one and onto. Lesson: invertible functions

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