Exercise 1.2 · Q10: $f(x)=\frac{x-2}{x-3}$ from $\R-\{3\}$ to $\R-\{1\}$

Basics you need

Solution

1. $f$ is defined on all of $A$ and never equals 1

   Defined on $A$⁠: for $x\in A$ we have $x\ne3$⁠, so the denominator $x - 3 \ne 0$⁠, and $f(x)$ is a real number.

   Never equal to 1: suppose some $x\in A$ had $f(x) = 1$⁠. Multiplying both sides by $x-3$ (non-zero):

   $$\frac{x-2}{x-3} = 1 \ \Longrightarrow\ x - 2 = x - 3 \ \Longrightarrow\ -2 = -3,$$

   which is false. So $f(x) \ne 1$ for every $x\in A$, and $f$ really maps $A$ into $B = \R-\{1\}$⁠.

   The same fact in one line: $f(x) = \dfrac{(x-3)+1}{x-3} = 1 + \dfrac{1}{x-3}$⁠, and a fraction with numerator 1 is never 0.

   The question declares the codomain to be $\R-\{1\}$⁠. That is only legitimate if no output equals 1, so we prove it. The contradiction argument covers every $x$ at once: whatever $x$ is, $x-2$ and $x-3$ differ by 1, so they cannot be equal. This step also explains, before we even start, why the question removed 1 from $B$⁠.
   To show an equation has no solution, assume a solution and reach a false statement (Basic 4). Multiplying both sides by a non-zero quantity keeps an equation equivalent.

2. One-one: cross-multiply and simplify

   Let $x_1, x_2 \in A$ with $f(x_1) = f(x_2)$⁠:

   $$\frac{x_1-2}{x_1-3} = \frac{x_2-2}{x_2-3}.$$

   Both denominators are non-zero (since $x_1, x_2 \ne 3$⁠), so we may cross-multiply:

   $$\begin{aligned} (x_1-2)(x_2-3) &= (x_2-2)(x_1-3)\\ x_1x_2 - 3x_1 - 2x_2 + 6 &= x_1x_2 - 2x_1 - 3x_2 + 6\\ -3x_1 - 2x_2 &= -2x_1 - 3x_2 &&\text{(cancel } x_1x_2 \text{ and } 6\text{)}\\ -x_1 &= -x_2 &&\text{(add } 2x_1 + 2x_2\text{)}\\ x_1 &= x_2 &&\text{(multiply by } -1\text{)}. \end{aligned}$$

   So $f$ is one-one.

   Cross-multiplying clears the fractions so we can compare polynomials. It is legal only because $x_1 - 3$ and $x_2 - 3$ are non-zero, which is guaranteed by $x_1, x_2 \in A$⁠. After expanding, the product $x_1x_2$ and the constant 6 appear on both sides and cancel, leaving a linear equation whose only solution is $x_1 = x_2$⁠. The form $f(x) = 1 + \frac{1}{x-3}$ gives a quicker check: equal values give $\frac{1}{x_1-3} = \frac{1}{x_2-3}$⁠, so $x_1 - 3 = x_2 - 3$⁠.
   Cross-multiplication with non-zero denominators (Basic 2). One-one: $f(x_1)=f(x_2)\Rightarrow x_1=x_2$ (Definition 5).

3. Onto: solve $f(x) = y$ for any $y\ne1$

   Take an arbitrary $y\in B$⁠, so $y$ is real and $y \ne 1$⁠. We look for $x\in A$ with $\dfrac{x-2}{x-3} = y$⁠. Multiply both sides by $x - 3$ and collect the $x$-terms:

   $$\begin{aligned} x - 2 &= y(x - 3) = yx - 3y\\ x - yx &= 2 - 3y\\ x(1 - y) &= 2 - 3y\\ x &= \frac{2-3y}{1-y} = \frac{3y-2}{y-1}. \end{aligned}$$

   The last division is allowed because $y\ne1$⁠, so $1 - y \ne 0$⁠. Now two checks.

   (a) The solution lies in the domain $A$⁠, that is, $x\ne3$⁠. Suppose $\dfrac{3y-2}{y-1} = 3$⁠. Then $3y - 2 = 3(y-1) = 3y - 3$⁠, so $-2 = -3$⁠, which is false. Hence $x \ne 3$ and $x\in A$⁠.

   (b) It really maps to $y$⁠. With $x = \frac{3y-2}{y-1}$⁠:

   $$\begin{aligned} x - 2 &= \frac{3y-2-2(y-1)}{y-1} = \frac{y}{y-1},\\ x - 3 &= \frac{3y-2-3(y-1)}{y-1} = \frac{1}{y-1},\\ f(x) &= \frac{x-2}{x-3} = \frac{y/(y-1)}{1/(y-1)} = y. \ ✓ \end{aligned}$$

   So every $y\in B$ is the image of some $x\in A$⁠: $f$ is onto.

   Solving the equation finds the only candidate; it does not by itself prove the candidate is valid. Check (a) is essential because the domain has a hole at 3: if the formula had produced $x = 3$ for some $y$⁠, that $y$ would have no preimage in $A$⁠. Check (b) confirms the algebra. Notice that $x - 3 = \frac{1}{y-1}$ is never zero, which is another way of seeing check (a).
   Onto: for an arbitrary $y$ in the codomain, find $x$ in the domain with $f(x)=y$ (Definition 6, Basic 3). Basics: one-one and onto

4. Why the codomain had to exclude 1

   The formula $x = \frac{3y-2}{y-1}$ breaks down at exactly one value, $y = 1$⁠, where it divides by zero. Step 1 showed this is not an accident of the method: no $x\in A$ gives $f(x) = 1$⁠. So the range of $f$ is exactly $\R - \{1\}$⁠.

   Consequently, if the codomain had been all of $\R$⁠, then $1$ would be an element of the codomain with no preimage, and $f$ would not be onto. Removing 1 from the codomain makes the codomain equal to the range, which is exactly what onto requires.

   Whether a function is onto depends on the codomain chosen, not only on the formula. The textbook's Remark after Definition 6 says $f$ is onto if and only if its range equals the codomain. Here the range is $\R-\{1\}$⁠, and the question chose $B = \R-\{1\}$⁠, so the two match.
   Onto ⇔ range = codomain (Remark after Definition 6). Lesson: types of functions

5. Conclude

   $f$ is one-one (Step 2) and onto (Step 3). So the answer is yes: $f$ is one-one and onto, that is, bijective. Its inverse is the formula from Step 3:

   $$f^{-1}(y) = \frac{3y-2}{y-1}, \qquad y \in \R - \{1\}.$$
   The onto step did more than prove existence: it found the unique input for each output, which is the inverse function. The inverse swaps the roles of the excluded points: it is undefined at $y = 1$ and never takes the value 3.
   A function is invertible exactly when it is one-one and onto. Lesson: invertible functions

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