Example 17: Inverse of $f(x)=4x+3$

Basics you need

Solution

1. See what $Y$ is

   Put $x = 1, 2, 3, 4, \dots$ into $4x+3$⁠:

   $$4(1)+3 = 7,\quad 4(2)+3 = 11,\quad 4(3)+3 = 15,\quad 4(4)+3 = 19,\ \dots$$

   So $Y = \{7, 11, 15, 19, \dots\}$⁠: the natural numbers that are 3 more than a positive multiple of 4. Numbers such as $1, 2, 3, 4, 5, 6, 8, 9, 10$ are not in $Y$⁠.

   By its very definition, $Y$ is the set of values of $f$⁠. So $f$ maps $\N$ into $Y$⁠, and every element of $Y$ is reached: $f$ is onto $Y$⁠.

   The codomain is deliberately not $\N$⁠. If it were $\N$⁠, the number 8 (for instance) would need a natural $x$ with $4x+3 = 8$⁠, that is $x = \frac54$⁠, which is not natural. So $f : \N\to\N$ would not be onto, and not invertible. Shrinking the codomain to exactly the set of outputs removes that problem.
   The range of a function is the set of its actual outputs. When the codomain is chosen equal to the range, the function is automatically onto. Basics: one-one and onto

2. Find the candidate inverse by solving for $x$

   Take an arbitrary $y\in Y$⁠. By the definition of $Y$⁠, $y = 4x + 3$ for some $x$ in $\N$⁠. Solve this for $x$⁠:

   $$y = 4x + 3 \ \Longrightarrow\ y - 3 = 4x \ \Longrightarrow\ x = \frac{y-3}{4}.$$

   So define $g : Y \to \N$ by

   $$g(y) = \frac{y-3}{4}.$$
   An inverse must take each output $y$ back to the input $x$ that produced it. Solving $y = f(x)$ for $x$ does exactly that. We subtract 3 first and divide by 4 second because $f$ did the reverse: it multiplied by 4 first and added 3 second. Undoing goes in the opposite order, like taking off shoes before socks.
   To undo a sequence of operations, apply the opposite operations in the reverse order. Subtracting the same number from both sides, or dividing both sides by the same non-zero number, keeps an equation true.

3. Check that $g$ really sends $Y$ into $\N$

   For $y \in Y$ there is an $x\in\N$ with $y = 4x + 3$⁠. Then

   $$g(y) = \frac{(4x+3)-3}{4} = \frac{4x}{4} = x \in \N.$$

   Also this $x$ is the only one: if $4x+3 = 4x'+3$⁠, then $4x = 4x'$⁠, so $x = x'$⁠. Hence $g(y)$ is a single, well-defined natural number for each $y\in Y$⁠, and $g$ is a genuine function from $Y$ to $\N$⁠.

   The formula $\frac{y-3}{4}$ could produce fractions or zero or negatives for some numbers ($\frac{8-3}{4} = \frac54$⁠, $\frac{3-3}{4} = 0$⁠). Before calling $g$ a function into $\N$⁠, we must show that for the allowed inputs, the elements of $Y$⁠, it always gives a natural number. It does, because every element of $Y$ has the form $4x+3$ with $x$ natural. The uniqueness line shows that $f$ is one-one, so there is never a choice to make.
   A function $g : Y \to \N$ must give exactly one output, lying in $\N$⁠, for every input in $Y$⁠. Basics: what a function is

4. Verify $gof = I_{\N}$

   Take any $x \in \N$⁠. Then

   $$gof(x) = g(f(x)) = g(4x+3) = \frac{(4x+3) - 3}{4} = \frac{4x}{4} = x.$$

   So $gof(x) = x = I_{\N}(x)$ for every $x \in \N$⁠, that is $gof = I_{\N}$.

   This says: start with a natural number, apply $f$⁠, then apply $g$⁠, and you are back where you started. Because $x$ was arbitrary, the calculation holds for every natural number at once.
   Two functions are equal when they agree at every point of the domain. Here $gof$ and $I_{\N}$ agree at every $x\in\N$⁠.

5. Verify $fog = I_Y$

   Take any $y \in Y$⁠. Then

   $$\begin{aligned} fog(y) &= f(g(y)) = f\!\left(\frac{y-3}{4}\right)\\ &= 4\left(\frac{y-3}{4}\right) + 3 = (y-3) + 3 = y. \end{aligned}$$

   So $fog(y) = y = I_Y(y)$ for every $y \in Y$⁠, that is $fog = I_Y$.

   Definition 9 asks for both identities, and they say different things. $gof = I_{\N}$ says $g$ undoes $f$⁠; $fog = I_Y$ says $f$ undoes $g$⁠. The second one needs $g(y)$ to be a legal input for $f$⁠, which is why Step 3 (showing $g(y)\in\N$⁠) had to come first.
   Definition 9: $f : X\to Y$ is invertible if some $g : Y \to X$ satisfies $gof = I_X$ and $fog = I_Y$⁠. Lesson: invertible functions

6. Conclude

   We have found $g : Y \to \N$ with $gof = I_{\N}$ and $fog = I_Y$⁠. By Definition 9, $f$ is invertible and $f^{-1}(y) = \dfrac{y-3}{4}$ for $y\in Y$⁠.

   The book also mentions a shortcut: a function is invertible exactly when it is one-one and onto. We have in fact seen both properties on the way: onto in Step 1 and one-one in Step 3. But finding $g$ and checking the two identities does more: it proves invertibility and gives the inverse in one go.

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