RELATIONS AND FUNCTIONS

Chapter 1

Relations and Functions

Big idea: a relation says which things are connected. A function is a relation with a promise: each input has exactly one output. That promise lets you predict, compare, and sometimes undo.

A shop lists three phone models. Each model comes in some of three colours. If you write down "model 1 comes in colour 2", you have written one connection between two things. A list of such connections is called a relation. Relations are everywhere: a timetable links classes to rooms, a menu links dishes to prices, a map links towns to roads.

An ordered pair $(a, b)$ is two things written in a fixed order. The order matters: $(1, 2)$ is not the same pair as $(2, 1)$. If $A$ and $B$ are sets, the Cartesian product $A \times B$ is the set of all ordered pairs with the first item from $A$ and the second from $B$.

Let $A = \{1, 2, 3\}$ be the phone models and $B = \{1, 2, 3\}$ the colours (1 = black, 2 = silver, 3 = blue). Then $A \times B$ has $3 \times 3 = 9$ pairs. In general, if $A$ has $m$ items and $B$ has $n$ items, then $A \times B$ has $m \times n$ pairs.

A relation $R$ from $A$ to $B$ is any subset of $A \times B$. Suppose the shop sells model 1 in black and silver, model 2 only in silver, and model 3 in all three colours. Then

$$R = \{(1, 1), (1, 2), (2, 2), (3, 1), (3, 2), (3, 3)\}$$

We write $(a, b) \in R$, or $a \, R \, b$, when the pair is in the relation. Because $A \times B$ has 9 pairs and each pair is either in $R$ or out of $R$, there are $2^{9} = 512$ possible relations from $A$ to $B$.

Look at model 1 in the relation above. It has two colours, so if someone asks "what is the colour of model 1?" there is no single answer. A function removes this problem. It is a relation in which every input has exactly one output.

Now suppose each model has one default colour: model 1 is black, model 2 is silver, model 3 is black. This is $F = \{(1, 1), (2, 2), (3, 1)\}$. Every model appears exactly once as a first item, so $F$ is a function. We write $F(1) = 1$, $F(2) = 2$, $F(3) = 1$.

In this chapter you will learn:

• the main types of relations: reflexive, symmetric, transitive, and the very useful equivalence relation;
• the main types of functions: one-one, onto, and bijective;
• composition: doing one function after another;
• inverse functions: when a function can be undone, and how to find the undo rule.

From now on we study relations inside one set: a relation $R$ in a set $X$ is a subset of $X \times X$. This is how we compare things of the same kind: numbers with numbers, lines with lines, words with words.

• The empty relation has no pairs: $R = \varnothing$. Nothing is related to anything.
• The universal relation has all pairs: $R = X \times X$. Everything is related to everything.

• Reflexive: every item is related to itself. $(a, a) \in R$ for every $a \in X$.
• Symmetric: if $a$ is related to $b$, then $b$ is related to $a$. $(a, b) \in R$ implies $(b, a) \in R$.
• Transitive: if $a$ is related to $b$ and $b$ is related to $c$, then $a$ is related to $c$. $(a, b) \in R$ and $(b, c) \in R$ imply $(a, c) \in R$.

Think of "is the same height as". Everyone is the same height as themselves (reflexive). If A is the same height as B, then B is the same height as A (symmetric). If A matches B and B matches C, then A matches C (transitive). Compare "is taller than": it is transitive, but it is not reflexive and not symmetric.

A relation that is reflexive, symmetric and transitive is called an equivalence relation. It says "these two items are the same in the way we care about". Same height, same colour, same remainder after dividing by 3: all of these are equivalence relations.

Take $X = \{1, 2, 3, 4, 5, 6\}$ and test ten relations. The answers below were computed pair by pair.

• To prove a property for a rule on all numbers, take general items ($a$, $b$, $c$) and use the rule with algebra.
• To disprove a property, one counterexample is enough. Find one pair (or two linked pairs) where the rule fails.

Let $R$ be the relation on the set $\mathbb{Z}$ of integers given by $a \, R \, b$ when $a - b$ is a multiple of $4$. Show that $R$ is an equivalence relation.

• Reflexive: $a - a = 0 = 4 \times 0$, which is a multiple of $4$. So $a \, R \, a$.
• Symmetric: suppose $a - b = 4k$ for some integer $k$. Then $b - a = 4(-k)$, which is also a multiple of $4$. So $b \, R \, a$.
• Transitive: suppose $a - b = 4k$ and $b - c = 4m$. Adding, $a - c = 4(k + m)$, a multiple of $4$. So $a \, R \, c$.

All three hold, so $R$ is an equivalence relation. As a check, the computer tested every pair of the integers from $-8$ to $8$ and agreed: reflexive Yes, symmetric Yes, transitive Yes. (The proof above covers all integers. The test covers this sample.)

On $X = \{-3, -2, -1, 0, 1, 2, 3\}$ let $a \, R \, b$ when $a \times b \ge 0$ (the product is zero or positive). It is reflexive, because $a \times a \ge 0$. It is symmetric, because $a \times b = b \times a$. But it is not transitive: $(-1, 0) \in R$ because $(-1)(0) = 0 \ge 0$, and $(0, 1) \in R$ because $0 \ge 0$. Yet $(-1, 1) \notin R$, because $(-1)(1) = -1 < 0$. The number $0$ links numbers that should not be linked.

An equivalence relation sorts a set into groups. The equivalence class of $a$, written $[a]$, is the set of all items related to $a$. Take $X = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$ and let $a \, R \, b$ when $a - b$ is a multiple of $3$ (same remainder when divided by $3$). The classes are

• $[1] = \{1, 4, 7\}$
• $[2] = \{2, 5, 8\}$
• $[3] = \{3, 6, 9\}$

Every item is in exactly one class, and different classes never overlap. We say the classes partition the set. This is why equivalence relations matter: they turn "same in some way" into clean groups. Days of the week work like this: every day number has a remainder when divided by 7, and the same remainder means the same weekday.

Decide which of the three properties each relation has.

1. On $\{1, 2, 3, 4\}$: $R = \{(1, 1), (2, 2), (3, 3), (4, 4), (1, 2)\}$.
2. On $\{1, 2, 3\}$: $R = \{(1, 2), (2, 1), (1, 1)\}$.
3. On $\{1, 2, 3, 4\}$: $a \, R \, b$ when $a + b = 5$.
4. On $\{1, 2, 3, 4\}$: $a \, R \, b$ when $a \le b$.

Answers

1. Reflexive and transitive, but not symmetric.
2. Symmetric, but not reflexive or transitive.
3. Symmetric, but not reflexive or transitive.
4. Reflexive and transitive, but not symmetric.

1. On $\{1, 2, 3\}$, let $R = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)\}$. Which of the three properties does $R$ have? Is it an equivalence relation?
2. On $X = \{1, 2, 3, 4, 5, 6\}$, let $a \, R \, b$ when $a \times b < 10$. Test the three properties.
3. On $X = \{1, 2, \dots, 8\}$, let $a \, R \, b$ when $a$ divides $b$. Test the three properties.
4. On the integers from $-6$ to $6$, let $a \, R \, b$ when $a - b$ is a multiple of $5$. Show that it is an equivalence relation and list the equivalence classes. What is the class of $0$?
5. How many relations on a set of $3$ items are (a) reflexive, (b) symmetric, (c) equivalence relations?
6. On the set of all straight lines in a plane, let "$l$ is perpendicular to $m$" be the relation. Which properties does it have?
7. On the words $\{\text{cat}, \text{dog}, \text{tree}, \text{bird}, \text{sun}, \text{moon}\}$, let two words be related when they have the same number of letters. Show it is an equivalence relation and find the classes.
8. On $\{-2, -1, 0, 1, 2\}$, let $a \, R \, b$ when $a^2 = b^2$. Is $R$ an equivalence relation? Find its classes.
9. On the real numbers, let $a \, R \, b$ when $a - b$ is an integer. Show that $R$ is an equivalence relation. Which of $0.25$, $1.25$, $3.25$, $0.5$, $2.5$ are in the same class?
10. How many equivalence relations on $\{1, 2, 3\}$ contain the pair $(1, 2)$?

Answers

1. Reflexive: $(1,1), (2,2), (3,3)$ are all in $R$. Symmetric: $(1,2)$ and $(2,1)$ are both there. Transitive: no chain breaks the rule. So $R$ is reflexive, symmetric and transitive, so it is an equivalence relation.
2. Not reflexive, because $4 \times 4 = 16$ is not less than $10$, so $(4, 4) \notin R$. Symmetric, because $a \times b = b \times a$. Not transitive: $(4, 1) \in R$ and $(1, 6) \in R$, but $(4, 6) \notin R$ because $24 \ge 10$.
3. Reflexive (every number divides itself) and transitive (if $a$ divides $b$ and $b$ divides $c$, then $a$ divides $c$). Not symmetric: $(1, 2) \in R$ but $(2, 1) \notin R$.
4. The proof is the same as Example 1, with $5$ in place of $4$. There are 5 classes: $\{-6, -1, 4\}, \; \{-5, 0, 5\}, \; \{-4, 1, 6\}, \; \{-3, 2\}, \; \{-2, 3\}$. The class of $0$ is $\{-5, 0, 5\}$.
5. (a) $64$. (b) $64$. (c) $5$. All three counts were found by testing every one of the $2^9 = 512$ relations.
6. Symmetric only. A line is not perpendicular to itself, so it is not reflexive. If $l \perp m$ and $m \perp n$, then $l$ and $n$ are parallel, not perpendicular, so it is not transitive.
7. Each word has the same length as itself; "same length" can be read in either direction; and if $A$ and $B$ have the same length and $B$ and $C$ have the same length, then $A$ and $C$ do. So it is an equivalence relation. The classes are $\{\text{cat}, \text{dog}, \text{sun}\}, \; \{\text{tree}, \text{bird}, \text{moon}\}$.
8. Yes: equal squares is a form of "the same value", so all three properties hold. The classes are $\{-2, 2\}, \; \{-1, 1\}, \; \{0\}$.
9. Reflexive: $a - a = 0$ is an integer. Symmetric: if $a - b$ is an integer, so is $b - a$. Transitive: the sum of two integers $(a - b) + (b - c) = a - c$ is an integer. In the list, $0.25$, $1.25$ and $3.25$ are in one class (they differ by whole numbers), and $0.5$ and $2.5$ are in another.
10. $2$. Either $1$ and $2$ form a class and $3$ is alone, or all three numbers are in one class.

A function $f$ from a set $X$ to a set $Y$ is written $f : X \to Y$. The set $X$ is the domain (the allowed inputs). The set $Y$ is the codomain (where outputs are allowed to live). The range is the set of outputs that actually happen, and it can be smaller than the codomain.

Example: let $f : \mathbb{R} \to \mathbb{R}$ be $f(x) = x^2$. The domain and codomain are both $\mathbb{R}$. But no square is negative, so the range is only $[0, \infty)$.

Two questions decide what kind of function we have. Both are about the outputs:

• One-one (injective): do different inputs always give different outputs? Nothing is shared.
• Onto (surjective): is every item of the codomain actually used as an output? Nothing is left out.

In symbols, $f$ is one-one when $f(a) = f(b)$ forces $a = b$. And $f$ is onto when for every $y$ in the codomain there is some $x$ in the domain with $f(x) = y$. A function that is both is called bijective: a perfect pairing of the domain with the codomain.

Let the domain be $\{1, 2, 3\}$ and the codomain $\{a, b, c\}$.

$f_1$ pairs everything perfectly, so it is bijective. In $f_2$, the inputs $1$ and $2$ both go to $a$ (not one-one), and $c$ is never used (not onto).

• One-one: assume $f(a) = f(b)$ and show that $a = b$. To show it is not one-one, find two different inputs with the same output.
• Onto: take any $y$ in the codomain and try to solve $f(x) = y$ for $x$. If you always find an $x$ in the domain, the function is onto. To show it is not onto, find one $y$ that no input can reach.
• Graph test: draw horizontal lines. If some horizontal line meets the graph twice, the function is not one-one. If some horizontal line in the codomain meets the graph never, it is not onto.

Show that $f : \mathbb{R} \to \mathbb{R}$, $f(x) = 3x + 2$ is bijective.

One-one. Suppose $f(a) = f(b)$. Then $3a + 2 = 3b + 2$, so $3a = 3b$, so $a = b$.

Onto. Take any real $y$. Solving $3x + 2 = y$ gives $x = \frac{y - 2}{3}$, which is a real number. And $f\!\left(\frac{y-2}{3}\right) = y$. So every $y$ is reached.

Is $f : \mathbb{R} \to \mathbb{R}$, $f(x) = x^2$ one-one? Is it onto?

Not one-one: $f(-2) = 4$ and $f(2) = 4$. Two different inputs share one output. Not onto: every square is $0$ or more, so no input gives $-1$. The graph shows both problems at once.

Fixing it. Whether a function is one-one or onto depends on its domain and codomain, so we can repair it. Let $g : [0, \infty) \to [0, \infty)$, $g(x) = x^2$. Now each output $y \ge 0$ comes from exactly one input, $x = \sqrt{y}$. So $g$ is bijective.

The function $f : \mathbb{R} \to \mathbb{R}$, $f(x) = x^3$ is bijective. It rises all the way along the real line, so no output repeats (one-one). And every real $y$ has a real cube root, so every $y$ is reached (onto).

Show that $f : \mathbb{R} - \{3\} \to \mathbb{R} - \{2\}$, $f(x) = \frac{2x + 1}{x - 3}$ is bijective.

One-one. Suppose $f(a) = f(b)$. Then $\frac{2a + 1}{a - 3} = \frac{2b + 1}{b - 3}$. Cross-multiply: $(2a + 1)(b - 3) = (2b + 1)(a - 3)$. Expand both sides: $2ab - 6a + b - 3 = 2ab - 6b + a - 3$. Cancel $2ab$ and $-3$: $-6a + b = -6b + a$, so $7b = 7a$, so $a = b$.

Onto. Let $y \ne 2$. Solve $y = \frac{2x + 1}{x - 3}$: $y(x - 3) = 2x + 1$, so $x(y - 2) = 3y + 1$, so $x = \frac{3y + 1}{y - 2}$. This is defined because $y \ne 2$. And it is never $3$ (if it were, $3y + 1 = 3y - 6$, which is impossible). So every $y \ne 2$ is reached.

Let the domain have $m$ items and the codomain $n$ items. Each input can go to any of $n$ outputs, so there are $n^m$ functions in all. For one-one functions, the first input has $n$ choices, the next has $n - 1$ (it cannot repeat), and so on: $n(n - 1)(n - 2) \cdots (n - m + 1)$ functions. The table was found by listing every function by computer.

• If $m > n$ there is no one-one function: too many inputs for too few outputs (the "pigeonhole idea").
• If $m < n$ there is no onto function: too few inputs to reach every output.
• A bijection needs $m = n$. When $m = n$, the number of bijections is $n!$ (the table shows $3! = 6$).

Domain $\{1, 2, 3, 4\}$. For each map, say if it is one-one, onto, both or neither.

1. $1 \to p, \; 2 \to q, \; 3 \to r, \; 4 \to s$, codomain $\{p, q, r, s\}$.
2. $1 \to p, \; 2 \to q, \; 3 \to r, \; 4 \to r$, codomain $\{p, q, r\}$.
3. $1 \to p, \; 2 \to q, \; 3 \to r, \; 4 \to s$, codomain $\{p, q, r, s, t\}$.
4. Every input goes to $p$, codomain $\{p, q, r\}$.

Answers

1. It is one-one and onto (bijective).
2. It is onto but not one-one.
3. It is one-one but not onto.
4. It is neither one-one nor onto.

1. Domain $\{1, 2, 3\}$, codomain $\{x, y, z\}$: $1 \to y, \; 2 \to z, \; 3 \to x$. Is it one-one? Onto? And the map from $\{1, 2, 3, 4\}$ to $\{x, y\}$ with $1 \to x, \; 2 \to y, \; 3 \to x, \; 4 \to y$?
2. Show that $f : \mathbb{R} \to \mathbb{R}$, $f(x) = 5x - 7$ is bijective. Which input gives the output $18$?
3. Show that $f : \mathbb{R} \to \mathbb{R}$, $f(x) = |x|$ is neither one-one nor onto.
4. Is $f : \mathbb{R} \to \mathbb{R}$, $f(x) = x^2 + 1$ one-one? Onto? Change the domain and codomain to make it bijective.
5. Show that $f : \mathbb{R} - \{-2\} \to \mathbb{R} - \{1\}$, $f(x) = \frac{x}{x + 2}$ is bijective.
6. Domain with $4$ items, codomain with $3$ items. How many functions are there? Then find: one-one functions from $3$ items to $5$ items; onto functions from $4$ items to $2$ items; bijections from $4$ items to $4$ items.
7. Is $f : \mathbb{Z} \to \mathbb{Z}$, $f(x) = 2x$ one-one? Onto?
8. Is $f : \mathbb{N} \to \mathbb{N}$, $f(n) = n + 1$ one-one? Onto? (Here $\mathbb{N} = \{1, 2, 3, \dots\}$.)
9. A thermometer converts Celsius to Fahrenheit with $F = 1.8C + 32$. Show that this is a bijection from $\mathbb{R}$ to $\mathbb{R}$. What is $F$ for $C = 37$? What is $C$ for $F = 50$?
10. Give an example of (a) a one-one function from $\{1, 2, 3\}$ to $\{1, 2, 3, 4\}$ that is not onto, and (b) an onto function from $\{1, 2, 3, 4\}$ to $\{1, 2, 3\}$ that is not one-one.

Answers

1. The first map is one-one and onto (bijective). The second map is onto but not one-one: it reaches both $x$ and $y$, but four inputs share two outputs.
2. If $5a - 7 = 5b - 7$ then $a = b$ (one-one). For any $y$, $x = \frac{y + 7}{5}$ gives $f(x) = y$ (onto). For $y = 18$: $x = \frac{25}{5} = 18$. Check: $5(18) - 7 = 18$.
3. Not one-one: $f(-3) = 3$ and $f(3) = 3$. Not onto: $|x| \ge 0$ for every $x$, so $-1$ is never an output.
4. Not one-one: $f(-2) = 5 = f(2)$. Not onto: $x^2 + 1 \ge 1$, so $0$ is never an output. Take $f : [0, \infty) \to [1, \infty)$. Then each $y \ge 1$ comes from exactly one input $x = \sqrt{y - 1}$, and $f$ is bijective.
5. One-one: $\frac{a}{a + 2} = \frac{b}{b + 2}$ gives $a(b + 2) = b(a + 2)$, so $2a = 2b$. Onto: solving $y = \frac{x}{x + 2}$ gives $x = \frac{2y}{1 - y}$, defined for $y \ne 1$, and it is never $-2$ (that would need $2y = -2 + 2y$).
6. $3^4 = 81$ functions. One-one from $3$ to $5$: $5 \times 4 \times 3 = 60$. Onto from $4$ to $2$: $14$ (all $2^4 = 16$ functions, minus the $2$ that send everything to one item). Bijections of $4$ items: $4! = 24$.
7. One-one: $2a = 2b$ gives $a = b$. Not onto: $2x$ is always even, so an odd number such as $1$ is never an output.
8. One-one: $a + 1 = b + 1$ gives $a = b$. Not onto: the smallest output is $2$, so $1$ is never reached.
9. It is a straight line with non-zero slope, so it is one-one and onto (as in Example 1). $F(37) = 1.8 \times 37 + 32 = 98.6$. Solving $C = \frac{F - 32}{1.8}$ gives $C = \frac{18}{1.8} = 10$.
10. (a) $1 \to 1, \; 2 \to 2, \; 3 \to 3$: one-one, and $4$ is never used. (b) $1 \to 1, \; 2 \to 2, \; 3 \to 3, \; 4 \to 3$: every output is used, but $3$ and $4$ share the output $3$. Both were confirmed by computer.

Imagine a two-step machine. The first machine $f$ takes a number and doubles it, then adds $1$. The second machine $g$ squares whatever it receives. If you put a number through both machines in order, you have built a new function. This is called the composition of $f$ and $g$.

The composition, written $g \circ f$ (read "$g$ after $f$") means: apply $f$ first, then $g$. In symbols, $(g \circ f)(x) = g(f(x))$.

With $f(x) = 2x + 1$ and $g(x) = x^2$: $(g \circ f)(x) = (2x + 1)^2$. At $x = 3$, the first machine gives $7$ and the second gives $49$. In the other order, $(f \circ g)(x) = 2x^2 + 1$, and at $x = 3$ it gives $19$.

For $g \circ f$ to make sense, every output of $f$ must be an allowed input of $g$. In other words, the range of $f$ must fit inside the domain of $g$.

Let $f : \{1, 2, 3\} \to \{a, b, c\}$ be $f = \{(1, \text{b}), (2, \text{a}), (3, \text{c})\}$, and $g : \{a, b, c\} \to \{10, 20\}$ be $g = \{(\text{a}, 10), (\text{b}, 20), (\text{c}, 10)\}$. Follow each input through both steps:

So $g \circ f = \{(1, 20), (2, 10), (3, 10)\}$. Notice that $g$ is not one-one (both $a$ and $c$ go to $10$), and neither is $g \circ f$.

• Composition of one-one functions is one-one. If $g(f(a)) = g(f(b))$, then $g$ being one-one gives $f(a) = f(b)$, and then $f$ being one-one gives $a = b$.
• Composition of onto functions is onto. Take any target $z$. Because $g$ is onto, some $y$ has $g(y) = z$. Because $f$ is onto, some $x$ has $f(x) = y$. Then $g(f(x)) = z$.
• So the composition of two bijections is a bijection. This was checked on the pair of functions $f = \{(1, b), (2, a), (3, c)\}$ and $g = \{(a, 10), (b, 20), (c, 30)\}$, giving $g \circ f = \{(1, 20), (2, 10), (3, 30)\}$.

Composition also has a tidy property: it is associative. $(h \circ g) \circ f = h \circ (g \circ f)$. With $f(x) = x + 1$, $g(x) = 2x$, $h(x) = x^2$, both ways give $(2(x + 1))^2$, which equals $64$ at $x = 3$. So we can write $h \circ g \circ f$ with no brackets.

Some functions can be undone. If $f$ turns $3$ into $7$ by "double and add $1$", the undo machine turns $7$ back into $3$ by "subtract $1$ and halve". We say $g$ is the inverse of $f$ when doing one and then the other gets you back where you started, in both orders:

• $g(f(x)) = x$ for every $x$ in the domain of $f$ (do $f$, then undo it);
• $f(g(y)) = y$ for every $y$ in the domain of $g$ (undo, then do $f$).

We write the inverse as $f^{-1}$. A function that has an inverse is called invertible.

1. Check that the function is a bijection between the given domain and codomain.
2. Write $y = f(x)$.
3. Solve this equation for $x$, so that $x$ is written using $y$.
4. Rename: the result is $f^{-1}(y)$. You may then call the variable $x$ again.
5. State the domain and range of $f^{-1}$. They are the range and domain of $f$, swapped.

Find the inverse of $f(x) = 4x - 7$ on $\mathbb{R}$. Write $y = 4x - 7$. Then $4x = y + 7$, so $x = \frac{y + 7}{4}$. Hence $f^{-1}(y) = \frac{y + 7}{4}$.

For $f = \{(1, \text{b}), (2, \text{a}), (3, \text{c})\}$, swapping each pair gives $f^{-1} = \{(\text{a}, 2), (\text{b}, 1), (\text{c}, 3)\}$.

In Example 4 of Section 1.3, solving $y = \frac{2x + 1}{x - 3}$ gave $x = \frac{3y + 1}{y - 2}$. So the inverse of $f : \mathbb{R} - \{3\} \to \mathbb{R} - \{2\}$ is $f^{-1} : \mathbb{R} - \{2\} \to \mathbb{R} - \{3\}$, $f^{-1}(y) = \frac{3y + 1}{y - 2}$. Notice how the missing input $3$ and the missing output $2$ trade places.

The function $f : [0, \infty) \to [2, \infty)$, $f(x) = x^2 + 2$ is a bijection. Solving $y = x^2 + 2$ with $x \ge 0$ gives $x = \sqrt{y - 2}$. So $f^{-1}(y) = \sqrt{y - 2}$. For instance $f(3) = 11$ and $f^{-1}(11) = 3$.

To undo "put on socks, then put on shoes" you must take off the shoes first, then the socks. In symbols, if $f$ and $g$ are invertible,

$$(g \circ f)^{-1} = f^{-1} \circ g^{-1}$$

Check with $f(x) = 2x + 1$ and $g(x) = 3x - 5$. Then $(g \circ f)(x) = 3(2x + 1) - 5 = 6x - 2$, whose inverse is $\frac{x + 2}{6}$. On the other side, $f^{-1}(x) = \frac{x - 1}{2}$ and $g^{-1}(x) = \frac{x + 5}{3}$, so $f^{-1}(g^{-1}(x)) = \frac{\frac{x + 5}{3} - 1}{2} = \frac{x + 2}{6}$. They agree. In the wrong order, $g^{-1}(f^{-1}(x)) = \frac{x + 9}{6}$, which is different.

1. Let $f(x) = 3x + 1$ and $g(x) = x^2$. Find $(g \circ f)(2)$ and $(f \circ g)(2)$.
2. Find the inverse of $f(x) = \frac{x - 4}{5}$.
3. Is $f = \{(1, \text{q}), (2, \text{q}), (3, \text{r})\}$, from $\{1, 2, 3\}$ to $\{p, q, r\}$, invertible?
4. Find the inverse of $f(x) = x^3 + 1$ on $\mathbb{R}$.

Answers

1. $(g \circ f)(2) = g(7) = 49$ and $(f \circ g)(2) = f(4) = 13$.
2. $y = \frac{x - 4}{5}$ gives $x = 5y + 4$, so $f^{-1}(y) = 5y + 4$.
3. No. It is not one-one ($1$ and $2$ both go to $q$), and it is not onto ($p$ is never used).
4. $y = x^3 + 1$ gives $x = \sqrt[3]{y - 1}$, so $f^{-1}(y) = \sqrt[3]{y - 1}$.

Take the ordered pairs $(a, b)$ with $a, b \in \{1, 2, 3, 4\}$. Say $(a, b) \sim (c, d)$ when $a \times d = b \times c$. Show that this is an equivalence relation and describe the classes.

Reflexive: $a b = b a$. Symmetric: if $a d = b c$ then $c b = d a$. Transitive: if $a d = b c$ and $c f = d e$, multiply the first by $f$ and the second by $b$ and compare: $a d f = b c f = b d e$, and cancelling $d$ gives $a f = b e$. The computer also checked all $256$ pairs of pairs.

The rule says "$\frac{a}{b}$ and $\frac{c}{d}$ are equal fractions". The $16$ pairs fall into 11 classes, one for each different fraction. The class of $(1, 2)$ is $\{(1, 2), (2, 4)\}$, because $\frac{1}{2} = \frac{2}{4}$. The class of $(1, 1)$ has 4 pairs: $\frac{1}{1} = \frac{2}{2} = \frac{3}{3} = \frac{4}{4}$. So a fraction is an equivalence class of pairs.

How many onto functions are there from $\{1, 2, 3, 4\}$ to $\{a, b, c\}$?

Start with all functions: $3^4 = 81$. Remove those that miss at least one output. There are $3$ ways to miss one chosen output, each leaving $2^4 = 16$ functions, so subtract $3 \times 16 = 48$. But functions that miss two outputs (everything goes to one item) were subtracted twice. There are $3$ of them, so add back $3$. Total: $81 - 48 + 3 = 36$. Listing every function by computer also gives 36.