Lesson 1

Numbers, and Sets

The number line, the sets built on it, the rules of arithmetic, and the algebra that follows from them.

Taught 11 Aug 2026

We note early that the notes below and onwards assume the basic algebra and arithmetic of a GCSE math course; everything past that point is built up here from the start.

The Number Line

Two ideas carry almost all of what follows: the number line, which is where numbers live, and the set, which is how we talk about a collection of them at once.

We start with the number line. Here is an infinite line.

Figure 1.1. An infinite line.

The arrowheads are there to say that the line is meant to extend in both directions, to infinity. Now let us fix a point on this line (can be any point, but we do normally like it being in the middle), and label it $0$ .

Figure 1.2. A zero point fixed on the line.

This point represents the number zero, and it is the point against which every other number is measured. Numbers sitting to the left of $0$ are called negative, numbers sitting to the right are called positive, and $0$ itself is neither.

One thing is still missing; we can now say which side of $0$ a point lies on, but not how far along it lies, so let us finally fix a unit of length.

Figure 1.3. A unit of length fixed on the line.

This unit of length is what we use, among other things, to compare how far the other numbers sit from zero.

Definition 1.1 (The Number Line).

The infinite line above, together with its fixed zero point and its fixed unit of length, is the (real) number line. The numbers represented by the points of the number line are called real numbers, and the collection of all of them is written $\RR$ .

A few real numbers, and where they sit:

Figure 1.4. Seven real numbers on the number line.

The Natural Numbers

The natural numbers are the numbers we count with. Counting is done by incrementing a quantity, one object at a time, and on the number line an increment is a move to the right by the unit length. The distance we have moved from zero is then the number of times we moved, which is exactly the number of objects counted.

Definition 1.2 (Natural Numbers).

The natural numbers (sometimes called whole numbers) are the numbers represented by those points of the number line that can be reached by starting at $0$ and moving right by a whole unit length some number of times.

Figure 1.5. The natural numbers. The point

0

is drawn hollow because we do not count it as one of them.

Remark (A Convention).

Whether $0$ counts as a natural number is a matter of convention; but my personal preference is we do not count it, so $\NN = \{1, 2, 3, \ldots\},$ and when we do want zero alongside them we write $\NN_0 = \{0, 1, 2, \ldots\}.$

The Integers

The natural numbers count, but they cannot measure the difference between two counts; if an increment in quantity is a move to the right by the unit length, then a decrement is a move to the left by the unit length, and allowing both directions gives the integers.

Definition 1.3 (Integers).

The integers are the numbers represented by those points of the number line that can be reached by starting at $0$ and moving in either direction by a whole unit length some number of times. The collection of all of them is written $\ZZ$ .

Figure 1.6. The integers.

Closure

One property that ordinary arithmetic quietly assumes has not been mentioned yet.

Definition 1.4 (Closure).

A collection of numbers is closed under an operation if carrying out that operation on any two of its members again produces a member of that same collection.

The real numbers are closed under addition, subtraction, multiplication and division: the sum, difference, product and ratio of any two real numbers is again a real number. Division is the one case needing a footnote, and we come back to it below.

Smaller collections need not behave so well: If $a$ and $b$ are natural numbers then so are $a + b$ and $ab$ , and if $a$ and $b$ are integers then so are $a + b$ , $a - b$ and $ab$ ; but by our definition the naturals are not closed under subtraction, since $1 - 2$ is not a natural number, and the integers are not closed under division, since $1/2$ is not an integer. Each failure is a reason to enlarge the collection we are working with.

Problem 1.1.

Which of the four operations is the collection of negative integers closed under?

The Rational Numbers

The integers were forced on us by subtraction; division forces the next enlargement. Where an integer is what we get by laying whole units end to end, a rational number is what we get by first cutting the unit into equal segments.

Definition 1.5 (Rational Number).

A rational number is a real number that can be written as a fraction whose numerator is an integer and whose denominator is a nonzero integer; that is, $x$ is rational if and only if there are integers $a$ and $b$ with $b \neq 0$ such that $x = a/b.$ The collection of all rational numbers is written $\QQ$ .

Placing them on the line takes a little more work than placing the integers did. Given a positive integer $b$ , the number $1/b$ is the length obtained by breaking the unit length into $b$ equal parts. For $b = 3$ :

Figure 1.7. The unit length broken into three equal parts.

The numbers of the form $a/b$ with $a$ an integer are then those reached by starting at $0$ and moving left or right by $1/b$ units some number of times, with $a$ deciding both the direction and the number of steps.

Figure 1.8. Thirds on the number line. Every integer is a whole number of thirds, three for each unit.

Nothing new happens when $b$ is negative, because then $-b$ is positive and $a/b = -a/-b,$ so the picture for a negative denominator is the picture for a positive one.

Remark (Why the denominator cannot be zero).

In terms of the picture above, $a/0$ asks us to break the unit into zero equal parts, which does nothing at all: there is no piece left to step by, so no point on the line gets named. The same objection can be made in one line. Dividing by $b$ is meant to undo multiplying by $b$ , so if $a/0$ were a number $q$ we would need $a = q \cdot 0 = 0$ . If $a \neq 0$ there is then no such $q$ at all, and if $a = 0$ then every $q$ qualifies. Either there is no answer or there is no way to choose between the answers, so we leave $a/0$ undefined.

The Rationals are Closed

The rationals are closed under addition, subtraction and multiplication as the integers are, and unlike the integers they are also closed under division. This is quick to see.

Let $x$ and $y$ be rational. By the definition we may write $x = a/b$ and $y = c/d$ , where $a$ , $b$ , $c$ , $d$ are integers and $b \neq 0$ , $d \neq 0$ . Then

x + y = \frac{a}{b} + \frac{c}{d} = \frac{ad + bc}{bd}.

The integers are closed under multiplication, so $ad$ , $bc$ and $bd$ are integers; they are closed under addition, so $ad + bc$ is an integer; and $bd \neq 0$ , since neither $b$ nor $d$ is zero. So $x + y$ is an integer over a nonzero integer, which is to say it is rational. The same follows for the rest. For the difference and the product,

x - y = \frac{ad - bc}{bd}, \qquad xy = \frac{ac}{bd},

and both are again an integer over a nonzero integer. For the quotient, suppose also that $y \neq 0$ , so that $c \neq 0$ ; then

\frac{x}{y} = \frac{a}{b} \cdot \frac{d}{c} = \frac{ad}{bc},

and $bc \neq 0$ , so $x/y$ is rational as well.

Problem 1.2.

Where exactly in the argument above was it used that $b$ and $d$ are nonzero? Would anything go wrong if $a$ or $c$ were zero?

Sets

The number line says what a single number is; to speak about many at once — the naturals, the integers, the solutions of an equation — we need the second idea.

Definition 1.6 (Set).

A set is a collection of distinguishable objects, mathematical or otherwise.

The objects need not be numbers. The set of all animals in this building called Bob is a perfectly good set, and so is the set of letters appearing in the word “banana”, which is the collection consisting of $b$ , $a$ and $n$ .

Definition 1.7 (Membership).

The objects collected in a set are its elements. If $a$ is one of the objects collected in the set $A$ , we say $a$ is an element of $A$ , or that $a$ belongs to $A$ , and write $a \in A.$ If $a$ is not one of them we write $a \notin A$ , read ” $a$ is not in $A$ ”.

Example 1.8.

Let $B$ be the set of animals in this building called Bob. If the cat is called Bob then $\text{the cat} \in B$ , and if the dog is called Rex then $\text{the dog} \notin B$ . Numbers behave the same way: $-2 \in \ZZ$ , while $-2 \notin \NN$ .

Subsets

In the example above, every animal called Bob is also an animal in the building. That situation is common enough to deserve a name: given two sets $A$ and $B$ , it may happen that every element of $A$ is also an element of $B$ , which is to say that whenever $x \in A$ we also have $x \in B$ .

Definition 1.9 (Subset).

A set $A$ is included in a set $B$ , or is a subset of $B$ , if every element of $A$ is also an element of $B$ . We write $A \subset B.$

If every element of $A$ lies in $B$ and every element of $B$ lies in $A$ , then the two collections have exactly the same members, and we call them equal: $A = B$ precisely when $A \subset B$ and $B \subset A$ .

Remark.

Sometimes we want to say that $A$ is contained in $B$ and that at least one element of $B$ is missing from $A$ , so that the two are not equal. Then we call $A$ a proper subset of $B$ and write $A \subsetneq B$ .

The collections of numbers built up in the first half of this chapter now line up in a chain,

\NN \subsetneq \NN_0 \subsetneq \ZZ \subsetneq \QQ \subsetneq \RR,

and every inclusion in it is proper: $0$ lies in $\NN_0$ but not in $\NN$ , $-1$ lies in $\ZZ$ but not in $\NN_0$ , $1/2$ lies in $\QQ$ but not in $\ZZ$ , and $\sqrt{2}$ lies in $\RR$ but not in $\QQ$ — the last of these being a claim we do not settle here, though Exercise 1.6 does.

Remark (Decorating the symbols).

Three decorations on these symbols come up constantly. A star removes zero, so $\ZZ^{*}$ , $\QQ^{*}$ and $\RR^{*}$ are the integers, rationals and reals apart from $0$ . A subscript plus keeps the non-negative ones and a subscript minus the non-positive ones, giving $\ZZ_{+}$ , $\QQ_{+}$ , $\RR_{+}$ and $\ZZ_{-}$ , $\QQ_{-}$ , $\RR_{-}$ . The two can be combined: $\RR_{+}^{*}$ is the set of strictly positive real numbers.

It is worth pointing out how much the symbol $\subset$ resembles the symbol $<$ , and where the resemblance stops. Given two numbers $a$ and $b$ , exactly one of $a < b$ , $a = b$ , $b < a$ holds; two numbers cannot be different and yet fail to be comparable. Sets are not like this. Two sets can differ without either one being contained in the other.

Example 1.10.

Let $A = \{1, 2\}$ and $B = \{2, 3\}$ . Then $A \neq B$ , but $A \not\subset B$ , since $1 \in A$ and $1 \notin B$ , and also $B \not\subset A$ , since $3 \in B$ and $3 \notin A$ . Neither set comes first.

Remark.

As in the example, a slash through the symbol denies it: $A \not\subset B$ says that $A$ is not a subset of $B$ , which is to say that at least one element of $A$ escapes $B$ . It is not used often.

Problem 1.3.

Is $\NN \in \NN_0$ ? Is $\NN \subset \NN_0$ ? These are not the same question.

Problem 1.4.

\frac{\sqrt{3} + \sqrt{4}}{\sqrt{3} - \sqrt{4}} \in \QQ\,?

Listing and Describing

Describing a set in a sentence, as we have been doing, is exact but slow. Two shorter notations do most of the work. The first is simply to list the elements, separated by commas and enclosed in curly brackets. This is called the roster method:

A = \{2, 3, 5, 7\}.

A list may be too long to write out, or infinite, or of a length depending on some variable. In those cases we write out enough of it to make the pattern plain and leave the rest implicit, using an ellipsis:

\NN = \{1, 2, 3, \ldots\}, \qquad \{2, 4, 6, \ldots, 100\}.

Remark.

There is some genuine ambiguity here, since the reader is expected to work out what pattern the " $\ldots$ " is meant to suggest. Usually that is obvious. Sometimes it is not: $\{0, n, 3, 2, \ldots\}$ is meaningless.

In contrast to a list, the second notation describes the members implicitly by a rule. This is set-builder notation: we write

\{\, x \mid x \text{ has some property} \,\},

read “the set of all elements $x$ such that $x$ has that property”, the bar being read as “such that”. So the rationals, defined earlier in words, can be written

\QQ = \left\{\, \frac{a}{b} \;\middle|\; a \in \ZZ \text{ and } b \in \ZZ \text{ with } b \neq 0 \,\right\}.

Each notation has its own strength. A list says exactly what is in the set and nothing about why those things and not others; a rule says exactly why and leaves you to work out what actually satisfies it. The set $\{2, 3, 5, 7\}$ names its four members at a glance and hides that they are the primes below $10$ .

Problem 1.5.

Write each of these sets both ways, as a list and by a rule: the even integers between $-6$ and $6$ ; the integers whose square is $9$ ; the real numbers whose square is $-1$ .

Two Special Sets

The Empty Set

Calling a set a collection of objects is slightly misleading, because it suggests there has to be something there. There does not.

Every set carries with it a test for membership: given an object, either it belongs or it does not. Nothing in that test requires any object to pass it. Consider the set of all real numbers greater than $5$ but less than $3$ . Any such number would belong to the set, and there is no such number, so the collection is empty — and it is still perfectly well defined, because we can still answer the membership question for every object we are handed.

Definition 1.11 (The Empty Set).

The set with no elements is called the empty set, written $\varnothing$ .

Remark.

A set is well defined when, given any object, there is an objective rule deciding whether the object belongs to it or not: one of the two answers must happen, and not both. The sets we study are always of this kind. “The set of all large numbers” is not a set in this sense, since nothing decides whether $10^6$ belongs to it.

Remark (The empty set is not zero).

The number $0$ and the empty set are not the same thing. Take the set of all numbers that are neither positive nor negative. Exactly one number qualifies, so the set is $\{0\}$ , which has one element and so is not empty. A box with a zero in it is not an empty box.

The Universe of Discourse

At certain times we wish to limit membership in a set not only by the rule but also by what is eligible for membership in the first place: when solving an equation about lengths, nobody intends “the colour blue” to be a candidate.

Definition 1.12 (Universe of Discourse).

The universe of discourse, usually written $U$ , is a set fixed in advance such that for every object $b$ and every set $A$ under discussion, $b \in U \text{ and } A \subset U.$

So, for instance, we may write $\{\, k \in \{1,2,3,4\} \mid k \geqslant 2 \,\} = \{2, 3, 4\},$ where the eligible objects are the four listed and the rule then keeps three of them.

Once a universe is fixed, every set we can speak of is caught between two bounds: no set can have fewer elements than the empty set, and none can contain an object the universe does not, so

\varnothing \subset A \subset U.

Figure 1.9. A set

A

drawn inside its universe

U

, which is the rectangle surrounding it. Here

x \in A

, while

y \in U

but

y \notin A

A picture of this kind, with sets drawn as regions and the universe as the rectangle around them, is called a Venn diagram. It is a way of seeing an argument about sets rather than a way of settling one, but it is a very good way of seeing one.

The Arithmetic of Sets

Just as we combine two numbers to get a third, we can combine two sets to get a third. Doing so gives us an arithmetic of sets, and it has its own operations, its own rules, and its own closure.

Throughout this section every set is a subset of a fixed universe of discourse $U$ , as above.

Union

Definition 1.13 (Union).

The union of sets $A$ and $B$ , written $A \cup B$ , is the set of all elements belonging to at least one of $A$ and $B$ :

A \cup B = \{\, x \mid x \in A \text{ or } x \in B \,\}.

Figure 1.10. The union

A \cup B

is the whole of the shaded region.

The word “or” here is the inclusive one: it means at least one, not exactly one. If Tom goes to the shop, or Jerry does, or both of them do, then in every one of those three cases somebody from the union went.

Intersection

Definition 1.14 (Intersection).

The intersection of sets $A$ and $B$ , written $A \cap B$ , is the set of all elements belonging to both $A$ and $B$ at once:

A \cap B = \{\, x \mid x \in A \text{ and } x \in B \,\}.

Figure 1.11. The intersection

A \cap B

is the overlap.

The intersection of two sets can be empty even when neither set is. Take $I = \ZZ$ , let $A$ be the even integers and $B$ the odd ones; no integer is both, so $A \cap B = \varnothing$ . This is one of the reasons the empty set has to exist at all: without it, the intersection of two perfectly good sets would sometimes fail to be a set.

Complement

Definition 1.15 (Complement).

The complement of a set $A$ , written $A'$ , is the set of all elements of the universe $U$ that are not in $A$ :

A' = \{\, x \mid x \in U \text{ but } x \notin A \,\}.

Figure 1.12. The complement

A'

is everything inside the universe but outside

A

The complement depends entirely on the universe. We never speak of all the non- $A$ ’s in the world; only of all the elements of $U$ which are not in $A$ . Since everything under discussion lies in $U$ anyway, we could equally have written $A' = U \cap A'$ , and it is sometimes useful to remember that the $I$ is silently there.

Relative Complement

Definition 1.16 (Relative Complement).

The relative complement of $A$ in $B$ , also called the difference and written $B - A$ , is the set of all elements of $B$ that are not in $A$ :

B - A = \{\, x \mid x \in B \text{ but } x \notin A \,\}.

Figure 1.13. The difference

B - A

: the part of

B

that does not overlap

A

The difference is not really a fourth operation, because it can be written with the three we already have:

B - A = B \cap A'.

An element of $B - A$ is one that is in $B$ and is not in $A$ , and “not in $A$ ” is exactly “in $A'$ ”.

Combining the Operations

Expressions can be built up out of these operations as freely as arithmetic expressions are built out of $+$ and $\cdot$ , with brackets settling the order. So $A \cap (B \cup C)$ is read ” $A$ intersect the union of $B$ and $C$ ”, and moving the bracket to give $(A \cap B) \cup C$ need not leave the set alone.

One combination is worth recording, because it is the picture of a union taken apart:

A \cup B = (A \cap B') \cup (A \cap B) \cup (B \cap A').

The three pieces on the right are disjoint — no element lies in two of them — and they name the three regions of the diagram: the part of $A$ outside $B$ , the overlap, and the part of $B$ outside $A$ .

Figure 1.14. A union broken into three disjoint parts.

What is True of All Sets

Not everything that looks natural is true. It is tempting to expect complementation to pass through a union the way a minus sign passes through a bracket, giving $(A \cup B)' = A' \cup B'$ , and that is false in general. What is true is

(A \cup B)' = A' \cap B',

which is one of De Morgan’s laws: to be outside both $A$ and $B$ is to be outside $A$ and also outside $B$ , and the union has turned into an intersection along the way.

The distinction matters more than the particular formula. What we are after in this arithmetic are the statements that hold for all sets, not the ones that happen to come out right for a convenient choice of $A$ and $B$ .

Problem 1.6.

Draw the two-circle diagram and shade $(A \cup B)'$ . Then shade $A' \cup B'$ on a fresh copy, and $A' \cap B'$ on another. Which two agree?

Problem 1.7.

Take $U = \{1, 2, 3, 4, 5, 6\}$ , $A = \{1, 2, 3\}$ and $B = \{3, 4\}$ . Write out $A \cup B$ , $A \cap B$ , $A'$ , $B - A$ and $(A \cup B)'$ as lists.

Problem 1.8.

Is $A - B$ the same set as $B - A$ ? Compare with $a - b$ and $b - a$ for numbers.

Finally, both of the operations we started with are closed: given two sets, $A \cup B$ is a set and $A \cap B$ is a set, so the results can be combined again in their turn. That is what makes this an arithmetic rather than a collection of one-off constructions.

The Inclusion–Exclusion Principle

Write $N(X)$ for the number of elements of a finite set $X$ . The question of this section is what $N$ does to a union, and the answer is not the one most people guess.

Counting a Union

The mistake to get out of the way first is treating $\cup$ as though it were $+$ . A union does combine two sets, but it is not numerically additive unless the sets are completely separate.

Suppose $20$ students take Math 209, making up a set $A$ , and $30$ students take Bio 232, making up a set $B$ . Intuition says $N(A \cup B) = 20 + 30 = 50$ , and that is right if and only if no student takes both courses. Suppose $7$ students take both. Those $7$ are among the $20$ and they are also among the $30$ , so adding $20 + 30$ counts each of them twice, although each of them is one person. That leaves $13$ taking only Math 209 and $23$ taking only Bio 232.

Figure 1.15. The three regions hold

13

7

and

23

students, so

43

students in all.

Counting the diagram region by region gives $13 + 7 + 23 = 43$ distinct students, and the general rule follows.

For finite sets $A$ and $B$ ,

N(A \cup B) = N(A) + N(B) - N(A \cap B),

and this is called the inclusion–exclusion principle.

Here $20 + 30 - 7 = 43$ , as counting the regions said it should be.

The intersection is the controlling factor. Every element of $A \cap B$ contributes $1$ to $N(A)$ and another $1$ to $N(B)$ although it is only one element, so the sum $N(A) + N(B)$ counts the intersection exactly twice, and subtracting it once puts things right. The wrong answer $50$ minus the correct answer $43$ is $7$ , which is the size of the intersection and nothing else.

What the Two Sizes Alone Will Tell You

If all we are told is $N(A) = 20$ and $N(B) = 30$ , we cannot recover $N(A \cup B)$ ; we can only bound it, since

N(A \cup B) = 50 - N(A \cap B).

At one extreme the sets are disjoint, $N(A \cap B) = 0$ and $N(A \cup B) = 50$ . At the other, $A$ sits entirely inside $B$ , so $N(A \cap B) = 20$ and $N(A \cup B) = 30$ . The union therefore has at least $30$ and at most $50$ elements, and by choosing the intersection we can arrange for any answer between the two.

Adding is Safe on Disjoint Regions

Behind all of this is one rule: numbers may be added when the regions they count are mutually exclusive. The three regions $A \cap B'$ , $A \cap B$ and $A' \cap B$ of Figure 1.14 do not overlap, so

N(A \cup B) = N(A \cap B') + N(A \cap B) + N(A' \cap B),

and this addition needs no correction, because no element has been counted twice. The subtraction in the two-set formula is the price of using $N(A)$ and $N(B)$ , which overlap, instead of the disjoint regions.

Remark (Finite sets only).

From here on every set is finite. The formula becomes troublesome for infinite sets, because we cannot reliably subtract one infinity from another. Take $A = \NN$ . If $B$ is the set of even natural numbers then $A - B$ is the odd ones, which is infinite; but if $B$ is instead the set of natural numbers greater than $10$ , then $A - B = \{1, 2, \ldots, 10\}$ , which has exactly $10$ elements. In both cases $A$ and $B$ are infinite, so no arithmetic on their sizes alone can tell those two answers apart.

Example 1.17 (A Hand of Cards).

What is the chance of drawing a card that is either a heart or a face card from a standard deck of $52$ ?

There are $13$ hearts, making a set $H$ , and $16$ face cards — aces, kings, queens and jacks — making a set $F$ . The trap is to answer $13 + 16 = 29$ . Four cards, the ace, king, queen and jack of hearts, lie in both sets, so $N(H \cap F) = 4$ and

N(H \cup F) = 13 + 16 - 4 = 25.

So the chance is $25/52$ , not the $29/52$ the intuitive count gives.

Three Sets

For finite sets $A$ , $B$ and $C$ the principle reads

\begin{aligned} N(A \cup B \cup C) = \;& N(A) + N(B) + N(C) \\ & - N(A \cap B) - N(A \cap C) - N(B \cap C) \\ & + N(A \cap B \cap C). \end{aligned}

The name says what is happening: include, then exclude, then include again. Follow an element that lies in all three sets. The first line counts it three times, once in each of $N(A)$ , $N(B)$ and $N(C)$ . The second line subtracts it three times, once for each pair, which leaves it counted zero times. The third line adds it back once. It ends up counted once, which is correct. An element in exactly two of the sets is counted twice on the first line, subtracted once on the second and untouched by the third, so it too is counted once.

Example 1.18 (Language Classes).

Every student in a graduating class must take at least one of French, German or Spanish, giving sets $F$ , $G$ and $S$ with

\begin{aligned} N(F) &= 90, & N(G) &= 80, & N(S) &= 110, \\ N(F \cap G) &= 35, & N(F \cap S) &= 40, & N(G \cap S) &= 50, \end{aligned}

and $N(F \cap G \cap S) = 20$ . How large is the class?

Straight from the formula,

N(F \cup G \cup S) = 90 + 80 + 110 - 35 - 40 - 50 + 20 = 280 - 125 + 20 = 175.

The diagram gets there another way, by filling the seven disjoint regions from the inside out. Put $20$ in the centre. Each pair intersection already contains that $20$ , so the pair-only regions hold $35 - 20 = 15$ , $40 - 20 = 20$ and $50 - 20 = 30$ . Each single set contains the three regions just filled, so the parts taken by one language alone hold

90 - (15 + 20 + 20) = 35, \qquad 80 - (15 + 20 + 30) = 15, \qquad 110 - (20 + 20 + 30) = 40.

Figure 1.16. The seven disjoint regions of

F \cup G \cup S

, filled from the centre outwards.

Adding the seven regions, which is safe because they do not overlap,

35 + 15 + 40 + 15 + 20 + 30 + 20 = 175,

agreeing with the formula. The diagram carries more than the total: it also shows, for instance, that $15$ students took German and neither of the other two.

More Than Three

The pattern continues, with the signs alternating. Add the sets taken one at a time; subtract the intersections taken two at a time; add those taken three at a time; subtract those taken four at a time; and carry on alternating until the intersection of all of them has been used.

Problem 1.9.

In a group of $50$ people, $30$ read the news online and $25$ read it on paper. What is the largest the overlap could be, and what is the smallest, given that everybody reads it one way or the other?

Problem 1.10.

Write out the four-set formula in full. How many terms does it have, and how many of them carry a minus sign?

The Real Numbers

The first section of this chapter named the real numbers, as the points of the number line, and left it at that. The reason for leaving it there is that the real numbers are notoriously awkward to pin down, and doing it properly is the business of an analysis course rather than this one. For now the following will serve.

Definition 1.19 (Real Numbers, Informally).

A real number is a number that can be written as a decimal expansion, possibly one that never terminates and never repeats. These are exactly the numbers represented by the points of the number line: every point is a real number, and every real number is a point.

There also exist irrational numbers, meaning real numbers not contained in $\QQ$ . One example is the ratio between the circumference of a circle and its diameter, which is known as $\pi$ ; another is $\sqrt{2}$ . The set of real numbers describes all physical quantities that can be represented on a line, and $\RR$ is the set of all rational and irrational numbers together.

The Rules of Arithmetic

The operations $+$ and $\cdot$ in $\RR$ — and so also in $\NN$ , $\ZZ$ and $\QQ$ — satisfy the following properties. Every manipulation in the rest of these notes is built out of them.

Let $a, b, c \in \RR$ .

Commutativity. $a + b = b + a$ and $a \cdot b = b \cdot a$ .
Associativity. $(a + b) + c = a + (b + c)$ and $(a \cdot b) \cdot c = a \cdot (b \cdot c)$ .
Distributivity. $(a + b) \cdot c = a \cdot c + b \cdot c$ and $a \cdot (b + c) = a \cdot b + a \cdot c$ .
Neutral elements. $a + 0 = 0 + a = a$ and $a \cdot 1 = 1 \cdot a = a$ .
Inverses. There is a number $-a$ with $a + (-a) = 0$ , and if $a \neq 0$ there is a number $1/a$ with $a \cdot 1/a = 1$ .

We call $0$ the neutral element with respect to $+$ and $1$ the neutral element with respect to $\cdot$ , and we call $-a$ the negative of $a$ and $1/a$ the reciprocal of $a$ . None of these properties is claimed for subtraction or division, and in general several of them fail there; that is the point of Exercise 1.9.

What Follows from the Rules

A surprising amount of ordinary algebra is nothing but these five rules used carefully. We record the pieces we shall need.

If $a + b = 0$ then $b = -a$ and $a = -b$ . Adding $-a$ to both sides of $a + b = 0$ gives $-a + a + b = -a$ , and the left-hand side is $0 + b = b$ , so $b = -a$ . Adding $-b$ to both sides instead gives $a = -b$ in the same way.

As a matter of convention, we shall write

a - b \qquad\text{instead of}\qquad a + (-b).

With that convention, associativity and commutativity mean that a sum involving three terms may be written in many ways, all of them the same number:

(a - b) + c = \bigl(a + (-b)\bigr) + c = a + (-b + c) = a + (c - b) = (a + c) - b,

so we may also write this sum as $a - b + c = a + c - b$ .

$-(-a) = a$ . By definition $-a$ is the number that adds to $a$ to give $0$ ; but $(-a) + a = 0$ says exactly that $a$ is the number which adds to $-a$ to give $0$ , and that number is $-(-a)$ .

$-(a + b) = -a - b$ . We show this by adding $-a - b$ to $a + b$ : rearranging, $(a + b) + (-a - b) = \bigl(a + (-a)\bigr) + \bigl(b + (-b)\bigr) = 0 + 0 = 0$ , so $-a - b$ is the negative of $a + b$ .

$0 \cdot a = 0$ . Here distributivity does the work:

0 \cdot a + a = 0 \cdot a + 1 \cdot a = (0 + 1) \cdot a = 1 \cdot a = a.

Adding $-a$ to both sides gives $0 \cdot a + a - a = a - a = 0$ , and the left-hand side is simply $0 \cdot a + 0 = 0 \cdot a$ . So $0 \cdot a = 0$ .

$(-1) \cdot a = -a$ . We show this the same way:

(-1) \cdot a + a = (-1) \cdot a + 1 \cdot a = (-1 + 1) \cdot a = 0 \cdot a = 0,

and a number that adds to $a$ to give $0$ is the negative of $a$ .

$-(ab) = (-a)b$ . What has to be shown is that $(-a)b$ is the negative of $ab$ , which amounts to showing that the two add to zero; and by distributivity $ab + (-a)b = \bigl(a + (-a)\bigr)b = 0 \cdot b = 0$ .

$-(ab) = a(-b)$ . We show this by the same computation with the factors the other way round: $ab + a(-b) = a\bigl(b + (-b)\bigr) = a \cdot 0 = 0$ .

$(-a)(-b) = ab$ . Applying the previous two facts in turn, $(-a)(-b) = -\bigl(a(-b)\bigr) = -\bigl(-(ab)\bigr) = ab$ .

Problem 1.11.

A number has only one negative. Show this: if $a + b = 0$ and $a + c = 0$ , then $b = c$ .

Problem 1.12.

A nonzero number has only one reciprocal. Show this: if $ab = 1$ and $ac = 1$ , then $b = c$ .

Problem 1.13.

Show that if $ab = 0$ then $a = 0$ or $b = 0$ .

Fractions

With the rules of arithmetic written down we can say precisely how fractions behave, since every rule about them comes out of those five.

Throughout this section $m$ , $n$ , $r$ , $s$ are integers, and any letter standing in a denominator is nonzero.

The rule for cross-multiplying says that

\frac{m}{n} = \frac{r}{s} \qquad\text{if and only if}\qquad ms = rn.

This is the rule that turns a question about two fractions into a question about two integers, and it is the one to reach for whenever two fractions have to be compared.

The cancellation rule says that for a nonzero integer $a$ ,

\frac{am}{an} = \frac{m}{n}.

To test the equality we apply the rule for cross-multiplying, so what has to hold is $(am)n = m(an)$ , and that is true by associativity and commutativity. For instance,

\frac{4}{5} = \frac{(-2)(4)}{(-2)(5)} = \frac{-8}{-10}.

In dealing with quotients of integers which may be negative, it is useful to observe that

\frac{-m}{n} = \frac{m}{-n},

which cross-multiplying turns into $(-m)(-n) = mn$ , something we already know. Both are equal to $-m/n$ , and for this reason we shall write

-\frac{m}{n} = \frac{-m}{n} = \frac{m}{-n}

without worrying about which of the three is meant.

Definition 1.20 (Lowest Form).

A fraction $r/s$ of positive integers is in lowest form if $r$ and $s$ have no common divisor other than $1$ .

Any positive rational number has an expression as a fraction in lowest form. Start from any way of writing it as a quotient of positive integers $m/n$ . We know $1$ is a common divisor of $m$ and $n$ , and any common divisor is at most equal to $m$ or $n$ , so among all common divisors there is a greatest one; call it $d$ and write $m = dr$ and $n = ds$ with $r$ and $s$ positive integers. Cancelling $d$ ,

\frac{m}{n} = \frac{dr}{ds} = \frac{r}{s},

and $r$ and $s$ have no common divisor left, since one would have made $d$ larger.

Cancellation also explains addition. Given $m/n$ and $r/s$ we have seen that

\frac{m}{n} = \frac{sm}{sn}, \qquad \frac{r}{s} = \frac{nr}{ns},

so both now have the common denominator $sn$ , and adding is then adding numerators:

\frac{m}{n} + \frac{r}{s} = \frac{ms + nr}{ns}.

Multiplication needs no such preparation, since

\frac{m}{n} \cdot \frac{r}{s} = \frac{mr}{ns}.

These are the formulas the closure argument earlier in the chapter used, and the two of them together are why $\QQ$ is closed under all four operations while $\ZZ$ is not.

Problem 1.14.

Put $84/126$ in lowest form, and check the answer by cross-multiplying against the original.

Rearranging a Relation

Suppose three numbers are related by

a + b = c.

Adding $-b$ to both sides gives $a = c - b$ : the $b$ has crossed the equals sign and changed sign. That single move, together with the corresponding one for multiplication, is the whole of elementary equation solving.

Example 1.21.

Solve $3x + 4 = 19$ . Adding $-4$ to both sides gives $3x = 15$ ; multiplying both sides by $1/3$ gives $x = 5$ .

Distributivity is what lets a bracket be expanded before the rearranging starts. Used twice,

(x + 2)^2 = (x + 2)(x + 2) = x(x + 2) + 2(x + 2) = x^2 + 2x + 2x + 4 = x^2 + 4x + 4.

Equations, Identities and Inequalities

Consider the following four expressions.

\begin{aligned} (x + 2)^2 &= 2x + 7 & &[1] \\ (x + 2)^2 &= x^2 + 4x + 4 & &[2] \\ x - 2 &> 1 & &[3] \\ (x + 2)^2 & & &[4] \end{aligned}

By inspection we see that these four expressions are all different in nature, and by investigating each of them in turn we shall identify the differences.

Equations

Substituting $1$ for $x$ in the left-hand side (LHS) and right-hand side (RHS) of $[1]$ separately, we find

\text{LHS} = (1 + 2)^2 = 3^2 = 9, \qquad \text{RHS} = 2 + 7 = 9,

so LHS $=$ RHS when $x = 1$ . Substituting $2$ for $x$ in the same way, however,

\text{LHS} = (2 + 2)^2 = 16, \qquad \text{RHS} = 4 + 7 = 11,

so LHS $\neq$ RHS when $x = 2$ . Now rearranging the original expression gives

x^2 + 4x + 4 = 2x + 7 \quad\Longrightarrow\quad x^2 + 2x - 3 = 0 \quad\Longrightarrow\quad (x + 3)(x - 1) = 0,

from which LHS $=$ RHS if and only if either $x + 3 = 0$ , that is $x = -3$ , or $x - 1 = 0$ , that is $x = 1$ . The equality holds for no other value of $x$ .

Definition 1.22 (Equation).

An expression of this type, in which the two sides are equal only for a number of distinct values of the unknown quantity, is called an equation. The process of finding those values is called solving the equation.

Sets give us a tidy way of stating the answer. The problem “solve $(x + 2)^2 = 2x + 7$ ” becomes “find the solution set of $(x + 2)^2 = 2x + 7$ ”, and we write $S = \{-3, 1\}.$

Remark.

The same question can also be answered in set-builder notation, where $S = \{\, x \in \RR \mid (x + 2)^2 = 2x + 7 \,\},$ which is a statement of a different kind: the list tells us what the set contains but not the property its members share, while the rule tells us the property but not the members. Solving the equation is exactly the work of turning the second description into the first.

Recalling that two sets are equal when each is contained in the other, this gives a rather meaningful interpretation of what it is for two equations to be the same problem: two equations are equivalent when they have the same solution set. Each rearrangement above replaced an equation by an equivalent one, and that is what entitled us to read the answer off the final line.

Problem 1.15.

Are $x^2 = x$ and $x = 1$ equivalent? Compare their solution sets.

Identities

Now take $[2]$ . Substituting $1$ for $x$ in both sides,

\text{LHS} = (1 + 2)^2 = 9, \qquad \text{RHS} = 1^2 + 4 \cdot 1 + 4 = 9,

and substituting $-1$ for $x$ as before,

\text{LHS} = (-1 + 2)^2 = 1, \qquad \text{RHS} = (-1)^2 + 4(-1) + 4 = 1.

Whatever other numerical value we substitute for $x$ we find that LHS $=$ RHS, so it appears that the two sides agree for all values of $x$ . Expanding the bracket, as we did above, says the same thing in algebra; the following geometric illustration says it a third way, without any algebra at all. Consider a square of side $x + 2$ units.

Figure 1.17. One square, counted two ways.

Since these two squares are identical, their areas are identical. The left one has area $(x + 2)^2$ , and the right one has been cut into four pieces of areas $x^2$ , $2x$ , $2x$ and $4$ . Hence $(x + 2)^2 = x^2 + 4x + 4$ for all values of $x$ , and we say that $(x+2)^2$ is identical to $x^2 + 4x + 4$ .

Definition 1.23 (Identity).

A relationship whose two sides are equal for any value of the unknown quantity is called an identity. Using the symbol $\equiv$ for “is identical to”, the relationship above is written $(x + 2)^2 \equiv x^2 + 4x + 4.$

The two sides of an identity are not two things that happen to agree; they are, in fact, two forms of the same expression. We shall use the identity symbol whenever we are dealing with an identity relationship, and we strongly recommend that the reader does the same.

Problem 1.16.

State which of the following are equations and which are identities.

$x^2 - 9 = (x - 3)(x + 3)$
$p^2 + 2p - 3 = 3 - 2p - p^2$
$y - 1 = \dfrac{1}{y}$
$\dfrac{2q}{q^2 - 1} = \dfrac{1}{q - 1} + \dfrac{1}{q + 1}$

Inequalities

The third relationship, $x - 2 > 1$ , is obviously different from the first two. Reading from left to right, the symbol $>$ means “is greater than”, and $<$ means “is less than”.

By inspection we see that for $x - 2$ to have a value greater than one, $x$ must have a value greater than three:

\text{if } x - 2 > 1, \quad\text{then}\quad x > 3.

Definition 1.24 (Inequality).

A relationship between two expressions using one of $<$ , $>$ , $\leqslant$ , $\geqslant$ is called an inequality.

Here the number line does the explaining. Consider a line as being made up of adjacent points; then all the real values the variable $x$ can take are represented by positions of points on that line, and the position of a point to the left of a second point corresponds to a value of $x$ less than the value at the second point:

-99 < 1, \qquad -10 < -5, \qquad \ldots

The values of $x$ given by the statement $x > 3$ can then be represented by a section of this line.

Figure 1.18. A hollow circle leaves the endpoint out; a solid one takes it in.

From this we see that not all values of $x$ satisfy the inequality, but that there is an infinite set of values which do: the solution of an inequality is a range, or several ranges, of values of the variable involved. Note that $x = 3$ is not included in the range, and this is what the hollow circle records. For $x \geqslant 3$ , which means $x$ is greater than or equal to $3$ , the value $x = 3$ is included in the range, and the circle is filled in.

Problem 1.17.

Find the range of values of $x$ for which the following inequalities are true, and illustrate the range on a number line.

$x + 1 \leqslant -1$
$0 \geqslant x - 4$
$3 < 4 - x$

Expressions

The fourth of the four, $(x + 2)^2$ , differs from the other three in the simplest possible way: it asserts nothing. It is a term, not a relationship, so there is nothing to solve and nothing to check. It has a value once $x$ is given, and that is all.

Powers

We have been writing $x^2$ and $(x + 2)^2$ without comment, on the strength of school algebra. The notation is worth setting down properly, because we are about to push it a long way past whole-number exponents.

Definition 1.25 (Powers with Natural Exponents).

Let $a \in \RR$ and $n \in \NN$ . Then ” $a$ raised to the power of $n$ ” is defined as

a^n = \underbrace{a \cdot a \cdots a}_{n \text{ occurrences of } a},

where $a$ is called the base and $n$ the exponent.

Power Rules

Let $a, b \in \RR$ and $m, n \in \NN$ . Then

a^n \cdot a^m = a^{n+m}, \qquad a^n \cdot b^n = (a \cdot b)^n, \qquad (a^n)^m = a^{nm}.

Each of the three is a matter of counting the factors. For the first, writing both powers out gives $n$ copies of $a$ followed by $m$ more, which is $n + m$ copies in all:

a^n \cdot a^m = \underbrace{a \cdots a}_{n \text{ copies}} \cdot \underbrace{a \cdots a}_{m \text{ copies}} = a^{n+m}.

For the second, $n$ copies of $a$ beside $n$ copies of $b$ can be paired off, one $a$ to one $b$ , by commutativity, giving $n$ copies of $a \cdot b$ :

a^n \cdot b^n = \underbrace{a \cdots a}_{n \text{ copies}} \cdot \underbrace{b \cdots b}_{n \text{ copies}} = \underbrace{(ab) \cdots (ab)}_{n \text{ copies}} = (a b)^n.

For the third, $(a^n)^m$ is $m$ copies of a block of $n$ copies of $a$ , which is $nm$ copies of $a$ altogether:

(a^n)^m = \underbrace{\underbrace{a \cdots a}_{n} \cdots \underbrace{a \cdots a}_{n}}_{m \text{ blocks}} = a^{nm}.

Remark.

We define $a^0 = 1$ , and the reason is the first power rule: that is the only value making $a^0 \cdot a^n = a^{0+n} = a^n$ come out right for all $n \in \NN$ .

Definition 1.26 (Powers with Negative Integer Exponents).

Let $a \in \RR^{*}$ and let $n \in \NN$ . Then ” $a$ raised to the power of $-n$ ” is defined as

a^{-n} = \frac{1}{a^n}.

Remark.

Again the definition is forced rather than chosen. If the power rules are to hold for integer exponents then we need

a^{-n} \cdot a^{n} = a^{-n+n} = a^0 = 1,

and this makes sense if and only if $a^{-n}$ is the reciprocal of $a^n$ , which requires $a \neq 0$ . With this definition the power rules above also hold for $m, n \in \ZZ$ .

Roots

How should the square root of a number $a$ be defined? Each of the obvious attempts has a problem.

“The number which, raised to the power $2$ , gives $a$ .” But there may be two such numbers, since $(-1)^2 = 1^2 = 1$ .
“A number which, raised to the power $2$ , gives $a$ .” Then $-2$ would be a square root of $4$ , and $\sqrt{4}$ would not name one number.
” $a^{1/2}$ .” But we have not yet said what raising to the power $1/2$ means.

The way out is to demand a single number by insisting that it be the non-negative one.

Definition 1.27 (Square Root).

For $a \geqslant 0$ , the square root of $a$ , written $\sqrt{a}$ or $a^{1/2}$ , is the non-negative real number such that $\left(\sqrt{a}\right)^2 = a$ .

Remark.

Every word of that definition is doing work, in particular the condition $a \geqslant 0$ and the word non-negative. Both $2$ and $-2$ square to $4$ ; only $2$ is $\sqrt{4}$ .

Definition 1.28 ( $n$ -th Root).

Let $n \in \NN$ . For $a \geqslant 0$ , the $n$ -th root of $a$ , written $\sqrt[n]{a}$ or $a^{1/n}$ , is the non-negative real number such that $\left(\sqrt[n]{a}\right)^n = a$ .

With roots in hand, an exponent may be any rational number at all.

Definition 1.29 (Powers with Rational Exponents).

For $a \geqslant 0$ and a rational exponent $r = p/q$ with $p \in \ZZ$ and $q \in \NN$ , ” $a$ raised to the power of $r$ ” is defined as

a^{r} = \left(a^{p}\right)^{1/q}.

The power rules above hold with $m$ and $n$ rational as well, and from here on we use them without comment.

Problem 1.18.

Simplify each of the following.

$\dfrac{y^{1/6} \cdot y^{-2/3}}{y^{1/4}}$
$\dfrac{\left(\sqrt{t}\right)^3 \cdot t^2}{\sqrt{t^5}}$
$\dfrac{x^2 + x^{5/2}}{x^{-1/2}}$
$\dfrac{m^{3/2} - m^{-1/2}}{m^{1/2} + m^{-1/2}}$

Remarkable Identities

Three expansions come up so often that they are worth knowing by sight rather than working out each time.

Let $a, b \in \RR$ . Then

(a + b)^2 = a^2 + 2ab + b^2, \qquad (a - b)^2 = a^2 - 2ab + b^2, \qquad (a + b)(a - b) = a^2 - b^2.

The first comes out of distributivity used twice, exactly as the expansion of $(x + 2)^2$ did:

(a + b)^2 = (a + b)(a + b) = a(a + b) + b(a + b) = a^2 + ab + ba + b^2 = a^2 + 2ab + b^2,

the last step by commutativity. The other two go the same way.

The first identity is also the picture we drew for $(x + 2)^2$ , with the $2$ replaced by $b$ : the big square has area $(a+b)^2$ , and it decomposes into smaller rectangles of areas $a^2$ , $ab$ , $ab$ and $b^2$ .

Figure 1.19. A visualisation of

(a + b)^2 = a^2 + 2ab + b^2

Problem 1.19.

Use the third identity to compute $101 \times 99$ in your head.

Surds

Expressions such as $\sqrt{4}$ and $\sqrt{25}$ have exact numerical values, namely $\sqrt{4} = 2$ and $\sqrt{25} = 5$ . Expressions such as $\sqrt{2}$ , $\sqrt{3}$ , $\sqrt{5}$ cannot be written as numerically exact quantities. We might say that

\sqrt{2} = 1.4 \quad\text{correct to 2 significant figures,}

or that

\sqrt{2} = 1.4142136 \quad\text{correct to 8 significant figures,}

but we can never write down an exact quantity equal to $\sqrt{2}$ . Such numbers are the irrational numbers met earlier, and it is often convenient to leave them in the form $\sqrt{2}$ , or $\sqrt{3}$ , and so on.

Definition 1.30 (Surd).

A root left in its root form, rather than replaced by a decimal approximation, is called a surd.

Remark.

Recall that $\sqrt{2}$ means the positive square root of $2$ . So although $x^2 = 4$ is solved by both $x = 2$ and $x = -2$ , we still have $\sqrt{4} = 2$ .

Surds occur frequently in solutions, so it is useful to be able to simplify them. The tool is the power rule $\sqrt{ab} = \sqrt{a}\sqrt{b}$ : look for a square factor and take it outside.

Example 1.31.

Express $\sqrt{48}$ as the simplest possible surd.

\sqrt{48} = \sqrt{16 \times 3} = \sqrt{16} \cdot \sqrt{3} = 4\sqrt{3}.

Example 1.32.

Expand and simplify $(2 - 3\sqrt{3})(3 + 2\sqrt{3})$ and $(5 - 2\sqrt{7})(5 + 2\sqrt{7})$ .

For the first,

\begin{aligned} (2 - 3\sqrt{3})(3 + 2\sqrt{3}) &= 6 - 9\sqrt{3} + 4\sqrt{3} - 6\left(\sqrt{3}\right)^2 \\ &= 6 - 5\sqrt{3} - 6 \times 3 \\ &= -12 - 5\sqrt{3}. \end{aligned}

For the second, the brackets are of the form $(a + b)(a - b)$ , so the middle terms cancel:

\begin{aligned} (5 - 2\sqrt{7})(5 + 2\sqrt{7}) &= 25 - 10\sqrt{7} + 10\sqrt{7} - 4\left(\sqrt{7}\right)^2 \\ &= 25 - 28 \\ &= -3. \end{aligned}

Problem 1.20.

Expand and simplify.

$(2\sqrt{2} + 1)(\sqrt{2} - 2)$
$(3\sqrt{3} - 2)(3\sqrt{3} + 2)$
$\left(2\sqrt{6} - 3\right)^2$

Rationalising the Denominator

When the solution to a problem comes out containing surds, it is accepted practice to leave the answer in surd form, simplified as far as possible, unless an approximation has been asked for. Simplifying a fractional answer can often be managed by removing the surds from the denominator, and this process is called rationalising the denominator.

Example 1.33.

Rationalise the denominator of $\dfrac{3}{\sqrt{2}}$ .

\frac{3}{\sqrt{2}} = \frac{3}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{3\sqrt{2}}{2}.

The expansion of $(5 - 2\sqrt{7})(5 + 2\sqrt{7})$ above is the one to remember, because it is what makes this work in general: a bracket of the form $(a + b)(a - b)$ gives $a^2 - b^2$ , and squaring kills any surd in $a$ or in $b$ . So a denominator $a + b$ is rationalised by multiplying above and below by $a - b$ .

Example 1.34.

Simplify $\dfrac{3 - \sqrt{5}}{1 + 3\sqrt{5}}$ .

We multiply numerator and denominator by $1 - 3\sqrt{5}$ :

\begin{aligned} \frac{3 - \sqrt{5}}{1 + 3\sqrt{5}} &= \frac{(3 - \sqrt{5})(1 - 3\sqrt{5})}{(1 + 3\sqrt{5})(1 - 3\sqrt{5})} = \frac{18 - 10\sqrt{5}}{1^2 - \left(3\sqrt{5}\right)^2} \\ &= \frac{18 - 10\sqrt{5}}{-44} = \frac{9 - 5\sqrt{5}}{-22} = \frac{5\sqrt{5} - 9}{22}. \end{aligned}

Problem 1.21.

Rationalise the denominator.

$\dfrac{1}{3 + 2\sqrt{5}}$
$\dfrac{1}{\sqrt{6} - \sqrt{5}}$
$\dfrac{1}{\sqrt{2} + 1} + \dfrac{1}{\sqrt{2} - 1}$

Even and Odd

The last idea of the chapter costs nothing new and will be used constantly. It is a way of splitting the integers in two.

Definition 1.35 (Even and Odd).

An integer is even if it is twice an integer, and odd if it is one more than twice an integer. In set notation, the even integers and the odd integers are

E = \{\, a \in \ZZ \mid a = 2k \text{ for some } k \in \ZZ \,\}, \qquad O = \{\, a \in \ZZ \mid a = 2k + 1 \text{ for some } k \in \ZZ \,\}.

Every integer belongs to exactly one of $E$ and $O$ : dividing by $2$ leaves a remainder of either $0$ or $1$ , and no integer can be written in both forms.

The Parity of a Sum

Let $a$ and $b$ be positive integers.

If $a$ is even and $b$ is even, then $a + b$ is even.
If $a$ is even and $b$ is odd, then $a + b$ is odd.
If $a$ is odd and $b$ is even, then $a + b$ is odd.
If $a$ is odd and $b$ is odd, then $a + b$ is even.

We show the first. If $a$ and $b$ are even then $a = 2k$ and $b = 2l$ for some integers $k$ and $l$ , so by distributivity

a + b = 2k + 2l = 2(k + l),

and $k + l$ is an integer, so $a + b$ is twice an integer. The other three are left as problems below; each is the same computation with a $+1$ carried along.

The Parity of a Square

Let $a$ be a positive integer. If $a$ is even then $a^2$ is even, and if $a$ is odd then $a^2$ is odd.

If $a = 2k$ then $a^2 = (2k)^2 = 4k^2 = 2(2k^2)$ , which is even. If $a = 2k + 1$ then

a^2 = (2k + 1)^2 = 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1,

which is one more than twice an integer, and so odd.

This can also be read backwards: if $a^2$ is even then $a$ is even. Every integer is either even or odd, and not both. If $a$ were odd, what we have just shown would make $a^2$ odd; but $a^2$ is even, so the odd case is ruled out, and $a$ must be even. This is the step Exercise 1.6 turns on.

Problem 1.22.

Show the three remaining parts of the parity-of-a-sum result.

Problem 1.23.

Show that if $n$ is even then $(-1)^n = 1$ , and that if $n$ is odd then $(-1)^n = -1$ .

Problem 1.24.

Show that if $m$ and $n$ are odd then the product $mn$ is odd.

Exercises

Exercise 1.1.

Compute

\left(\frac{4}{3} + \frac{5}{2}\right) \cdot \frac{6}{5} - \frac{2}{5} \cdot \frac{5}{2}.

Exercise 1.2.

Simplify:

A = 2 + \cfrac{8}{9 + \cfrac{5}{19}}, \qquad B = \frac{1 + \frac{1}{8} + \frac{3}{4} - \frac{1}{12}}{\frac{5}{6} + \frac{1}{3} + 2}, \qquad C = \frac{\frac{5}{12}}{\frac{7}{3} \cdot \frac{8}{15} - \frac{5}{6}}.

Exercise 1.3.

Bigger, smaller, or equal?

\text{(i) } \frac{5}{14} \text{ and } \frac{6}{21}, \qquad \text{(ii) } \frac{27}{5} \text{ and } \frac{16}{3}, \qquad \text{(iii) } \frac{3}{13} \text{ and } \frac{21}{91}.

Exercise 1.4.

Compute

\text{(i) } \frac{1}{\left(\dfrac{2}{\frac{3}{4}}\right)}, \qquad \text{(ii) } \frac{1}{\left(\dfrac{\frac{2}{3}}{4}\right)}, \qquad \text{(iii) } \frac{\frac{1}{2}}{\frac{3}{4}}.

Exercise 1.5.

Let $x, y, z \in \RR$ be such that the following formulas are defined. Simplify:

\text{(i) } \left(\frac{4}{xy} + \frac{3}{yz}\right)\left(\frac{4z}{xy} - \frac{2}{y}\right), \qquad \text{(ii) } \frac{\frac{x}{y} - z}{\frac{x^2}{x} + \frac{z}{y}}, \qquad \text{(iii) } \frac{x}{1 - \frac{1}{1 - x}}.

Exercise 1.6.

Let us assume $\sqrt{2}$ is a rational number, that is, that there exist $a, b \in \ZZ$ with $a/b = \sqrt{2}$ . Suppose this fraction is reduced as much as possible.

By squaring the equation, show that $a$ is even.
Is $b$ even or odd?
Show that the two answers cannot both hold.
What do you conclude about $\sqrt{2}$ ?

Exercise 1.7.

Taking inspiration from the previous exercise, show that $\sqrt{p}$ is not a rational number when $p$ is prime. (Reminder: an integer is called prime when it has exactly two divisors.)

Exercise 1.8.

Let $x \in \QQ$ and $y \notin \QQ$ . Show that $x + y \notin \QQ$ .

Exercise 1.9.

Show that division in $\RR$ is neither commutative, nor associative, nor distributive over $+$ .

Exercise 1.10.

Let $a, b \in \RR_{+}^{*}$ .

If $a = b$ , can we say $\sqrt{a} = \sqrt{b}$ ?
If $\sqrt{a} = \sqrt{b}$ , can we say $a = b$ ?
Can we say $\sqrt{a + b} = \sqrt{a} + \sqrt{b}$ ?

Exercise 1.11.

Show that $1 + 2\sqrt{3} = \sqrt{13 + 4\sqrt{3}}$ .

Exercise 1.12.

Let $a, b \in \RR^{*}$ . Simplify the following expressions, where that is possible:

\text{(i) } \left(a^{-2} b^3\right)^2, \qquad \text{(ii) } \frac{a^2 + ab}{a^3}, \qquad \text{(iii) } \frac{a^2 + b^2}{ab}.

Exercise 1.13.

Let $a, b, c \in \RR$ . What is $(a + b + c)^2$ ?

Exercise 1.14.

Let $a > 0$ and $b > 0$ with $a \neq b$ . Show that

\frac{1}{\sqrt{a} - \sqrt{b}} = \frac{\sqrt{a} + \sqrt{b}}{a - b}.

Exercise 1.15.

Compute

\sqrt{\sqrt{6} - \sqrt{2}} \cdot \sqrt{\sqrt{6} + \sqrt{2}}.

Exercise 1.16.

Simplify:

\text{(i) } (-1 + a)^2 - (1 - a)^2, \qquad \text{(ii) } \left(a^2 + b^2\right)^2 - \left(a^2 - b^2\right)^2.

Exercise 1.17.

Simplify:

\text{(i) } \frac{3a^{n+1} \cdot 6x^{n+7} \cdot 9b^{x+1}}{3x^{n} \cdot 2b^{x+1} \cdot 3a}, \qquad \text{(ii) } \sqrt[3]{a^3} \cdot \sqrt{a^8}.

Exercise 1.18.

Let $A$ be the set whose elements are $a$ , $b$ , $c$ and $d$ .

Write $A$ in roster form.
How many subsets does $A$ have? How many of them have exactly two elements? List those by the roster method.
If a fair coin is tossed four times, is there a fifty-fifty chance of getting two heads and two tails? How is that question related to the previous part?
Now let $A$ have six members instead. How many subsets does it have, and how many of them have exactly two elements, exactly three, exactly four? Two of those three counts agree; say why.

Exercise 1.19.

The universe of discourse decides what an answer looks like.

Write the solution set of $x^4 = 1$ in set-builder notation, and then in roster form when the universe is (i) the complex numbers, (ii) the real numbers, (iii) the positive real numbers, (iv) the even integers. (Reminder: the complex numbers are the numbers $a + b\,i$ with $a$ and $b$ real and $i^2 = -1$ .)
Write the solution set of $(2x - 3)(x + 1)(x + 7) = 0$ in set-builder notation, and then in roster form when the universe is (i) the rational numbers, (ii) the real numbers, (iii) the integers, (iv) the positive integers.

Exercise 1.20.

Brackets matter in set expressions just as they do in arithmetic.

Show that $(A \cap B) \cup C$ and $A \cap (B \cup C)$ need not be equal. What is the most general condition on $A$ , $B$ and $C$ under which they are equal?
Why is it ambiguous to write $A \cap B \cup C$ ? Is it ambiguous to write $A \cap B \cap C$ ?
Show that $(A \cup B) \cap C$ and $A \cup (B \cap C)$ need not be equal, and that $(A \cup B) \cap C$ and $(A \cap C) \cup (B \cap C)$ always are.

Exercise 1.21.

Cancelling is not as free for sets as it is for numbers.

Show that it is possible to have $X \cup B = X \cup C$ and yet $B \neq C$ .
Show that if $X \cup B = X \cup C$ and also $X \cap B = X \cap C$ , then $B = C$ .

Exercise 1.22.

For real numbers $a$ and $b$ we say $a$ is less than $b$ , and write $a < b$ , if and only if $b - a$ is positive, that is, if $b = a + h$ for some positive $h$ . If $a < b$ we also say $b$ is greater than $a$ and write $b > a$ . Using only this definition and the rules of arithmetic, show the following.

If $a < b$ then $a + c < b + c$ , and also $a - c < b - c$ .
If $a < b$ and $c < d$ then $a + c < b + d$ . Is $a - c < b - d$ also true in this case? Explain.
If $a < b$ and $c$ is positive then $ac < bc$ . What happens if the restriction that $c$ be positive is dropped?
If $a < b$ and $a$ and $b$ are either both positive or both negative, which is to say $ab > 0$ , then $1/b < 1/a$ .

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