1. The language of algebra¶

Arithmetic answers one question at a time: what is \(7 \times 8\)? Algebra asks about all numbers at once: what happens to every number when you double it, add ten and halve it? To ask such questions you need a way to write "any number", and rules for working with what you wrote. This chapter builds that language: letters for numbers, the words for the parts of an expression, the handful of laws every calculation obeys, the order in which to read a line of symbols, and the two everyday skills of simplifying and substituting. It ends by proving, for every number at once, why a party trick always works.

In this chapter

You will learn:

why mathematicians write letters for numbers, and the three different jobs a letter can do
the names of the parts of an expression: terms, factors, coefficients, constants, like terms
the laws of arithmetic (commutative, associative, distributive, identities, inverses) and why they hold
the order of operations, where it comes from and where it is ambiguous
how to simplify and expand expressions without the classic mistakes
how to substitute into formulas, and how to prove a statement about every number

You need: only arithmetic with whole numbers, fractions and negative numbers. For a gentler version, see the Topics lessons Letters and shapes, A mind-reading trick, Variables and expressions, Like terms and The distributive law.

Levels: 1.1 to 1.6 are beginner; 1.3's proofs from the laws and 1.7 are intermediate.

1.1 Why letters?¶

Try this on a friend, or on yourself.

Think of a number.
Double it.
Add 10.
Halve the result.
Take away the number you first thought of.

Whatever number you started with, you now have 5. Here it is for a few starting numbers:

Start	Double	Add 10	Halve	Take away start
3	6	16	8	5
12	24	34	17	5
100	200	210	105	5
\(-4\)	\(-8\)	2	1	5
\(\tfrac{1}{2}\)	1	11	\(\tfrac{11}{2}\)	5

Five rows, five 5s. That is evidence, but it is not a reason. There are infinitely many numbers, and no table can list them all. Maybe the trick fails for 7,341, or for \(-\tfrac{2}{9}\), and we never tried them. To be sure, we need to follow any number through the steps, without saying which one it is. That is exactly what a letter lets us do.

Call the starting number \(n\). We do not know what \(n\) is, and we do not need to. Doubling gives \(2n\) (which means \(2 \times n\)). Adding 10 gives \(2n + 10\). Halving gives \(n + 5\), because half of \(2n\) is \(n\) and half of 10 is 5. Taking away the start gives \(n + 5 - n = 5\). The \(n\) has vanished: the answer is 5 whatever \(n\) was. Section 1.7 comes back to this argument and makes each step of it watertight, but already you can see the point of algebra. A letter lets you calculate with a number you have not chosen yet, and so lets you reason about all numbers at once.

The three jobs of a letter¶

Letters turn up in mathematics doing three different jobs, and it helps to tell them apart.

A variable stands for any number from some collection, and the statement is meant to hold for each of them. In "\(n + 5 - n = 5\)", \(n\) is a variable: the statement is true for every number \(n\). A statement true for every value of its variables is called an identity.
An unknown stands for one particular number (or a few) that we are trying to find. In "\(3x + 2 = 17\)", \(x\) is an unknown, and the job is to find which number makes the statement true (\(x = 5\)). Such a statement is an equation, the subject of Chapter 2.
A parameter (or constant) is a number that is fixed in a given situation but could change from one situation to another. In the formula for the distance fallen by a dropped stone, \(d = \tfrac{1}{2} g t^2\), the time \(t\) is a variable, while \(g\) (the strength of gravity) is a parameter: about \(9.81\) metres per second per second on Earth and about \(3.71\) on Mars.

The same letter can change jobs. Nothing in the symbol \(x\) says which job it is doing; the sentence around it does. That is why good algebra is written in sentences: "for every number \(n\)", "find \(x\) such that", "where \(g\) is the acceleration due to gravity".

Why not just use words?¶

People did, for most of history. The Babylonian scribes who solved quadratic problems around 1800 BCE wrote them as stories about lengths and widths. The ninth-century Baghdad scholar Muhammad ibn Musa al-Khwarizmi, whose book gave algebra its name (Chapter 5 tells that story), wrote everything in words, including the numbers. Here is our trick written that way:

A thing is doubled, and ten is added to it. Half of this is taken, and the thing is taken away from what remains. What is left is five.

It can be done, but try following ten steps of that. Symbols do three things words cannot. They are short, so a whole argument fits in your eye at once. They are mechanical, so you can push them around by rules without re-thinking the meaning at each step (and a computer can do it too). And they are general: once you have shown \(n + 5 - n = 5\), you have shown it for every number at the same time.

Symbols arrived slowly. Diophantus of Alexandria (probably third century CE) used abbreviations for an unknown and its powers. In 1591 the French lawyer François Viète used letters systematically, vowels for unknowns and consonants for known quantities, so that he could write a whole class of problems as one. René Descartes, in La Géométrie (1637), used the late letters \(x, y, z\) for unknowns and the early letters \(a, b, c\) for known quantities, and wrote powers as \(x^2, x^3\), which is very nearly the notation we use today. The equals sign itself was introduced by the Welsh doctor Robert Recorde in The Whetstone of Witte (1557), who chose two parallel lines because, as he wrote, no two things could be more equal.

1.2 Expressions, terms and coefficients¶

An expression is a meaningful string of numbers, letters and operation signs, such as

\[3x^2 - 5xy + 7, \qquad \frac{a + b}{2}, \qquad 4(n - 1) + 2.\]

An expression has no equals sign; it is a name for a number, the way "half the sum of \(a\) and \(b\)" is a name for a number. Once you know what the letters stand for you can evaluate it, that is, work out which number it names. A statement that two expressions are equal, such as \(4(n-1) + 2 = 4n - 2\), is an equation (if it is meant to be true for particular values) or an identity (if it is meant to be true for all of them).

The parts of an expression have names, and you will hear them in every later chapter.

A term is a piece that is added or subtracted. The expression \(3x^2 - 5xy + 7\) has three terms: \(3x^2\), \(-5xy\) and \(7\). We count the sign with the term that follows it, so the second term is \(-5xy\), not \(5xy\).
A factor is a piece that is multiplied. The term \(-5xy\) has factors \(-5\), \(x\) and \(y\). Terms are separated by \(+\) and \(-\); factors are joined by multiplication. Mixing the two words up is the root of many algebra errors.
The coefficient of a term is its number factor: in \(3x^2\) it is \(3\), in \(-5xy\) it is \(-5\), and in \(x\) alone it is \(1\) (because \(x = 1 \cdot x\)). In \(-x\) it is \(-1\).
A constant term has no letters: \(7\) in our example.
Like terms have exactly the same letters raised to exactly the same powers. \(3x^2\) and \(-8x^2\) are like terms; \(3x^2\) and \(3x\) are not; \(5xy\) and \(2yx\) are (since \(yx = xy\), see 1.3).
The degree of a term is the total number of letter factors: \(3x^2\) has degree 2, \(-5xy\) has degree 2 (one \(x\) and one \(y\)), \(7\) has degree 0.

We also use a few writing conventions. A number written next to a letter means multiplication: \(5y\) means \(5 \times y\), and \(xy\) means \(x \times y\). The number is written first (\(5y\), not \(y5\)). The power \(x^2\) means \(x \cdot x\) and \(x^3\) means \(x \cdot x \cdot x\); in general \(x^n\), for a whole number \(n \geq 1\), means \(n\) factors of \(x\) multiplied together. Here \(x\) is the base and \(n\) is the exponent. Division is usually written as a fraction, \(\frac{a}{b}\) or \(a/b\), rather than with \(\div\).

Example 1. Name the parts of \(\dfrac{x}{2} - 4x^3 + x - 9\).

There are four terms: \(\frac{x}{2}\), \(-4x^3\), \(x\) and \(-9\). The coefficients are \(\frac{1}{2}\) (since \(\frac{x}{2} = \frac{1}{2}x\)), \(-4\), \(1\), and the constant term is \(-9\). The terms \(\frac{x}{2}\) and \(x\) are like terms; together they make \(\frac{3}{2}x\). The highest degree of any term is 3.

1.3 The laws of arithmetic¶

Every move you make in algebra rests on a short list of laws. They are so familiar from arithmetic that most people never notice they are using them. Here they are, for any numbers \(a\), \(b\), \(c\).

Law	For addition	For multiplication
Commutative (order does not matter)	\(a + b = b + a\)	\(ab = ba\)
Associative (grouping does not matter)	\((a + b) + c = a + (b + c)\)	\((ab)c = a(bc)\)
Identity	\(a + 0 = a\)	\(a \cdot 1 = a\)
Inverse	\(a + (-a) = 0\)	\(a \cdot \frac{1}{a} = 1\) if \(a \neq 0\)

and one law that links the two operations:

\[\text{Distributive law:}\qquad a(b + c) = ab + ac.\]

What they say. The commutative laws say you may swap the two numbers in a sum or a product. The associative laws say that in a sum of three numbers it does not matter which pair you add first, so we can just write \(a + b + c\) without brackets; the same for products. The identity laws say that adding 0 and multiplying by 1 change nothing; 0 and 1 are called the identity elements. The inverse laws say that every number \(a\) has a negative \(-a\) that brings it back to 0, and every number except 0 has a reciprocal \(\frac{1}{a}\) that brings it back to 1. The distributive law says multiplying a sum is the same as multiplying each piece and adding.

Subtraction and division are not on the list. They are defined from the others: \(a - b\) means \(a + (-b)\), and \(\frac{a}{b}\) means \(a \cdot \frac{1}{b}\). They obey none of the commutative or associative laws: \(5 - 3 \neq 3 - 5\), \((8 - 4) - 2 = 2\) but \(8 - (4 - 2) = 6\), and \((8 \div 4) \div 2 = 1\) but \(8 \div (4 \div 2) = 4\). Turning every subtraction into "adding a negative" is the cure for most sign errors: it turns a non-commutative operation into a commutative one.

Why they hold for counting numbers¶

For whole numbers, each law is a fact about counting.

\(a + b = b + a\): put a pile of \(a\) stones next to a pile of \(b\) stones. Counting the whole lot from the left or from the right gives the same total.
\(ab = ba\): lay out \(a\) rows of \(b\) stones in a rectangle. Counting by rows gives \(a \times b\); turn the rectangle a quarter-turn and the same stones are \(b\) rows of \(a\). Nothing was added or removed.
\((ab)c = a(bc)\): a box of stones \(a\) wide, \(b\) deep and \(c\) high, counted layer by layer two different ways.
\(a(b + c) = ab + ac\): a rectangle \(a\) rows tall and \(b + c\) columns wide splits into an \(a\) by \(b\) rectangle and an \(a\) by \(c\) rectangle. The same picture works for lengths and areas:

Why they hold for all real numbers¶

Negative numbers, fractions and numbers like \(\sqrt{2}\) are not piles of stones, so the counting pictures stop working. There are two honest answers to "why do the laws still hold?"

The first is that these number systems were built so that the laws keep holding. When mathematicians extended whole numbers to integers, integers to fractions, and fractions to the real numbers, at each stage they defined the new operations in the only way that keeps the old laws true, and then proved that the laws really do hold for the new definitions. For fractions, for example, multiplication is defined by \(\frac{a}{b} \cdot \frac{c}{d} = \frac{ac}{bd}\), and the commutative law for fractions then follows from the commutative law for whole numbers: \(\frac{ac}{bd} = \frac{ca}{db}\).

The second answer is the modern one: take the laws as the starting rules (axioms) of the number system and derive everything else from them. A collection of things with an addition and a multiplication obeying all the laws in the table plus the distributive law (and \(0 \neq 1\)) is called a field. The rational numbers, the real numbers and the complex numbers are fields; so are the "clock arithmetics" modulo a prime that power modern cryptography. You will meet fields properly in Chapter 9 and Chapter 19. The point for now is that anything we prove using only these laws is true in every field at once.

Facts that follow from the laws¶

Many "rules" you learned at school are not extra laws at all; they are consequences of the ones above. Proving a few of them is good practice in reading algebra as a chain of reasons, and it answers a question every student asks: why is a negative times a negative positive?

Fact 1. For every number \(a\), \(a \cdot 0 = 0\).

Proof. By the identity law, \(0 + 0 = 0\). Multiply both sides by \(a\) and use the distributive law: \(a \cdot 0 + a \cdot 0 = a \cdot 0\). Now add \(-(a \cdot 0)\) to both sides. On the left, \(a\cdot 0 + a\cdot 0 + (-(a \cdot 0)) = a \cdot 0 + 0 = a \cdot 0\) (by associativity, then the inverse law, then the identity law). On the right we get \(0\). So \(a \cdot 0 = 0\). \(\square\)

Fact 2. For every \(a\), \((-1) \cdot a = -a\).

Proof. We show \((-1)a\) does the job of the negative of \(a\), that is, adding it to \(a\) gives 0: \(a + (-1)a = 1 \cdot a + (-1) a = (1 + (-1))a = 0 \cdot a = 0\), using Fact 1 at the end. A number has only one negative (if \(a + x = 0\) and \(a + y = 0\), then \(x = x + (a + y) = (x + a) + y = 0 + y = y\)), so \((-1)a = -a\). \(\square\)

Fact 3. \((-1)(-1) = 1\), and so \((-a)(-b) = ab\).

Proof. By Fact 2 with \(a = -1\), \((-1)(-1) = -(-1)\), the number which added to \(-1\) gives 0. That number is 1, since \(-1 + 1 = 0\). So \((-1)(-1) = 1\). Then \((-a)(-b) = (-1)a \cdot (-1)b = (-1)(-1) \cdot ab = 1 \cdot ab = ab\), using the commutative and associative laws to move the factors around. \(\square\)

So "a negative times a negative is a positive" is not a convention someone voted for. It is forced by the distributive law. Any other answer would break \(a(b + c) = ab + ac\), and with it most of algebra.

Fact 4. If \(ab = 0\), then \(a = 0\) or \(b = 0\).

Proof. Suppose \(ab = 0\) and \(a \neq 0\). Then \(a\) has a reciprocal, and \(b = 1 \cdot b = (\tfrac{1}{a} a) b = \tfrac{1}{a}(ab) = \tfrac{1}{a} \cdot 0 = 0\). \(\square\)

Fact 4, the zero product property, is the reason factoring solves equations (Chapter 4). It also explains why you can never divide by zero: if \(\frac{1}{0}\) existed, then \(1 = 0 \cdot \frac{1}{0} = 0\) by Fact 1, which is false.

1.4 The order of operations¶

What is \(2 + 3 \times 4\)? If you work left to right you get \(5 \times 4 = 20\). If you do the multiplication first you get \(2 + 12 = 14\). Both readings are reasonable, so we need a rule, and everyone has to use the same one. The agreed rule is:

Brackets first, innermost first. A fraction bar and a square-root sign act as brackets around what they cover.
Powers (exponents) next.
Multiplication and division, from left to right.
Addition and subtraction, from left to right.

So \(2 + 3 \times 4 = 14\). This rule is a convention, not a law of nature, but it is not arbitrary. It is chosen so that the most common kind of expression in algebra, a sum of terms such as \(3x^2 + 5x + 2\), can be written with no brackets at all. Without the convention it would have to be written \((3(x^2)) + (5x) + 2\).

A few cases catch people out.

\(-3^2 = -9\), not \(9\). The power applies to 3 only, and then the negative is taken; the expression means \(-(3^2)\). If you mean the square of \(-3\), write \((-3)^2 = 9\). This matters when you substitute: if \(x = -3\), then \(x^2\) is \((-3)^2 = 9\).
\(2^{3^2}\) means \(2^{(3^2)} = 2^9 = 512\), not \((2^3)^2 = 64\). Towers of powers are read from the top down.
\(\dfrac{6 + 4}{2} = 5\), because the fraction bar groups the top. Typed on one line, it must become \((6+4)/2\); \(6 + 4/2\) is \(8\).
\(12 \div 3 \times 2 = 8\), read left to right as \((12 \div 3) \times 2\), not \(12 \div 6 = 2\).

Some expressions are genuinely ambiguous. The famous one is \(6 \div 2(1 + 2)\). Strictly left to right it is \(3 \times 3 = 9\); but many people (and some calculators and older textbooks) treat the "joined" multiplication \(2(1+2)\) as binding more tightly, giving \(6 \div 6 = 1\). There is no universal agreement, and that is the lesson: if an expression could be read two ways, add brackets or use a fraction bar. Professional mathematicians never write \(a \div bc\); they write \(\frac{a}{bc}\) or \(\frac{a}{b}c\), whichever they mean.

Example 2. Evaluate \(\;4 - 2(3 - 5)^2 \div 8 + 1\).

The bracket gives \(3 - 5 = -2\). The power gives \((-2)^2 = 4\). Then multiplication and division from left to right: \(2 \times 4 = 8\), then \(8 \div 8 = 1\). Finally \(4 - 1 + 1 = 4\).

1.5 Simplifying and expanding¶

To simplify an expression is to rewrite it in a shorter, equivalent form: one that gives the same value for every value of the letters. To expand is to remove brackets by multiplying out. Both use only the laws of 1.3.

Collecting like terms¶

\(5x + 3x = 8x\) is the distributive law read backwards: \(5x + 3x = (5 + 3)x = 8x\). Only like terms combine: \(5x + 3y\) is already as simple as it gets, and \(x^2 + x\) is not \(x^3\) or \(2x^2\) (try \(x = 3\): \(9 + 3 = 12\), while \(x^3 = 27\) and \(2x^2 = 18\)).

Example 3. Simplify \(4a - 7b + 2 - a + 3b - 9\).

Group the like terms (allowed by the commutative and associative laws, carrying each sign with its term): \((4a - a) + (-7b + 3b) + (2 - 9) = 3a - 4b - 7\).

Expanding brackets¶

A number outside a bracket multiplies every term inside: \(3(2x - 5) = 6x - 15\). A minus sign outside is a multiplication by \(-1\), so it changes the sign of every term inside:

\[-(4x - 3y + 1) = -4x + 3y - 1.\]

Forgetting to change the last sign, writing \(-4x + 3y + 1\), is perhaps the single most common error in school algebra. The fix is to treat the minus as \((-1) \times\) and multiply each term in turn.

Example 4. Simplify \(5(x - 2) - 3(2x - 4)\).

\(5x - 10 - 6x + 12 = -x + 2\). Note \(-3 \times (-4) = +12\), by Fact 3. Check with \(x = 1\): the original is \(5(-1) - 3(-2) = -5 + 6 = 1\), and the answer gives \(-1 + 2 = 1\). ✓

Multiplying two brackets¶

To multiply \((a + b)(c + d)\), use the distributive law twice. Treat \((c + d)\) as one number first: \((a + b)(c + d) = a(c + d) + b(c + d) = ac + ad + bc + bd\). Every term of the first bracket multiplies every term of the second. The picture is a rectangle cut into four:

Example 5. Expand \((2x + 3)(x - 4)\).

\(2x \cdot x + 2x \cdot (-4) + 3 \cdot x + 3 \cdot (-4) = 2x^2 - 8x + 3x - 12 = 2x^2 - 5x - 12\). Check with \(x = 2\): \((7)(-2) = -14\) and \(8 - 10 - 12 = -14\). ✓

Example 6. Expand \((x + 3)^2\).

\((x + 3)^2\) means \((x + 3)(x + 3) = x^2 + 3x + 3x + 9 = x^2 + 6x + 9\). It is not \(x^2 + 9\). With \(x = 1\): \((1+3)^2 = 16\), while \(1 + 9 = 10\). The two "cross terms" \(3x + 3x\) are exactly the two rectangles that are missing if you only square the pieces. In general

\[(a + b)^2 = a^2 + 2ab + b^2,\]

and the false "\((a + b)^2 = a^2 + b^2\)" has earned the nickname "the freshman's dream". Chapter 4 collects this and other identities.

Rules for powers¶

For whole-number exponents, the power rules follow from counting factors.

\(x^m \cdot x^n = x^{m+n}\): \(m\) factors of \(x\) followed by \(n\) more make \(m + n\) factors.
\((x^m)^n = x^{mn}\): \(n\) groups of \(m\) factors each.
\((xy)^n = x^n y^n\): rearrange \(n\) copies of \(xy\) (commutative and associative laws).
\(\dfrac{x^m}{x^n} = x^{m-n}\) when \(m > n\) and \(x \neq 0\): \(n\) of the factors on top cancel the \(n\) on the bottom.

The pattern does not extend to sums: \((x + y)^2 \neq x^2 + y^2\) and \(x^2 + x^3 \neq x^5\). Powers distribute over products, not over sums. Chapter 7 extends powers to zero, negative and fractional exponents and explains why \(x^0 = 1\).

Example 7. Simplify \(\dfrac{(3a^2b)^3 \cdot a}{9a^4b^2}\), for \(a, b \neq 0\).

The top is \(27a^6b^3 \cdot a = 27a^7b^3\). Dividing, \(\frac{27}{9} a^{7-4} b^{3-2} = 3a^3b\).

1.6 Substitution and formulas¶

To substitute is to replace each letter by a number and evaluate. Two habits prevent nearly all mistakes: put every substituted number in brackets, and then follow the order of operations.

Example 8. Find \(3x^2 - 2xy + y\) when \(x = -2\) and \(y = 5\).

\(3(-2)^2 - 2(-2)(5) + (5) = 3(4) - (-20) + 5 = 12 + 20 + 5 = 37\). Without brackets, "\(3 \cdot -2^2\)" invites the wrong value \(-12\).

A formula is an equation that says how to compute one quantity from others. Formulas are algebra's most practical product: they package a calculation once so anyone can repeat it.

Example 9 (temperature). Temperatures in degrees Fahrenheit \(F\) and Celsius \(C\) are linked by \(F = \frac{9}{5}C + 32\). On a warm Martian afternoon near the equator the ground can reach about \(20\) degrees Celsius, and on a winter night near the poles the air can fall to about \(-125\) degrees Celsius. In Fahrenheit these are \(\frac{9}{5}(20) + 32 = 68\) and \(\frac{9}{5}(-125) + 32 = -225 + 32 = -193\).

Example 10 (weight). The weight of an object, the force with which a planet pulls on it, is \(W = mg\), where \(m\) is its mass in kilograms and \(g\) is the planet's surface gravity. A 70 kg person weighs about \(70 \times 9.81 \approx 687\) newtons on Earth and \(70 \times 3.71 \approx 260\) newtons on Mars, about \(38\%\) as much, since \(3.71 / 9.81 \approx 0.378\). Their mass, 70 kg, is the same on both planets.

Example 11 (falling). A stone dropped from rest falls \(d = \frac{1}{2}gt^2\) metres in \(t\) seconds (ignoring air). After 3 seconds on Earth, \(d = \frac{1}{2}(9.81)(3)^2 = 44.145\) metres; on Mars, \(d = \frac{1}{2}(3.71)(9) = 16.695\) metres. Notice that the time is squared: three times as long means nine times as far.

Formulas also let you check work by units. In Example 11, \(g\) is in metres per second squared and \(t^2\) in seconds squared, so \(gt^2\) is in metres, as a distance should be. If the units on the two sides of a formula do not match, the formula is wrong.

Try it in Python¶

A computer substitutes faster than we do, and it lets you test a claim on many values. This checks the trick of 1.1 on a thousand random numbers, including fractions (exact arithmetic, so no rounding):

from fractions import Fraction
import random

def trick(n):
    return (2*n + 10) / 2 - n

tests = [Fraction(random.randint(-10**6, 10**6), random.randint(1, 1000)) for _ in range(1000)]
print(all(trick(n) == 5 for n in tests))   # True

A thousand successes is still not a proof; only the algebra of the next section is.

1.7 Proving the trick¶

Level: intermediate.

Here is the argument of 1.1 again, this time with the law that justifies each step. Let \(n\) be any number.

\[ \begin{aligned} \text{double it:}\quad & 2n \\ \text{add 10:}\quad & 2n + 10 \\ \text{halve it:}\quad & \tfrac{1}{2}(2n + 10) = \tfrac{1}{2}\cdot 2n + \tfrac{1}{2}\cdot 10 = n + 5 \quad\text{(distributive law)} \\ \text{take away the start:}\quad & (n + 5) - n = (n + (-n)) + 5 = 0 + 5 = 5 \quad\text{(commutative, associative, inverse, identity)} \end{aligned} \]

Every step uses only the laws of 1.3, which hold for every real number. So the result is 5 for every number \(n\), not just the ones in the table. This is what a proof is: a chain of steps, each justified by a rule already accepted, that reaches the conclusion for every case at once.

Algebra also shows you how to design such tricks. If you add \(k\) instead of 10, the answer is \(\frac{k}{2}\); if you then multiply by 4 instead of halving, the "\(n\)" no longer cancels (\(4(2n + k) - n = 7n + 4k\)), and the trick fails. Seeing exactly why something works also tells you exactly when it would stop working.

More proofs about all numbers¶

The same idea, name a general number with a letter and compute, proves many facts that patterns only suggest.

Odd and even. An even number is one of the form \(2k\) for a whole number \(k\); an odd number has the form \(2k + 1\). These are definitions, and they turn words into algebra.

Claim: the sum of two odd numbers is even. Take two odd numbers \(2j + 1\) and \(2k + 1\) (different letters, because they need not be the same number). Their sum is \(2j + 2k + 2 = 2(j + k + 1)\), which is 2 times a whole number, so it is even. \(\square\)

Claim: the product of two odd numbers is odd. \((2j + 1)(2k + 1) = 4jk + 2j + 2k + 1 = 2(2jk + j + k) + 1\). \(\square\)

Three in a row. Claim: the sum of any three consecutive whole numbers is divisible by 3. Call the middle one \(n\); then the three are \(n - 1\), \(n\), \(n + 1\), and their sum is \(3n\). \(\square\) Choosing to name the middle number rather than the first made the algebra shorter: a small instance of the art of picking good letters.

The 1089 trick. Take a three-digit number whose first and last digits differ by at least 2, say \(752\). Reverse it (\(257\)) and subtract the smaller from the larger: \(752 - 257 = 495\). Reverse the answer (\(594\)) and add: \(495 + 594 = 1089\). Try \(831\): \(831 - 138 = 693\), and \(693 + 396 = 1089\) again.

Proof. A three-digit number with digits \(a, b, c\) is \(100a + 10b + c\). Suppose \(a > c\) (otherwise swap the roles of the two numbers). The difference is

\[(100a + 10b + c) - (100c + 10b + a) = 99(a - c).\]

Write \(d = a - c\); then \(2 \leq d \leq 9\), and \(99d = 100(d - 1) + 10 \cdot 9 + (10 - d)\). (Check: \(100d - 100 + 90 + 10 - d = 99d\).) So the difference has digits \(d - 1\), \(9\), \(10 - d\), each between 0 and 9. Its reverse is \(100(10 - d) + 90 + (d - 1)\), and adding the two:

\[100\big((d - 1) + (10 - d)\big) + 180 + \big((10 - d) + (d - 1)\big) = 100 \cdot 9 + 180 + 9 = 1089. \quad\square\]

The digit \(b\) disappeared in the first subtraction and \(d\) disappeared in the sum. You can also see why the digits must differ by at least 2: if \(d = 1\), the difference is \(99\), a two-digit number, and the trick (read as 099 plus 990) needs care.

One more than a product of four. \(1 \cdot 2 \cdot 3 \cdot 4 + 1 = 25 = 5^2\), \(2 \cdot 3 \cdot 4 \cdot 5 + 1 = 121 = 11^2\), \(3 \cdot 4 \cdot 5 \cdot 6 + 1 = 361 = 19^2\). Claim: one more than the product of four consecutive numbers is always a perfect square.

Proof. Pair the outer two and the inner two: \(n(n + 3) = n^2 + 3n\) and \((n + 1)(n + 2) = n^2 + 3n + 2\). Write \(m = n^2 + 3n + 1\), the number halfway between them. Then the product is \((m - 1)(m + 1) = m^2 - 1\) (multiply out to check), so the product plus one is \(m^2 = (n^2 + 3n + 1)^2\). For \(n = 3\), \(m = 9 + 9 + 1 = 19\). ✓ \(\square\)

The step "write \(m\) for the middle value" is a typical algebraic move: give a name to a piece that keeps appearing, and the structure becomes visible.

What a proof is not¶

A table of examples, however long, is not a proof, and some patterns that hold for a long time do fail. The expression \(n^2 + n + 41\) gives a prime number for \(n = 0, 1, 2, \ldots, 39\), forty primes in a row; Euler noticed it in 1772. But for \(n = 40\) it gives \(40^2 + 40 + 41 = 40 \cdot 41 + 41 = 41^2\), which is not prime. Algebra found the failure as well: factor out 41. Examples suggest; only proofs settle.

A short history thread¶

The word algebra comes from the title of al-Khwarizmi's book, written in Baghdad around 820 CE: al-Kitab al-mukhtasar fi hisab al-jabr wal-muqabala ("The compendious book on calculation by restoring and balancing"). Al-jabr, "restoring", was the move of taking a subtracted quantity to the other side of an equation, where it becomes added. When the book was translated into Latin in the twelfth century, al-jabr became algebra. (His own name, Latinised, became the word algorithm.)

Al-Khwarizmi's algebra was entirely rhetorical: written in words. Diophantus's Arithmetica, centuries earlier, was syncopated: words with abbreviations for the unknown and its powers. Symbolic algebra, in which letters stand for both unknown and known quantities and are manipulated by rules, arrived in Europe between Viète (1591) and Descartes (1637). The laws of 1.3 were used for millennia before anyone listed them; the names "commutative" and "distributive" were introduced in 1814 by the French mathematician François-Joseph Servois, and the idea of studying the laws for their own sake, which leads to the abstract algebra of Part IV of this book, belongs to the nineteenth century.

Check yourself¶

1. In \(5x^3 - x + \frac{2}{3} - 7x^2y\), list the terms, their coefficients and their degrees.

Terms: \(5x^3\) (coefficient 5, degree 3), \(-x\) (coefficient \(-1\), degree 1), \(\frac{2}{3}\) (a constant term, degree 0), \(-7x^2y\) (coefficient \(-7\), degree 3, from two \(x\)'s and one \(y\)).

2. Evaluate \(10 - 4 \div 2 \times 3^2 + (1 - 4)^2\).

Brackets: \(1 - 4 = -3\). Powers: \(3^2 = 9\) and \((-3)^2 = 9\). Multiplication and division left to right: \(4 \div 2 = 2\), then \(2 \times 9 = 18\). So \(10 - 18 + 9 = 1\).

3. Simplify \(3(2a - b) - 2(a - 4b) + a\).

\(6a - 3b - 2a + 8b + a = 5a + 5b\). Check with \(a = 1, b = 1\): \(3(1) - 2(-3) + 1 = 3 + 6 + 1 = 10\), and \(5 + 5 = 10\). ✓

4. Expand and simplify \((3x - 2)(2x + 5) - (x - 1)^2\).

\((3x - 2)(2x + 5) = 6x^2 + 15x - 4x - 10 = 6x^2 + 11x - 10\). \((x - 1)^2 = x^2 - 2x + 1\). The difference is \(6x^2 + 11x - 10 - x^2 + 2x - 1 = 5x^2 + 13x - 11\). Check with \(x = 1\): \((1)(7) - 0 = 7\), and \(5 + 13 - 11 = 7\). ✓

5. Find \(a^2 - 3ab + 2b^2\) when \(a = -1\) and \(b = -3\).

\((-1)^2 - 3(-1)(-3) + 2(-3)^2 = 1 - 9 + 18 = 10\).

6. Which of these are identities (true for every \(x\))? (a) \(2(x + 3) = 2x + 6\); (b) \((x + 2)^2 = x^2 + 4\); (c) \(x - (3 - x) = 2x - 3\); (d) \(\frac{x + 4}{2} = x + 2\).

(a) Yes, by the distributive law. (b) No: the left side is \(x^2 + 4x + 4\); they agree only when \(x = 0\). (c) Yes: \(x - 3 + x = 2x - 3\). (d) No: \(\frac{x + 4}{2} = \frac{x}{2} + 2\); try \(x = 2\): \(3 \neq 4\). To show a statement is not an identity, one counterexample is enough.

7. Using only the laws of 1.3 and Facts 1 to 3, prove that \((-a)b = -(ab)\).

\((-a)b = ((-1)a)b = (-1)(ab) = -(ab)\), using Fact 2 twice and the associative law in the middle.

8. Prove that the sum of five consecutive whole numbers is always divisible by 5.

Name the middle one \(n\): the numbers are \(n - 2, n - 1, n, n + 1, n + 2\), and the sum is \(5n\).

9. A trick: think of a number, add 3, multiply by 4, subtract 8, divide by 4, subtract your number. Prove the answer is always the same, and find it.

\(n \to n + 3 \to 4n + 12 \to 4n + 4 \to n + 1 \to 1\). The answer is always 1.

10. Prove that the square of every odd number is one more than a multiple of 8.

An odd number is \(2k + 1\), and \((2k + 1)^2 = 4k^2 + 4k + 1 = 4k(k + 1) + 1\). Of the two consecutive numbers \(k\) and \(k + 1\), one is even, so \(k(k + 1) = 2j\) for some whole number \(j\), and \(4k(k+1) = 8j\). So the square is \(8j + 1\). (Check: \(3^2 = 9 = 8 + 1\), \(5^2 = 25 = 24 + 1\), \(7^2 = 49 = 48 + 1\).)

11. Show that \((a + b)(a - b) = a^2 - b^2\) and use it to compute \(103 \times 97\) in your head.

\((a + b)(a - b) = a^2 - ab + ba - b^2 = a^2 - b^2\), since \(ab = ba\). Then \(103 \times 97 = (100 + 3)(100 - 3) = 10000 - 9 = 9991\).

12. Why does the 1089 trick need the first and last digits to differ by at least 2? What happens with 554?

With \(d = a - c = 1\) the difference is \(99\), which has only two digits. For 554: \(554 - 455 = 99\). Read as 99, its reverse is 99, and \(99 + 99 = 198\). If you insist on three digits, \(099 + 990 = 1089\) still works, because the proof's digit formula \(d - 1, 9, 10 - d\) gives \(0, 9, 9\).

What you learned¶

A letter lets you calculate with a number you have not chosen, so one calculation covers every number at once.
Letters act as variables, unknowns or parameters; the sentence around them says which.
Expressions are built from terms (added) and factors (multiplied); only like terms combine.
Five laws (commutative, associative, identity, inverse, distributive) govern all of algebra; rules such as "negative times negative is positive" and "if \(ab = 0\) then \(a = 0\) or \(b = 0\)" are consequences of them.
The order of operations is a convention chosen so that sums of terms need no brackets; when in doubt, add brackets.
Expanding uses the distributive law: every term multiplies every term, and \((a + b)^2 = a^2 + 2ab + b^2\).
A proof covers every case at once; a table of examples never does.

Next: Chapter 2, how to find the unknown that makes an equation or an inequality true, and why the methods are safe.