The rules of algebra

Solving $x + 7 = 3$ meant subtracting seven from both sides, and nobody said why that was allowed. The move is so familiar that the question sounds pedantic, but the whole of algebra is built out of a handful of such moves, and if they are taken on faith then every result downstream rests on faith too.

This lesson writes the licence down. It assumes only the number systems of the previous lesson, the naturals inside the integers inside the rationals inside the reals, and it produces a list short enough to memorise from which everything else follows, including the rule about two negatives that every student is told to accept.

The moves nobody justified

School algebra is taught as a set of permissions handed out one at a time. You may add the same thing to both sides. You may multiply out a bracket. You may cancel a common factor. You may not divide by zero. Each arrives as a separate instruction, and the impression left is of a large body of rules held together by nothing.

The truth is the reverse. There are about eight facts about addition and multiplication, all of them things you would guess, and every legitimate manipulation is a consequence of them. Nothing else is needed and nothing else is allowed. When a manipulation goes wrong, and one will go wrong later in this lesson in a way that produces the proof that $1 = 2$ , the fault is always that some step used a law outside the conditions the law actually carries.

The name for a system carrying these laws is a field, and $Q$ and $R$ are both fields. The integers are not, because they lack multiplicative inverses: $3 x = 7$ has no integer solution, which is exactly the failure that forced the rationals into existence.

The list

Addition is commutative, $a + b = b + a$ , and associative, $(a + b) + c = a + (b + c)$ . It has an identity, the number $0$ with $a + 0 = a$ , and every $a$ has an additive inverse $- a$ with $a + (- a) = 0$ . Multiplication has the same four properties: $a b = b a$ , $(a b) c = a (b c)$ , the identity $1$ with $a \times 1 = a$ , and for every $a$ except zero a multiplicative inverse $a^{- 1}$ with $a \times a^{- 1} = 1$ .

That is four laws twice over, and they say nothing about how the two operations interact. One law does: distributivity,

a (b + c) = a b + a c

This is the only bridge between addition and multiplication in the whole list, and it is the busiest law in mathematics. Expanding brackets is distributivity read left to right; factoring is the same law read right to left; long multiplication of ordinary numbers is distributivity applied to place value.

The laws about equations follow from these plus one fact about equality: if $a = b$ , then $a$ and $b$ are the same number, so anything true of one is true of the other. Adding $c$ to both sides is legal because $a + c$ and $b + c$ are the same number when $a$ and $b$ are. That is the licence that was missing in the previous lesson, and it costs one sentence.

Subtraction and division are shorthand

Notice what the list does not contain. There is no law of subtraction and no law of division, and this is not an oversight. Subtraction is defined out of what is there: $a - b$ means $a + (- b)$ , the addition of the additive inverse. Division likewise means multiplication by the multiplicative inverse, $a \div b = a \times b^{- 1}$ .

This is worth more than tidiness, because it explains a fact every student meets as a nuisance. Addition is commutative and associative; subtraction is neither. $7 - 3$ is not $3 - 7$ , and $(10 - 4) - 3 = 3$ while $10 - (4 - 3) = 9$ . The rules did not fail. Subtraction was never on the list, and the operation it abbreviates is commutative only in the sense that $a + (- b) = (- b) + a$ , which is not the statement that swapping the visible numbers is safe.

The same holds for division, which is why $12 \div 4 \div 2$ is ambiguous without a convention, while $12 \times 4^{- 1} \times 2^{- 1}$ is not ambiguous at all. Reading a subtraction as the addition of a negative, and carrying the sign with the number rather than leaving it stranded between two terms, removes most sign errors permanently.

What the laws force

The rule that a negative times a negative gives a positive is presented to most people as a convention, and Brahmagupta's fortunes and debts do read like a convention. They are not. Given the laws above, the rule is the only possibility, and here is the argument.

First, anything times zero is zero. Since $0 + 0 = 0$ , multiply both sides by $a$ and distribute: $a \times 0 = a (0 + 0) = a \times 0 + a \times 0$ . Now add $- (a \times 0)$ to both sides, which is legal, and the left becomes $0$ while the right becomes $a \times 0$ . So $a \times 0 = 0$ , derived rather than assumed.

Now take $(- 1) \times (- 1)$ . Start from $(- 1) + 1 = 0$ and multiply through by $- 1$ :

(- 1) big [(- 1) + 1 big] = (- 1) \times 0 = 0

Distribute the left side: $(- 1) (- 1) + (- 1) (1) = 0$ , which is $(- 1) (- 1) + (- 1) = 0$ . Add $1$ to both sides and $(- 1) (- 1) = 1$ . If a negative times a negative were anything else, distributivity would fail, and with it every bracket anyone has ever expanded. The rule is not a choice made for convenience; it is the price of keeping the arithmetic consistent.

Example. Use distributivity to compute $97 \times 103$ mentally.

Write it as $(100 - 3) (100 + 3)$ and expand: $100 \times 100 + 100 \times 3 - 3 \times 100 - 3 \times 3$ . The two middle terms cancel exactly, leaving $10000 - 9 = 9991$ . This is the difference of two squares, $(a - b) (a + b) = a^{2} - b^{2}$ , and it is nothing but distributivity applied twice.

Now you. Compute $58 \times 62$ the same way.

Answer

Both numbers sit six from $60$ , so this is $(60 - 2) (60 + 2) = 3600 - 4 = 3596$ .

Why zero has no inverse

The one prohibition in school algebra is the ban on dividing by zero, and it is usually taught as a rule with a vague warning attached. It is a theorem. The laws say every number except zero has a multiplicative inverse, and the exception is forced, because granting zero an inverse contradicts a result already proved.

Suppose $0^{- 1}$ existed, meaning some number $c$ with $0 \times c = 1$ . But it was proved two sections ago that $a \times 0 = 0$ for every $a$ , so $0 \times c = 0$ , and therefore $0 = 1$ . Once zero and one are the same number, every number is zero, since $a = a \times 1 = a \times 0 = 0$ , and arithmetic collapses into a single point. Division by zero is not undefined out of squeamishness. Defining it destroys the system.

This is also where the famous fake proof that $1 = 2$ hides. Start with $a = b$ , multiply by $a$ to get $a^{2} = a b$ , subtract $b^{2}$ to get $a^{2} - b^{2} = a b - b^{2}$ , factor both sides into $(a - b) (a + b) = b (a - b)$ , then cancel the common factor $(a - b)$ to get $a + b = b$ . With $a = b = 1$ that reads $2 = 1$ . Every step is legal except the cancellation, which is multiplication by $(a - b)^{- 1}$ , and $a - b$ is zero. Cancelling a common factor is only ever multiplication by an inverse, so it always carries the condition that the factor is not zero, and forgetting that condition is the commonest way to lose a solution to an equation.

Distributivity in both directions

Expanding is mechanical: every term in the first bracket multiplies every term in the second. Factoring is the same law used backwards, and it is harder because it is a search rather than a procedure. It is worth the effort because a factored expression displays its zeros: a product is zero exactly when one of its factors is, which is the fact the whole of equation solving rests on.

For a quadratic $x^{2} + b x + c$ the search has a shape. If it factors as $(x + p) (x + q)$ , then expanding gives $x^{2} + (p + q) x + p q$ , so $p$ and $q$ are two numbers that multiply to $c$ and add to $b$ . Since there are finitely many integer pairs multiplying to $c$ , the search terminates.

Example. Expand $(2 x + 3) (x - 5)$ , then factor $x^{2} - x - 12$ .

Expanding: $2 x \times x + 2 x \times (- 5) + 3 \times x + 3 \times (- 5) = 2 x^{2} - 10 x + 3 x - 15 = 2 x^{2} - 7 x - 15$ . Factoring: two numbers multiplying to $- 12$ and adding to $- 1$ are $- 4$ and $3$ , so $x^{2} - x - 12 = (x - 4) (x + 3)$ . Check by expanding: $x^{2} + 3 x - 4 x - 12 = x^{2} - x - 12$ .

Now you. Expand $(3 x - 4) (2 x + 5)$ , then factor the result back.

Answer

$6 x^{2} + 15 x - 8 x - 20 = 6 x^{2} + 7 x - 20$ . Factoring back means finding the same pair of brackets, $(3 x - 4) (2 x + 5)$ , which is the only factorisation with integer coefficients.

Order of operations, and what is actually a convention

Some of what looks like a law is only notation. The agreement that multiplication binds tighter than addition, so $2 + 3 \times 4$ is $14$ and not $20$ , is a convention about reading, not a fact about numbers. It could have gone the other way, and expressions would be written with more brackets. Associativity, by contrast, is a fact: $(2 + 3) + 4$ and $2 + (3 + 4)$ are equal whatever anyone agrees.

Distinguishing the two matters when a manipulation feels forbidden. Rearranging $a + b c$ into $b c + a$ is licensed by commutativity of addition. Rewriting it as $(a + b) c$ is licensed by nothing and is false, as $2 + 3 \times 4 = 14$ against $(2 + 3) \times 4 = 20$ shows. When in doubt, put in the brackets the convention allowed you to omit, and check which law the intended step is appealing to.

Example. Simplify $5 (2 x - 3) - 2 (4 x - 7)$ .

Distribute each bracket, taking the second sign with the $2$ : $10 x - 15 - 8 x + 14$ . The $- 2 \times - 7 = + 14$ is the negative-times-negative rule proved above. Collecting like terms, which is distributivity backwards on $10 x - 8 x = (10 - 8) x$ , gives $2 x - 1$ .

Now you. Simplify $3 (4 x + 5) - 4 (2 x - 3)$ .

Answer

$12 x + 15 - 8 x + 12 = 4 x + 27$ .

Where the laws stop

Two honest limits are worth stating. First, the laws describe addition and multiplication only. Inequalities carry their own rules, and one of them famously breaks the pattern: multiplying both sides of $a < b$ by a negative number reverses it, since $2 < 3$ but $- 2 > - 3$ . Nothing in the field laws covers order, and an ordered field is a stronger structure than a field.

Second, commutativity is not a property of everything worth multiplying. It fails for rotations in space, for the matrices met in linear algebra, and for the operators of quantum mechanics, where the failure is the physics. That such systems exist is the reason the laws are stated explicitly rather than assumed: writing them down is what makes it possible to notice which ones a new system keeps and which it abandons.

Within $R$ , though, the list holds without exception, and it licences every rearrangement the rest of this course performs. Rearranging is only half of what algebra does. The other half begins by taking two expressions and asserting they are equal, which turns a manipulation into a question with an answer, and answering that question is the next lesson.