Why the rules must be true

In the earlier layers you used the properties and tested them. This layer asks the deepest question: how do we know they are true for every number, even numbers nobody will ever write down? And then: what do these properties let us build?

You will meet short, real proofs, some using pictures (arrays and boxes of cubes) and some using letters. You will see that long multiplication, divisibility tests and the whole of school algebra are built on the distributive property. You will find out why defining a value for 1 ÷ 0 would break arithmetic, trace how Indian mathematicians argued about zero 1,400 years ago, and invent your own operations to see which properties they keep. Take it slowly; the ideas here are the ones that make mathematics feel solid.

We checked 4 × 6 = 6 × 4 with chairs. But to be sure that 3,417 × 98,006 = 98,006 × 3,417 without calculating, we need an argument that does not depend on the particular numbers.

The array proof. Take any whole numbers a and b. Build an array of dots with a rows and b dots in each row. By the meaning of multiplication, it holds a × b dots. Now rotate the whole array a quarter turn. No dot is added or removed, so it still holds a × b dots. But now it has b rows with a dots in each row, which is b × a dots by the meaning of multiplication. The same set of dots was counted two ways, so a × b = b × a. Nothing in the argument depended on what a and b were, so it covers every pair.

The addition proof is even simpler: a red beads followed by b blue beads on a string. Read left to right: a + b. Turn the string round: b + a. Same beads.

This style of proof, counting the same collection in two ways, is one of the most powerful ideas in mathematics. You will meet it again in combinatorics and in the handshake problem in Extend.

Build a box out of small cubes: 2 cubes deep, 3 wide and 4 tall. How many cubes?

• Count by horizontal layers: each layer is 2 × 3 = 6 cubes, and there are 4 layers: (2 × 3) × 4 = 24.
• Count by vertical slices from front to back: each slice is 3 × 4 = 12 cubes, and there are 2 slices: 2 × (3 × 4) = 24.

Same box, same cubes, so (2 × 3) × 4 = 2 × (3 × 4) = 24. With a cubes by b cubes by c cubes, the same two counts show (a × b) × c = a × (b × c) for all whole numbers. Combined with commutativity, it means a product of many numbers can be rearranged in any order and grouping. A box of a × b × c cubes can be counted from any side.

The rectangle proof. A rectangle a units tall and (b + c) units wide can be cut by a vertical line into an a × b rectangle and an a × c rectangle. Cutting does not change area. So a × (b + c) = a × b + a × c for all whole numbers. This picture is in Euclid's Elements (Book II, Proposition 1), written in Greece around 300 BCE, long before letters were used for numbers.

Two brackets. Cut both sides: (a + b) × (c + d) is a rectangle split into four pieces, so (a + b) × (c + d) = a × c + a × d + b × c + b × d. Every piece of the first bracket multiplies every piece of the second. For example:

Adding the four pieces: 800 + 140 + 120 + 21 = 1,081, and indeed 23 × 47 = 1,081.

Squaring near round numbers. When both brackets are the same, (a + b) × (a + b) gives a × a + 2 × a × b + b × b. Written with squares: (a + b)² = a² + 2ab + b². This makes squares of numbers near a round number easy:

• 21² = 400 + 2 × 20 × 1 + 1 = 441.
• 99² = 10,000 − 2 × 100 × 1 + 1 = 9,801 (using (a − b)² = a² − 2ab + b²).
• 105² = 10,000 + 1,000 + 25 = 11,025.

And one more beautiful one: (a + b) × (a − b) = a² − b². So 52 × 48 = 50² − 2² = 2,500 − 4 = 2,496, and 103 × 97 = 10,000 − 9 = 9,991.

Why does "add the digits" test for divisibility by 9? Take 7,452. By place value it is 7 × 1,000 + 4 × 100 + 5 × 10 + 2. Now write each power of ten as a multiple of 9 plus 1: 1,000 = 999 + 1, 100 = 99 + 1, 10 = 9 + 1. By the distributive law:

7,452 = 7 × 999 + 4 × 99 + 5 × 9 + (7 + 4 + 5 + 2).

The first three parts are multiples of 9. So 7,452 is a multiple of 9 exactly when the leftover part, the digit sum 7 + 4 + 5 + 2 = 18, is a multiple of 9. It is, so 7,452 ÷ 9 = 828. Since 999, 99 and 9 are also multiples of 3, the same argument gives the test for 3.

In Understand we said a ÷ 0 has no answer because nothing times 0 gives a. Here is a stronger argument: if we insisted on giving 1 ÷ 0 a value, arithmetic would collapse.

Suppose 1 ÷ 0 = k for some number k. Division is defined so that (1 ÷ 0) × 0 = 1, so k × 0 = 1. But the zero property says k × 0 = 0. Therefore 0 = 1. And if 0 = 1, then adding 1 to both sides gives 1 = 2, and so on: every number equals every other number. The whole of arithmetic becomes meaningless.

So the choice is stark: either keep the zero property, the distributive law and the rest, or allow division by zero. Mathematicians keep the properties and leave division by zero undefined. (Some specialised systems, used in parts of higher mathematics and computing, add a special symbol "∞" or "NaN", but they pay for it by breaking some ordinary rules.)

Could there be a second additive identity, some number z other than 0 with a + z = a for every a? Here is a two-line proof that there cannot be.

Suppose both 0 and z are additive identities. Look at the single number 0 + z.

• Because 0 is an identity, 0 + z = z.
• Because z is an identity, 0 + z = 0.

So z = 0 + z = 0. The "second" identity was 0 all along. The additive identity is unique. The same argument with × shows 1 is the only multiplicative identity.

Inverses. For 5, is there a whole number w with 5 + w = 0 (the identity)? No: adding a whole number never takes you below 5. We need a new number, −5, the additive inverse. With it, subtraction becomes adding the inverse: 8 − 5 = 8 + (−5). Similarly, 5 × w = 1 needs the fraction ⅕, the multiplicative inverse or reciprocal, and division becomes multiplying by it: 8 ÷ 5 = 8 × ⅕. This is how mathematicians "fix" subtraction and division: they are not new operations at all, just + and × with inverses. And 0 has no multiplicative inverse, since 0 × w = 0, never 1. That is division by zero, seen from a new angle.

The last line gives a surprising fact: every square leaves remainder 0 or 1 when divided by 4, never 2 or 3. Check: 36 = 4 × 9, 49 = 4 × 12 + 1, 81 = 4 × 20 + 1. So a number like 1,234,567 (remainder 3 when divided by 4, since its last two digits 67 leave 3) can never be a perfect square. You do not need a calculator to know it.

Also, odd squares are always 1 more than a multiple of 8: m² + m = m × (m + 1) is a product of consecutive numbers, hence even, so 4 × (m² + m) is a multiple of 8. Check: 9 − 1 = 8, 25 − 1 = 24, 49 − 1 = 48, 81 − 1 = 80. All multiples of 8.

Mathematicians often ask: which properties does this operation have? Let's invent some operations on whole numbers and investigate. We use a new symbol for each.

• a ⊕ b = a + b + 1 ("add, then one more").
• a ⊡ b = the larger of a and b (write max(a, b); max(3, 7) = 7).
• a ⊖ b = the difference between a and b, bigger minus smaller (5 ⊖ 8 = 3).
• a ⊘ b = the average of a and b ((a + b) ÷ 2).
• a ⊲ b = a (just keep the first number).
• a ↑ b = a multiplied by itself b times (a power: 2 ↑ 3 = 8).

In algebra, letters stand for numbers. Everything you are allowed to do with letters comes from the properties, because the letters could be any numbers.

• Collecting like terms is distributive. 5x + 3x = (5 + 3) × x = 8x. There is no separate rule; it is the distributive law read backwards.
• Rearranging is commutative and associative. x + 7 + 2x + 3 = (x + 2x) + (7 + 3) = 3x + 10.
• Expanding brackets is distributive. 4(x + 5) = 4x + 20, and (x + 3)(x + 2) = x² + 2x + 3x + 6 = x² + 5x + 6.
• Solving equations uses identities and inverses. From x + 7 = 12, add −7 to both sides: x + 0 = 5, so x = 5. From 3x = 21, multiply by ⅓: 1 × x = 7.
• Factorising is distributive backwards. 6x + 9 = 3(2x + 3).

Whenever an algebra step seems mysterious, ask: which property is this? There is always an answer.