Add $3+5$ and $5+3$: you get $8$ both times. Try $12+7$ and $7+12$, $100+1$ and $1+100$. It always works. But how many examples do you need to be sure? A thousand? A million? No number of examples is enough, because there are infinitely many pairs of numbers.

The sentence

$$a+b=b+a$$

solves the problem in another way. It speaks of no particular pair: it speaks of all of them at once. The letters $a$ and $b$ here do not hide a value, as the unknown does; they stand for any value. This is what Usiskin called generalised arithmetic.

Think of a number

An old party trick: think of a number, double it, add $6$, divide by $2$ and take away the number you thought of. The result is always $3$.

With examples you can gain confidence: with $5$ you get $10$, $16$, $8$, $3$; with $20$ you get $40$, $46$, $23$, $3$. With a letter you can be certain. Call the number $n$:

$$n\ \to\ 2n\ \to\ 2n+6\ \to\ \frac{2n+6}{2}=n+3\ \to\ (n+3)-n=3.$$

The letter did what the examples could not: it ran through all the numbers at once. And it showed why the trick works: the number you thought of goes in and comes out again, and only the $6$ divided by $2$ survives.

The trick step by step (click to advance)

  1. Think of a number: $n$.
  2. Double it: $2n$.
  3. Add $6$: $2n+6$.
  4. Divide by $2$: $n+3$.
  5. Take away the number you thought of: $(n+3)-n=3$. The $n$ has disappeared; the result does not depend on it.

The sum of the odd numbers

Add the first odd numbers: $1=1$, $1+3=4$, $1+3+5=9$, $1+3+5+7=16$. Squares appear. To say this for every $n$, we need a letter:

$$1+3+5+\cdots+(2n-1)=n^2.$$

The expression $2n-1$ is a portrait of the $n$-th odd number: for $n=1$ it gives $1$, for $n=4$ it gives $7$. The proof can be done with another letter and a sum written backwards, as the young Gauss did: adding the list to itself reversed, each pair gives $2n$, and there are $n$ pairs, so twice the sum is $2n\cdot n$ and the sum is $n^2$.

When examples deceive

Examples can be dangerous. In 1772 Euler noticed that the expression

$$n^2+n+41$$

gives prime numbers for $n=0, 1, 2, 3, \dots$: $41$, $43$, $47$, $53$, $61$, $71$… Test up to $n=39$ and all of them are prime. Forty examples in a row! But for $n=40$:

$$40^2+40+41=1681=41^2,$$

which is not prime. Written with letters, the failure can be seen without calculating: for $n=40$, $n^2+n+41=n^2+n+(n+1)=n(n+1)+(n+1)=(n+1)^2$. The letter does not only generalise: it also shows where the generalisation fails.

Viète’s species

It was François Viète, in 1591, who turned this idea into a method. He called calculation with letters logistica speciosa, calculation with species, and set it against logistica numerosa, calculation with numbers. A species, for Viète, was a general form, as the species “horse” covers all horses. His great innovation was to use letters also for the known quantities: in this way an equation could stand for a whole family of problems, and solving it was solving all of them at once.

The title page of Viète's Introduction to the Analytic Art (Tours, 1591), the book of the logistica speciosa.
The title page of Viète’s Introduction to the Analytic Art (Tours, 1591), the book of the logistica speciosa.Marcello Amadeo, from the copy in the Bibliothèque nationale de France · Wikimedia Commons · CC BY-SA 4.0

Semiotically, the letter that generalises is a symbol in Peirce’s sense, but of a special kind: it stands not for an object but for a class of objects, and any member of the class can take its place. The sentence with letters works like a promise: “choose whatever number you like; the equality will remain true”.

Exercise: prove with letters that the sum of two odd numbers is even

Any odd number can be written $2m+1$, and another one $2n+1$, with $m$ and $n$ any integers. The sum is $2m+2n+2=2(m+n+1)$, which is twice an integer: it is even. Note that we need two different letters. With just one, $2n+1$ and $2n+1$, we would only have proved that twice an odd number is even.

Exercise: invent a trick that always ends in 5

One solution: think of a number $n$, multiply by $4$, add $20$, divide by $4$ and take away the number you thought of: $\frac{4n+20}4-n=n+5-n=5$. The general recipe is to multiply by $k$, add $5k$, divide by $k$ and take away $n$; the letter shows that the trick works for any $n$ and for any $k\ne0$.


Chapter index · ← The unknown: a place for what is being sought · The variable: the letter that moves →