You bundle at ten because you have ten fingers. Nothing else about arithmetic depends on it. Chapter 1 showed that a numeral never has two readings as long as no digit reaches ten. This chapter shows that the same is true for any bundle size, and then asks what each size costs you and what it buys.
On the table: make bundles of five instead of ten. Seven stones become one bundle and two loose, written 12 in fives. Nothing goes wrong, provided you never leave five loose stones unbundled. Five loose would be a bundle you forgot to tie.
For the teacher: Bundling.lean proves this for numerals of any length, in any base. If two digit lists of the same length, with every digit below the base, name the same number, they are the same list (unique_same_length). The one-line version for two digits was already in Numerals.lean; this is the general one. The script also checks it by brute force: 9,083 numerals in bases 2 to 12, and no two collide COMPUTED. And it keeps Chapter 1's break in view: let a digit reach the base, and [10, 1] and [0, 2] both name twenty.
Small bundles mean few symbols to learn but long numerals. Big bundles mean short numerals but many symbols, and a multiplication table that grows as the square.
| bundle | symbols to learn | digits for a million | times-table facts |
|---|---|---|---|
| 2 | 2 | 20 | 4 |
| 3 | 3 | 13 | 9 |
| 10 | 10 | 7 | 100 |
| 12 | 12 | 6 | 144 |
| 20 | 20 | 5 | 400 |
| 60 | 60 | 4 | 3,600 |
One standard way to weigh the trade is symbols times length. Measured as b / ln b, it is smallest at three, with two close behind and ten well back (2.73, 2.89, 4.34) COMPUTED. By that measure the cheapest bundle is small. Ten is a compromise that hands happened to make.
A fraction 1/n ends, instead of repeating forever, exactly when every prime in n divides the base. So the base decides which shares are clean. Of the nineteen fractions 1/2 to 1/20:
| base | end | which |
|---|---|---|
| 10 | 7 | 1/2 1/4 1/5 1/8 1/10 1/16 1/20 |
| 12 | 9 | 1/2 1/3 1/4 1/6 1/8 1/9 1/12 1/16 1/18 |
| 60 | 13 | all of those, and 1/5 1/10 1/15 1/20 |
In tens a third never ends. In twelves it is 0;4, four twelfths COMPUTED. That is the honest case for twelve, and a stronger one for sixty, which handles fifths as well.
This series argued the case for twelve in WP-45 · Dividing Unity. The script checks two of that paper's claims and both hold: twelve is the smallest number with six divisors, and the smallest that divides evenly into halves, thirds, quarters and sixths. It also records what the paper's title, “minimal sufficient arithmetic”, leaves out. Twelve is sufficient for those four cuts, not for fifths, and by this chapter's count sixty ends more fractions still.