Part I, Chapter 3A Story of Numbers

Since when have humans been counting? Since the Stone Age.

The Lebombo bone, with 29 notches, is about 44,000 years old.

What was their need for counting? What were they counting? They counted food, animals in their livestock, goods traded and offerings in rituals.

They also counted the passing days, to know when the new moon, the full moon or a season would come.

Since when have people been writing numbers in the modern form? The ten digits including 0 were developed in India, around 2000 years ago.

The first known use is in the Bakhshali manuscript (c. 3rd century CE), where 0 was written as a dot.

How would the Mesopotamians have written 20? 50? 100? They used the symbol for 1 and the symbol for 10, in base 60.

\(20 = 2\times10\), so 20 is two symbols for 10.

\(50 = 5\times10\), so 50 is five symbols for 10.

\(100 = (1)\times60 + 40\), so 100 is one symbol for 1 in the 60s place and four symbols for 10 in the 1s place.

Q1: Keep one stick for every cow (one-to-one mapping).

In the evening match each returning cow with one stick.

If every stick gets a cow and no stick is left, all the cows have returned.

Each stick left over stands for a missing cow.

Q2: Make a collection of sticks for our herd and another for the neighbour’s herd.

Pair the sticks one-to-one: take one stick from each collection and put them aside together.

Repeat this till one collection finishes.

The collection that finishes first is the smaller one.

So we have fewer cows if our sticks finish first.

Q3: After the pairing, the sticks left in the neighbour’s collection show how many more cows we need.

Take that many new sticks and add them to ours.

Now both collections can be paired exactly.

Each object is matched to one letter, in the order of the letters.

So the count stops when the letters finish.

English: the letters a to z give 26 numbers.

Hindi: the 13 vowels and 33 consonants of the varnamala give \(13 + 33 = 46\) numbers.

The number of numbers is the number of letters in the alphabet, so it is always limited.

Table 1 uses I for 1, V for 5 and X for 10.

Other numbers are made by putting these symbols together, like XXVII for 27.

To go on, we can introduce a new symbol for a bigger group, just as V was made for 5 ones and X for 10.

For example, a new symbol for 50 (instead of XXXXX), then for 100, and so on.

Every bigger number is then written as a sum of these symbols, taking the biggest symbols first.

Addition: put the two collections of sticks together.

The new collection is the sum.

Example:

| | | and | | | | together give | | | | | | |.

Subtraction: pair each stick of the smaller collection with a stick of the bigger one and remove the pairs.

The sticks left in the bigger collection are the difference.

Multiplication: to multiply a collection by another, make as many groups as the sticks in the first collection.

Each group has as many sticks as the second collection.

Put all the groups together.

This is the product.

Example:

| | times | | | gives 2 groups of 3 sticks, which is | | |  | | |.

Division: to divide a collection by another, take out groups with as many sticks as the second collection.

Keep taking out groups till the sticks are finished (or fewer than one group are left).

The number of groups made is the quotient.

The sticks left over are the remainder.

The single letters a, b, c, …, z represent 1 to 26.

Method A: use two copies of a letter for the next 26 numbers: aa = 27, bb = 28, cc = 29, …, zz = 52.

Then use three copies: aaa = 53, bbb = 54, …, zzz = 78, and so on.

A string with \(k\) copies of a letter covers 26 numbers, so the strings never finish and every number gets a name.

Method B: write the letters like columns of a spreadsheet: a to z (1 to 26), then aa, ab, ac, …, az (27 to 52), then ba, bb, …

This gives \(26\times26 = 676\) numbers with two letters, \(26^{3} = 17576\) with three letters, and so on.

Here is one possible system.

Count in groups of 6.

The landmark numbers are 1, 6 and 36.

Use a triangle for 1, a square for 6 and a hexagon for 36.

Example 1: \(50 = 36 + 6 + 6 + 1 + 1\), which is 1 hexagon, 2 squares and 2 triangles.

Example 2: \(100 = 36 + 36 + 6 + 6 + 6 + 6 + 1 + 1 + 1 + 1\), which is 2 hexagons, 4 squares and 4 triangles.

No symbol appears 6 or more times, because 6 triangles make a square and 6 squares make a hexagon.

The dog (1 object) and the hens (2 objects) are seen at once, without counting.

The pyramids (3 objects) and the nesting dolls (4 objects) can also be seen at a glance.

The apples (5 objects) and the pencils (9 objects) are already difficult to take in at once, and we start counting.

For the bunch of flowers, the bunch of grapes and the steps of the staircase, we have to count one by one.

Most humans find it difficult to count groups having 5 or more objects in a single glance.

Representing 1345: count the number of 5s in 1345.

\(1345 \div 5 = 269\), with remainder 0.

So 1345 is 269 groups of 5.

We would have to write the symbol for 5 as many as 269 times.

Difficulties: the number of symbols is large, and it is hard to read and count 269 symbols.

This is the same problem as with tally marks, only 5 times shorter.

Numbers that are not multiples of 5 also need extra symbols for the leftover ones.

For very big numbers we would need to count the groups of 5 again, in a bigger group size.

LXXXVII has 1 L, 3 Xs, 1 V and 2 Is.

LXXVIII has 1 L, 2 Xs, 1 V and 3 Is.

Total: 2 Ls, 5 Xs, 2 Vs and 5 Is.

Regroup from the smallest symbol: 5 Is make 1 V.

Now there are 3 Vs.

2 Vs make 1 X.

So there is 1 V left and the Xs become 6.

6 Xs: 5 Xs make 1 L, so there is 1 X left, and the Ls become 3.

2 Ls make 1 C, so there is 1 L left.

The sum is C + L + X + V = CLXV.

Check: \(87 + 78 = 165\) and CLXV = 100 + 50 + 10 + 5 = 165.

There is no easy rule.

We can repeat the number again and again, and then regroup the symbols.

V × L: 5 Ls = L L L L L.

Two Ls make a C, so L L L L L = C C L.

V × L = CCL.

L × D: 50 Ds.

Two Ds make an M, so 50 Ds = 25 Ms.

There is no symbol bigger than M.

L × D = M written 25 times (MMMM…M).

V × D: 5 Ds = D D D D D.

Two Ds make an M, so this is M M D.

V × D = MMD.

VII × IX: VII = V + I + I and IX = X − I.

Multiply each part and regroup.

VII × X = LXX (7 tens).

VII × I = VII.

So VII × IX = LXX − VII.

LXX − VII = LXIII.

Check: \(7\times9 = 63\) and LXIII = 50 + 10 + 3 = 63.

Daredevil contest: multiply CCXXXI and MDCCCLII.

There is no easy rule to do this directly in Roman numerals.

For checking only: CCXXXI = 231 and MDCCCLII = 1852.

\(231\times1852 = 427812\).

The biggest Roman symbol is M (1000), and \(427812 = 427\times1000 + 812\).

So the product is M written 427 times, followed by DCCCXII (812).

Different objects are usually counted, traded or shared in different group sizes.

For example, we also say “a pair of shoes”, “a dozen bananas”, “a score”, or “a ream of paper”.

A separate sequence of names for each object can carry the group size that suits that object.

It also tells the listener what is being counted, and avoids mistakes when different goods are traded.

Another reason may be that counting in the natural groups of an object is faster.

In this system urapon = 1 and ukasar = 2.

A number is written by counting in 2s, so 5 = ukasar-ukasar-urapon.

Addition: write the two names one after the other.

Every pair of urapons becomes one ukasar.

Subtraction: cancel the same words from both names.

If an urapon has to be taken from an ukasar, break the ukasar into two urapons.

Multiplication by ukasar: double the name.

Each ukasar becomes ukasar-ukasar, and each urapon becomes ukasar.

To multiply by ukasar-ukasar, double twice.

Division by ukasar: halve the name.

Pair up the ukasars and keep one of every pair (an urapon cannot be halved).

To divide by ukasar-ukasar, halve twice.

(i) ukasar-ukasar-ukasar-ukasar-urapon is 9 and ukasar-ukasar-ukasar-urapon is 7.

Join them: 7 ukasars and 2 urapons.

The 2 urapons make 1 ukasar, so there are 8 ukasars.

The sum is ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar (8 ukasars, which is 16).

(ii) 9 − 6: cancel 3 ukasars from both.

Left: ukasar-urapon.

The difference is ukasar-urapon (3).

(iii) 9 × 4: double 9 once to get 18 (9 ukasars).

Double again to get 36.

The product is ukasar written 18 times (ukasar-ukasar-…-ukasar), which is 36.

(iv) 16 ÷ 4: halve 8 ukasars to get 4 ukasars.

Halve again to get 2 ukasars.

The quotient is ukasar-ukasar (4).

Start with the system made in Set 1 (landmark numbers 1, 6 and 36, with a triangle, a square and a hexagon).

Idea 1 (landmark numbers): group using landmark numbers, and never write a symbol 6 or more times.

Six triangles are replaced by one square.

Idea 2 (a base): make each landmark number 6 times the previous one: \(1, 6, 36, 216, \dots\) (powers of 6).

Then addition and multiplication become easy.

Idea 3 (place value): instead of new symbols for 6, 36, 216, …, use only the digits 0, 1, 2, 3, 4, 5 and let the position show the landmark number.

Idea 4 (zero): use the symbol 0 for an empty position.

For example, \(36 + 1 = 101\) in this system.

Each landmark number is a power of 10.

Multiplying by 10 increases the power by 1, which gives the next landmark number.

(i) heel bone × heel bone: \(10\times10 = 10^{2}\), which is the coil.

(ii) coil × heel bone: \(10^{2}\times10 = 10^{3}\), which is the lotus.

(iii) lotus × heel bone: \(10^{3}\times10 = 10^{4}\), which is the finger.

(iv) finger × heel bone: \(10^{4}\times10 = 10^{5}\), which is the tadpole.

Multiplying a power of 10 by \(10^{2}\) increases the power by 2.

(i) heel bone × coil: \(10\times10^{2} = 10^{3}\), the lotus.

(ii) coil × coil: \(10^{2}\times10^{2} = 10^{4}\), the finger.

(iii) lotus × coil: \(10^{3}\times10^{2} = 10^{5}\), the tadpole.

(iv) finger × coil: \(10^{4}\times10^{2} = 10^{6}\), the astonished man.

Add the powers of 10: \(10^{a}\times10^{b} = 10^{a+b}\).

(i) heel bone × tadpole: \(10^{1}\times10^{5} = 10^{6}\), the astonished man.

(ii) coil × lotus: \(10^{2}\times10^{3} = 10^{5}\), the tadpole.

(iii) lotus × lotus: \(10^{3}\times10^{3} = 10^{6}\), the astonished man.

(iv) finger × astonished man: \(10^{4}\times10^{6} = 10^{10}\).

There is no symbol for \(10^{10}\) in the Egyptian system (the last symbol is \(10^{7}\)).

In the base-5 system the landmark numbers are \(5^{0}, 5^{1}, 5^{2}, \dots\)

\(5^{a}\times5^{b} = 5^{a+b}\), which is again a power of 5, that is, a landmark number.

For example, \(25\times125 = 5^{2}\times5^{3} = 5^{5} = 3125\), which is the arrow.

In a base-\(n\) system the landmark numbers are the powers of \(n\), and \(n^{a}\times n^{b} = n^{a+b}\) is again a power of \(n\).

Take a number such as \(522 = 5\) coils, 2 heel bones, 2 strokes.

Multiplying by 10 multiplies each landmark number in it by 10, which gives the next landmark number.

So each coil becomes a lotus, each heel bone becomes a coil and each stroke becomes a heel bone.

The number becomes 5 lotuses, 2 coils, 2 heel bones, which is 5220.

(i) The numeral has 5 coils, 2 heel bones and 2 strokes, which is 522.

Multiply each part by 10 (distributive property).

\(522\times10 = (500 + 20 + 2)\times10 = 5000 + 200 + 20\)

This is 5 lotuses, 2 coils and 2 heel bones, which is 5220.

(ii) The numeral has 1 lotus and 1 heel bone, which is \(1000 + 10 = 1010\).

\(1010\times10 = 10000 + 100\)

This is 1 finger and 1 coil, which is 10100.

Every landmark number multiplied by 10 gives the next landmark number.

So replace each symbol of the number by the next higher symbol: stroke → heel bone, heel bone → coil, coil → lotus, lotus → finger, and so on.

The number of times each symbol appears does not change.

Example: 5 coils, 2 heel bones, 2 strokes becomes 5 lotuses, 2 coils, 2 heel bones.

Each line of the abacus is a power of 10.

A counter above a line is worth 5 on that line.

Bring the counters of each line together.

Line of 1000: 2 counters.

Line of 100: 1 upper counter and 4 counters (900).

Line of 10: 4 counters.

Line of 1: the 7 ones (1 upper counter and 2 counters) and the 3 ones make 10 ones.

When a line has 10 ones, remove them and put 1 counter on the line of 10.

This gives 5 counters on the line of 10.

5 counters on the same line are replaced by 1 upper counter (worth 5).

Check: \(2907 + 43 = 2950\).

No.

Suppose a symbol occurs 10 times.

Its value is \(10\times\) (the landmark number).

But 10 times any landmark number is the next landmark number, which has its own symbol.

So 10 copies of a symbol are always replaced by 1 copy of the next symbol.

Hence every symbol occurs at most 9 times.

In a base-4 system the landmark numbers are \(4^{0} = 1\), \(4^{1} = 4\) and \(4^{2} = 16\).

Let a triangle stand for 1, a square for 4 and a hexagon for 16.

Group each number into 16s, 4s and 1s.

A symbol never occurs 4 or more times, because 4 triangles make a square and 4 squares make a hexagon.

For example, \(11 = 4 + 4 + 1 + 1 + 1\), which is 2 squares and 3 triangles.

\(60 = (1)\times60\) is written with one symbol for 1 in the 60s place.

The 1s place is empty.

\(3600 = (1)\times3600\) is also written with one symbol for 1, in the 3600s place.

The 60s place and the 1s place are empty.

The empty places come after the symbol, so nothing is written for them.

So 3600 looks exactly like 60 (and like 1).

We cannot tell them apart without knowing the spacing.

The landmark numbers are 1, 20 and 360.

A dot is 1 and a bar is 5, so numbers from 1 to 19 are made from dots and bars.

The shell stands for 0.

The places are written one above the other.

The lowest place is the 1s, the one above is the 20s, and the next above is the 360s.

(i) \(77 = (3)\times20 + (17)\times1\).

The 20s place has 3 dots.

The 1s place has \(17 = 5 + 5 + 5 + 2\), which is 3 bars and 2 dots.

(ii) \(100 = (5)\times20 + (0)\times1\).

The 20s place has 1 bar (5).

The 1s place has the shell (0).

(iii) \(361 = (1)\times360 + (0)\times20 + (1)\times1\).

The 360s place has 1 dot, the 20s place has the shell, and the 1s place has 1 dot.

(iv) \(721 = (2)\times360 + (0)\times20 + (1)\times1\).

The 360s place has 2 dots, the 20s place has the shell, and the 1s place has 1 dot.

Where does it figure? It is the last stage of the ideas we have seen: counting in groups, landmark numbers, a base, place value, and 0 as a digit and as a number.

It uses a base (10), place value, and the symbol 0.

Landmark numbers: the powers of 10, that is \(1, 10, 10^{2}, 10^{3}, \dots\)

Place value: yes. The position of each digit shows which landmark number it goes with.

For example, in 375 the digit 3 is in the \(10^{2}\) place, 7 is in the 10s place and 5 is in the 1s place.

\(375 = (3)\times10^{2} + (7)\times10 + (5)\times1\)

Because 0 is used as a digit, there is no ambiguity in reading a numeral.

Why alternate? The Zong symbols are vertical and the Heng symbols are horizontal.

Two neighbouring places then look different.

So the digits of the numeral can be separated even when the spacing between the places is not clear.

41 with only Zong symbols: 4 vertical strokes for the tens, a small gap, then 1 vertical stroke for the unit.

Other readings: if the gap is not clear, the numeral is just 5 vertical strokes.

It can be read as 23 (2 + 3 strokes), 32 (3 + 2), or 122 (1 + 2 + 2).

More readings: 5, 14, 41, 113, 131, 311, 212, 221, 1112, 1121, 1211, 2111 and 11111.

In all, a group of 5 strokes can be read in 16 ways.

With alternating symbols: 41 is 4 Heng bars (tens) followed by 1 Zong stroke (unit).

This can be read only as 41.

With 8 fingers we would probably count in groups of 8, with a base-8 system.

The digits would be 0, 1, 2, 3, 4, 5, 6, 7 (no 8 and no 9).

The landmark numbers would be \(1, 8, 64, 512, \dots\) and the digit in each place would show how many of that landmark number there are.

In the same way, a base-5 system would use the digits 0, 1, 2, 3, 4 and the landmark numbers \(1, 5, 25, 125, \dots\)

25 in base 8: \(25 = 3\times8 + 1\), so 25 is written as 31.

25 in base 5: \(25 = 1\times25 + 0\times5 + 0\times1 = 1\times5^{2} + 0\times5^{1} + 0\times5^{0}\), so 25 is written as 100.

25 in base 2: \(25 = 16 + 8 + 1 = 1\times2^{4} + 1\times2^{3} + 0\times2^{2} + 0\times2^{1} + 1\times2^{0}\), so 25 is written as 11001.