Positional Number Systems

When we write a number like 363, we are using our positional base ten number system. It is a positional system, because the first 3 means something quite different that the second 3. It is a base ten system, because the digits tell us how many given powers of ten we have. We all understand that 363= 3(10²)+6(10¹)+3(10⁰); more popularly, we talk of the ones place, the tens place, the hundreds place, and so on. The positional system is a much more efficient and compact than earlier tally-based systems of numeration.

Although the base ten system undoubtedly arose because we have ten fingers (digits!), it is interesting to note that first real positional system invented by the Sumerians in Mesopotamia was actually a base 60 system (representing the numbers from 1 to 59 was done using a ten-based tally system). The digit places then represented ones, sixties, sixty-squareds = 3600s, sixty-cubes = 216,000, etc. This is called a sexigesimal system. It seems reasonably clear than this system arose as a consequence of the fact that there are roughly 30 days in a month, and 360 days in a year. Indeed, the base sixty system survived in astronomical calculations into the time of Copernicus and Kepler. And when we measure the degrees, minutes and seconds of an angle, or minutes and seconds of time, we persist in using it to this day.

We will refrain here from using the Babylonian symbols for the “digits”. We will then interpret 12,43,17,0,5 as the base sixty number which is equal to

12(60⁴)+43(60³)+17(60²)+0(60¹)+5(60⁰)=164,869,205.

It is thus straightforward arithmetic to translate a base sixty number into a familiar base ten number.

For the reverse process, we have the right tools available. We use the division algorithm and modular arithmetic, modulo 60.

There are really two ways to do this, top-down and bottom up. In the first, we get the sexigesimal “digits” from left to right, and in the second from right to left.

We will illustrate here by an example. Suppose we wish to translate N=165,471 into base sixty notation.

To do this in the top-down fashion, we need to note first the values of the powers of 60:

Power of 60	Numerical Value
0	1
1	60
2	3600
3	216,000
4	12,960,000

The largest lump of 60s that is relevant is 60². If we divide 165,471 by 3600 we get 45, with a remainder of 3471. That is

165,471=45(60²)+3471.

We then divide 3471 by 60 to get

3471=57(60¹)+51.

Thus,

165,471=45(60²)+ 57(60¹)+51,

and so 165,471 in sexigesimal is really just 45,57,51.

The bottom-up method is a little slicker. We just repeatedly divide by 60. As a first step, we get

165,471=(2757)(60)+51.

The remainder 51 is precisely what N is equal to modulo 60, and it is precisely the ones place for the base sixty representation of 165,471. But now divide the quotient 2757 by 60. We get 2757=(45)(60)+57. We then have

N=(2757)(60)+51=(45)(60²)+(57)(60¹)+51.

Thus the second remainder in this process gives us the next sexigesimal digit. In this example, we get the last digit as the quotient 45, because this is less than 60. In principle, we could move on from here, in an efficient algorithmic way.

Special notice should be made of the importance the number zero plays here. In a positional system, it is vital to indicate that there may be zero of that particular power of the base. The Babylonians early on just left a space (which caused some ambiguity). Introducing zero into number systems was historically very important – earlier Babylonian calculations just left a space (which was confusing). Later calculations actually used a special symbol for zero – a first in the history of mathematics. Zero did not become a regular part of arithmetic until the Arabic adoption of the Hindu-Arabic numerals during the middle ages.

In our decimal system, we also include decimal places to the right of the decimal point. These digits represent tenths, hundredths, etc. The Babylonians did the same thing with sixtieths, (3600)ths, etc. We won’t actually pursue fractional numeration here (whew!).

There is no reason to stop with base 10 and base 60. Computer scientists are particularly enamored of base two (or binary) arithmetic. The binary places represent powers of two: ones, twos, fours, eights, etc. The advantage is that we need only two digits: 0 and 1 – or on and off, if we wish to translate numbers into electronic circuitry.

Homework problems on numeration, due on Monday, February 25^th:

1. Consider the integer 5,23,0,7,9 in sexigesimal notation. Translate this into a base ten integer.
2. Consider the integer 579,231 in base ten. Translate it into sexigesimal notation, using the top-down method. Now translate it into sexigesimal notation, using the bottom-up method. (Of course, you should get the same answer in either direction J).
3. Consider the integer 1001110 in binary notation. Translate this into a base ten integer.
4. Consider the integer 577 in base ten. Translate this into binary notation.
5. Consider the integer 45,691. Translate this into base 20.

Mayan Numbers and Calendars

The Mayans used a base 20 system (with an apparently bizarre twist, discussed below). The following symbols for the integers from 1 to 19; here a dot represents 1 and a dash represents 5. In addition, they had a symbol for zero.

These numbers gave them the ability to represent any integer using what is basically a base twenty system, with one important exception. If a Mayan were to write

this would represent the sequence of integers 10, 1, 4, 0, which the Mayan would interpret as . The funky 18 was used for calendar calculations. The number of days in the 365 day year (called the haab) is then just

.

That is, one 360=(18)(20), together with an extra five days. Now the haab is often called the civil year (because it was used for day-to-day, commercial purposes). Ascher prefers to call it the vague year – it is clearly based on the astronomical solar year, but it drifts away from the sun pretty quickly, because it is about a quarter-day off. The Mayans were well aware of the disparity, but did not seem to care. It is interesting that the ancient Egyptians used a 365 year; they too were well aware that such a year very quickly gets out of synchronization with the sun. Such years are sometimes called wandering years.

The haab has 18 “months” (called uinals), each with 20 days – we’ll call these months, even though they clearly have nothing to do with the moon. Each uinal (and the special five day period at the end of the year) has a name and glyph, which we will not bother to learn. For example, the Mayans would represent the 12^th day of the uinal called Mol as

.

Following Ascher, we will just denote this as 12A8: the 12^th day of the 8^th month. There is a special five-day period at the end of the year called the Uayeb.

More generally, we could denote the Cth day of the Dth month by CAD, where C is an integer between 1 and 20, and D is an integer between 1 and 18.

It is then easy to determine N, the day number in the haab corresponding to CAD. We have had D-1 complete months with 20(D-1) days. Then we have to add the extra C days to get 20(D-1)+C.

Conversely, suppose that N is the number of the day of the haab. To find the number of the month and the day of the month, we need to divide N by 20. If we do so, we get N=20Q+R, where Q and R are the quotient and remainder, respectively. Then C=R, and D=Q+1. Why do we have to add the 1?

Actually, there is a funny special case regarding C. If the remainder R is 0 when we divide by 20, this would say that we are at the 0^th day of the Q+1 month. We would then interpret this to mean the 20^th day of the Qth month. This complication arises because remainders when we divide by 20 consist of the number 0, 1, 2, …, 19, while day count numbers are 1, 2, 3, …, 20.

There is also the Sacred Round or tzolkin, or the counting of the days. For unknown reasons, Ascher prefers the French spelling of this word -- you will find it called the tolzin in her text; I prefer tzolkin, since it is the more commonly used word. This is a 260 day calendar, used for religious purposes. There are 13 day and 20 day cycles running simultaneously. Each day of the 20 day cycle (veintena) has a god name associated with it, and each day in the trecena (13 day cycle) has a number associated with between 1 and 13. We let A and B be the trecena and veintena counts respectively; we then represent a typical day of the tzolkin by ADB. For example, if we write 5D7, this means that we are at the 5^th day of the 13-cycle, and the 7^th day of the 20 cycle.

Suppose now that we have the 93th day of the tzolkin. We will always call this count M. At what day in the tricena and vientena are we? In other words, if we know M, what are A and B? Let’s divide M by 13. We then get 93=13Q+R. Now Q gives us how many completed 13-cycles we have, while R gives us the number of the day in the next cycle; this R=A (with the usual last-day exception). In this case, 93=13(7)+2, and so A=2. Similarly, if we divide M by 20, we get 93=20(4)+13. We thus have 4 complete 20-cycles, and we are in the 13^th day of the next one. That is, B=13.

We can thus record that in general M=A(mod 13) and M=B(mod 20). In our specific example above where M=93, A=2, B=13, we have that 93=2(mod 13), and 93=13(mod 20).

The converse problem is more difficult. Suppose that the tzolkin date is 7D11. This means that we are looking for a day in the tzolkin which is at the 7^th spot in the 13-cycle, and the 11^th spot in the 20-cycle. We must solve M=7(mod 13) and M=11(mod 20), simultaneously. Here are two solutions to this, in abbreviated form:

First solution (plug M=7(mod 13) into M=11(mod 20):

M=7(mod 13)

M=7+13Q

7+13Q=11(mod 20)

13Q=24(mod 20) [add 13 to both sides to eliminate the 7)

13Q=4(mod 20)

Q=68(mod 20) [multiply by 17, which Euclid gives as the mult. Inverse]

Q=8(mod 20)

M=7+13(8)=111. It is the 111^th day of the tzolkin.

Second solution (plug M=11(mod 20) into M=7(mod 13):

M=11(mod 20)

M=11+20Q [different Q than above!]

11+20Q=7(mod 13)

11+7Q=7(mod 13) [20=7(mod 13)]

7Q=9(mod 13) [add 2 to both sides to eliminate the 11]

Q=18(mod 13) [multiply by 2, the mult. Inverse of 7]

Q=5(mod 13)

M=11+20(5)=111. It is the 111^th day of the tzolkin.

The same answer both ways!!

The Calendar Round puts the 260 and 365-day cycles together. It lasts about 52 years. If R is the number in a calendar round, then R=M (mod 260) and R=N (mod 365). We’ll do examples in class.

The Long Count was a system designed to locate a day in much longer time spans. The terminology is as follows:

1 day = 1 kin

20 kins = 1 uinal = 1 “month”

18 uinals = 1 tun = 1 “year”

20 tuns = 1 katun = 1 “score of years”

20 katuns = 1 baktun = 1 “400 years”

13 katuns = Great Cycle = ‘’5200 years”

Note that the Great Cycle is 5200 360-day years, which translates to about 5130 of our years. The Great Cycle ends on December 23^rd, 2012.

Mayan Homework, Part One (due February 26^th)

1. Translate the Mayan number as an ordinary integer.
2. Translate the ordinary integer 784 into Mayan notation.
3. Consider the day 5A11 in the haab. What day of the year is it? How about for the day 11A5?
4. Suppose that it is the day N = 184 of the haab. What is the Ascher date for this?
5. Suppose that it is day M = 117 in the tzolkin. What are the tricena and vientena numbers for this day (that is, what is the Ascher date for this)?
6. Consider the Ascher date 9D7. What day of the tzolkin is this?
7. Redo exercise 6, making your substitution in the opposite order!
8. Consider the Long Count number 5.7.12.4.9 (we read from left to right: 5 baktun, 7 katuns, 12 tuns, 4 uinals, 9 days. How many days does this represent?
9. Express 1,000,000 days in terms of katun, tun, uinal and days. Turn this into a Mayan numeral.

Mayan Homework, Part One (due February 27^th)

1. Consider the day 2D4, 14A5. How many days after 4D5, 5A3 could this take place?

2. Suppose that M and N are 112 and 37; that is, it is the 112^th day of the tzolkin, and the 37^th day of the haab. What is the day R in the Calendar Round?

3. Suppose it is day R=1200 in a calendar round. Give the Ascher dates in both the tzolkin and the haab.