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History of Math
for the Liberal Arts
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CHAPTER 3
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Babylonian Mathematics
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Lawrence Morales
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Seattle Central
Community College
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MAT107 Chapter 3, Lawrence Morales, 2001; Page 1
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Table of Contents
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Part 1: Introduction to Babylonian Numbers ...................................................................................... 4
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Background & Historical Information.................................................................................................. 4
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Early Mathematical Development ........................................................................................................ 5
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The Babylonian Writing System and Scribal Schools........................................................................... 5
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Babylonian Number Symbols................................................................................................................ 7
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Part 2: The Sexagesimal System............................................................................................................ 8
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The Sexagesimal System as Used by the Babylonians .......................................................................... 8
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Difficulties With Babylonian Numbers ............................................................................................... 10
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Babylonian Fractions.......................................................................................................................... 11
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Going the Other Direction. ................................................................................................................. 14
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Why Base−60? .................................................................................................................................... 18
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Part 3: Babylonian Arithmetic ............................................................................................................ 19
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Addition............................................................................................................................................... 19
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A Primer on Multiplication................................................................................................................. 21
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The Babylonian Multiplication Table ................................................................................................. 28
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Babylonian Multiplication with Tables............................................................................................... 30
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Babylonian Multiplication on Tablets ................................................................................................ 34
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Babylonian Division............................................................................................................................ 34
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Division on Babylonian Tablets.......................................................................................................... 37
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Part 4: Babylonian Root Approximations .......................................................................................... 38
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Babylonians and Square Roots ........................................................................................................... 38
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An Alternate Method of Estimating Roots .......................................................................................... 41
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Plimpton 322....................................................................................................................................... 43
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Part 5: Babylonian Algebra ................................................................................................................. 44
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Systems of Equations........................................................................................................................... 44
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Babylonian Quadratics ....................................................................................................................... 50
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General Approaches to Quadratics .................................................................................................... 54
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Part 6: Homework Problems ............................................................................................................... 57
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Conversions......................................................................................................................................... 57
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A Babylonian Translation Problem .................................................................................................... 57
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Multiplication...................................................................................................................................... 58
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Division ............................................................................................................................................... 58
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Tablet Problems .................................................................................................................................. 58
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Root Approximations .......................................................................................................................... 59
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The Alternate Method of Estimating Roots......................................................................................... 60
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The Babylonians and Pythagorean Triples......................................................................................... 60
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A Famous Babylonian Tablet.............................................................................................................. 61
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Babylonian Algebra − Systems of Equations ...................................................................................... 61
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Babylonian Algebra − Solving Quadratic Equations ......................................................................... 62
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Practice and Application of the Quadratic Equation ......................................................................... 63
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New Quadratic Equations................................................................................................................... 63
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Yet Another Quadratic Formula ......................................................................................................... 64
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Real Babylonian Algebra Problems From Tablets............................................................................. 65
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Writing ................................................................................................................................................ 66
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Appendix: Blank Gelosia Grids........................................................................................................... 67
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Specific Gelosia Grids ........................................................................................................................ 71
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Part 6: Endnotes.................................................................................................................................... 73
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Part 1: Introduction to Babylonian Numbers
The next civilization that we will explore in this course is ancient Mesopotamia. While this is its
official, scholarly title, it is often referred to as Babylon. However, Babylon was only a small
part of Mesopotamia. In this reading, I will use the terms interchangeably, despite the fact that
they are not the same. As we did with the Egyptians, we will explore the symbols for their
numbers, their base system, and their basic
methods of doing arithmetic. We will need to
come up with new methods of operations for
multiplication and division, as the
Babylonians did not give us as much
information on how they did their calculations
as the Egyptians did. We will also look at
some of their algebraic techniques,
particularly for solving quadratic equations.
Background & Historical Information1
The Babylonian civilization was made up of
people who lived in the alluvial plain between
the Tigris and Euphrates rivers. This is the
area from Baghdad south to the Persian Gulf. The Greeks were originally those who called it
“Mesopotamia,” which means, “land between the rivers.” (See the map of the region. 2)
Cities, writings, and metallurgy were necessary to form this civilization.3 The cities had
extensive irrigation systems, codes of law, postal services, and an administrative bureaucracy.
Besides writing, they utilized other forms of technology such as wheeled vehicles, boats, plows,
weaving, and brick towers known as ziggurats. To build tools (and weapons), they used copper
smelting and bronze smelting techniques.
Knowledge about their political and cultural history is somewhat limited4 but is growing. There
are many tablets that have been discovered with writings from this time period, but translation is
slow and a relatively recent process. It wasn’t until 1846 that the deciphering code was broken,
which then allowed the translation process to begin. However, there are so many tablets that the
task is still far from complete.5
From what we do know, a series of invasions and sporadic warfare served to awaken and
stimulate the culture. The geography of the land made cities of the river plain open to blatant
attack from many groups. Many groups of invaders battled for control in the region for about
three thousand years.
One dynasty, the Amorite dynasty, achieved its greatest might under the rule of Hammurabi
(circa 1792-1750 B.C.E.). Hammurabi is now known for his code of law, “The Hammurabi
Code” which was very logical but also very harsh. One of its injunctions is the familiar “an eye
for an eye and a tooth for a tooth.”
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Early Mathematical Development
In terms of mathematics, influences of imperialism and trade over long distances served as the
catalysts of mathematical development6. These were both relatively new to human society at the
time and encouraged the establishment of news schools and temple scribes who could record and
manage the new collections of information needed to support these endeavors. However, everyday needs such as religion, commerce, and agriculture were even stronger influences on the
development of mathematics.7 Grain supplies had to be tracked and distributed among an
increasingly large population. Daily business transactions and the use of wills encouraged the
creation of numerical tables. The building of dams, irrigation canals, granaries and other
buildings necessitated calculations be made while the religious practices of the time were heavily
reliant on having a dependable calendar. This means keeping careful records of astronomical
data.
In the following table, you can see a general breakdown of periods of history in the context of
their mathematics.
Period
5000−3000BCE
3000−2000BCE
2000−1600BCE
1600−0BCE
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133
134
135
136
137
138
139
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Mathematical Development
Development of number concepts in prehistoric Middle East
Evolution of the sexagesimal place value system in southern Iraq
Arithmetic in Old Babylonian scribal schools
Old Babylonian mathematics
As the table indicates, the Babylonians had what is called a sexagesimal place system, which
means they had a base−60 system. Also, notice that there is an entry in the table dealing with
scribal schools. We’ll briefly look at the role of such schools in the development of Babylonian
mathematics.
The Babylonian Writing System and Scribal Schools
There are about 400,000 Babylonian tablets that have survived to the present day. Many of them
are hidden in the library collections of older universities and have not been looked at in many
years. (In the summer of 1999, Dr. Eleanor Robson discovered several at Catholic University of
America while we were attending an NSF Institute on the Use of History of Math in the
Classroom.) Of these, about 400 are related to mathematics. Most are what are called “Old
Babylonian,” (1800 to 1600 B.C.E.) and they contain less information than the Egyptian
mathematics sources that we have available to us. However, from what we do have, it appears
that the Babylonians were more “advanced” in mathematics than the Egyptians were.
The tablets come in a variety of shapes and sizes:
−Large multi−columned tablets: 2 to 6 columns, “printed” on both sides.
−Large student−teacher tablets: The teacher would use large script while student writing
would be smaller and to the right.
−Small one-column tablets.
−Round “buns”: 2 to 4 lines, usually used for student−teacher exercises.
MAT107 Chapter 3, Lawrence Morales, 2001; Page 5
A “bun” tablet, which is a school exercise in
repeated multiplication and division.8
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A meteorological calculation
From the list of types of tables above you see a hint that a big source of tablets, math, and the
start of the sexagesimal system comes from school tablets. These were tablets used in scribal
schools that were used to train young boys/men to do the work of scribes. The curriculum of the
school was progressive9:
1. Learn the basics of writing letters, names, and forms.
2. Learn lists of trees, reeds, vessels, animals, stones, plants, and other nouns by theme.
3. Learn how to represent meteorological data, weights, multiplication, and reciprocals.
4. Learn how to write contracts and proverbs.
Shortly after 3000 B.C.E., the Babylonians developed a system of “pictographs” (sort of like
hieroglyphics) to represent their numbers. These are called cuneiform.10 While the Egyptians
used a form of ink to write on their papyrus, the Babylonians used a reed and later a stylus with a
triangular head to make pictographic impressions into clay tablets. These tablets were then baked
so they could become hardened to preserve the writing.
The limitation of the writing stylus that they used prevented them from having a wide variety of
symbols that they could use to represent numbers. In particular, the creation of curved lines was
pretty much impossible so they resorted to a series of vertical, horizontal, and oblique marks, as
you can see above. As time passed, they also found a way to draw a wedge that looks like an
angled bracket opening to the right. (Holding the stylus so that its sides were inclined on the
tablet could do this.)
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Babylonian Number Symbols
The symbols that we will be using (and which they used) are the following:
Symbol
Name
Decimal Value
or
Wedge
1
or
Corner
10
None. It specifies when a place is
Empty Placeholder empty and was not used until 300
B.C.E.
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185
As you can see, there are only two main symbols for numbers in the Babylonian system, the
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for ones and the
for tens. Compare this with the Egyptian system and even our own system
that needs many more symbols to represent some arbitrary number. The Babylonians would use
just these two symbols within a base−sixty system to represent any number they wanted. In the
picture below11, you can see the symbols for various numbers:
1=
7=
15 =
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192
193
194
195
196
197
198
199
200
201
202
2=
3=
8=
9=
20 =
30 =
4=
5=
6=
10 =
11 =
12 =
40 =
50 =
They also has certain special symbols for some common fractions
which we will not use much in this text, but they are interesting to note
nonetheless.
60 =
Think About It
In the list of symbols
above, the symbol for
1 and for 60 is the
same? Why would
that be true?
Note the absence of a symbol for zero or for the equivalent of a decimal point (some symbol that
separates whole numbers from fractional numbers). This will cause some problems and
ambiguity in interpreting the values of Babylonian numbers, but for now it’s just interesting to
note the limited number of symbols they had available to them to write numbers.
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Part 2: The Sexagesimal System
In the first topic of this course, we learned about different base systems. In particular, we looked
at the Mayan base−20 system and how to convert between their system and our own. The
process of converting between base−60 and base−10 is essentially the same as it was with the
Mayans, except that we replace the base of 20 with the base of 60. Let’s first start by converting
from base−sixty to base−ten, which is generally easier.
The Sexagesimal System as Used by the Babylonians
Unlike the Egyptians’ system, the sexagesimal system is a positional system. That means that the
position of a symbol will be important in determining the value of that symbol. In the
sexagesimal system, every new place is a sequential power of 60. Let’s look at a table to
compare our decimal system to the Babylonian’s.
Decimal (Base−10) System Sexagesimal (Base−60) System
100
1
600
1
1
1
10
10
60
60
102
100
602
3,600
103
1000
603
216,000
Etc
Etc
Etc
Etc
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218
219
220
221
222
223
224
225
226
227
228
229
230
231
232
233
234
235
236
237
238
239
240
241
As you can see, both systems start off with the ones place, but then they diverge pretty quickly
from there. The sexagesimal system place values increase much more quickly, for obvious
reasons.
We will need a standard way to represent a sexagesimal number so that it’s easy for us to see
what number is being discussed. Let’s take the following example.
Example 1
Convert 21,40,5960 to base 10.
(The subscript, once again, gives the base in use.)
Solution:
This number may look odd to you. The commas appear to be in the wrong
places. However, this is a sexagesimal number, not a decimal number, so you
can’t read it the way you are used to. This number has the following meaning.
21,40,59 =
59 in the ones place
40 in the sixties place
21 in the thirty−six hundreds place
Note that the commas serve to separate the places from each other, which is a
different role than they play in the decimal system. Because we have 60 as our
base, each place value can have as many as 59 in it before we have to carry to
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the next place up. In the sexagesimal system, no place can have more than 59
in it, just as in the decimal system no place can have more than 9 in it before
we carry up to the next place. If we want to convert this number to base−10,
we would to the following:
21, 40,59 = ( 21× 602 ) + ( 40 × 601 ) + ( 59 × 600 )
= ( 21× 602 ) + ( 40 × 60 ) + ( 59 × 1)
247
= 75600 + 2400 + 59
= 78059
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267
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270
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272
273
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276
277
Hence, 21,40,59 in base 60 is equal to 78,059 in base 10. ♦
251
Of course, 21,40,59 is a modern
252
way of representing sexagesimal
253
numbers. We’ll call it modern
21 thirty-six hundreds
40 sixties
59 ones254
sexagesimal notation for
21 times 3600
40 times 60
59 times 255
1
representing Babylonian numbers.
256
It is used for our own convenience.
The Babylonians, on the other hand, would write this number as shown here.
Note some key points: (1) The ones places starts on the right, as ours does. (2) Each place value
is separated by a space to give the reader a clue about how many symbols are in each position.
(3) The symbols generally go to the right of the
symbols. (4) There is no zero symbol. To
represent 40, four corners (each of which has a value of 10) are recorded and no ones symbols
are recorded (like the Egyptians). The four corners in the sixties represent a total of 2,400 in
base 10 because of their position…40 sixties is 2,400.
Example 2
Write 13,7,34 in Babylonian notation and determine its decimal value.
Solution:
In Babylonian notation, it looks like this:
Its decimal value is:
(13 × 602 ) + ( 7 × 60 ) + ( 34 × 1) = 46800 + 420 + 34
= 47254
♦
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Check Point A
What is the decimal value of the following Babylonian number? Write it in
modern notation. Be careful to note where the spaces are to distinguish place
values.
Solution:
See endnote to check your answer. 12
Check Point B
Write 1,13,5,11 in Babylonian notation and then find the decimal value of the
number by converting to base 10.
Solution:
See endnote to check your answer. 13
Difficulties With Babylonian Numbers
As you can imagine, these numbers can be very hard to read if you’re
not carefully trained. The spaces between positions may not be large and
may make distinguishing the positions difficult. As you can see in the
picture, the clay tables that the Babylonians used were usually not much
larger than the palm of your hand and could have a lot of writing on
them. So reading them is not easy. (This table is only about 6cm long,
or about 2.5 inches, and look at the detail on it!)
Another major difficulty with reading the numbers accurately is the fact
that the Babylonians did not generally have a symbol for zero.
Therefore, if a place has zero in it, that place is just skipped and the
reader of the tablet has to determine this fact from the context of the rest
of the tablet. For example the following number could have many
different values:
It could be 34×60 + 2×1, the most straightforward value.
It could be 34×602 + 2×1 if the 60’s place is empty.
It could be 34×602 + 2×60 if the ones place is empty.
It could be 34×603 + 2×60 if the 602 and ones places are empty.
And so on.
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This is certainly a drawback to their system, at least from our own modern viewpoint. In our
modern notation, we will go ahead and use our symbol for zero if there is something missing
from a place value. Hence, the number 18,0,32 means that there are no 60’s. In about 300 B.C.E.
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326
327
328
329
symbol as a placeholder to indicate that a position
(relatively late), they did start using the
was blank. However, as far as we know, they did not view this as a number as we do the number
zero. Hence, to write the number 18,0,32 in Babylonian notation, they would mark the following
into a clay tablet:
330
331
332
333
334
335
336
337
338
339
340
341
342
343
344
345
346
347
348
349
350
351
352
Babylonian Fractions
What about non−whole numbers? We know that in the decimal system, 39.7 means thirty−nine
whole parts plus seven tenths (fractional parts). The decimal point tells us this. The Babylonians,
however, did not have a symbol that accomplishes the same thing. This meant that the tablet
reader had to determine the value of the number from the context of the tablet. In modern
sexagesimal notation we will use the semicolon to separate whole numbers from fractions. For
example, we might see a number that looks like this in modern works that study Babylonian
mathematics.
23,16;30
This is how this number might appear on a Babylonian tablet:
There is no marker to tell you where the whole numbers end and where the fraction part starts.
To help us keep track of what is what, we’ll use the non−standard notation of drawing a dotted
line where the whole numbers and fractions are split. Thus, we will write the number above as:
353
354
355
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However, assuming we know that a group of symbols is not in a
whole−number place, how do we interpret their values? In our
decimal system, once you move to the right of the decimal point,
every place is one tenth of the previous place. In the sexagesimal
system, each place to the right of the sexagesimal point is
one−sixtieth of the previous place. Here is a table that illustrates
this.
The fractional place values in the sexagesimal system get
smaller much more quickly than they do in the decimal
system…again, for obvious reasons.
Sexagesimal
(Base−60)
1’s
1
1
=
2
10 100
1
1
=
2
60
3600
1
1
=
3
10 1000
1
1
=
3
60
216000
1
60
So now we can interpret the 30 after the dotted line. It means
that we have 30 sixtieths, which is the same as 30/60. Therefore, the decimal value of the number
is as follows:
1 

23,16;30 = ( 23 × 60 ) + (16 ×1) +  30 × 
60 

1
= 1380 + 16 +
2
= 1396.5
Example 3
Find the value of the following Babylonian number:
Solution:
In modern notation, this would be represented by 6,10;40. Its decimal value is:
1
( 6 × 60 ) + (10 ×1) +  40 ×  =
60 

2
= 360 + 10 +
3
≈ 370.666
380
381
382
383
384
385
386
Decimal
(Base−10)
1’s
1
10
Note that 40/60 has been reduced to 2/3. ♦
Example 4
Find the value of the following Babylonian number:
387
388
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Solution:
In modern notation, this number would be represented by 10;10,2. Here is its
decimal value:
1  
1 

10 + 10 ×  +  2 × 2  =
60   60 

1
1
= 10 + +
6 1800
≈ 10.167222
♦
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401
402
403
404
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407
408
409
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413
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415
416
417
Check Point C
Write the following Babylonian number in modern notation and find its
decimal value:
Solution:
See the endnote to check your answer. 14
Check Point D
Write the following Babylonian number in modern notation and find its
decimal value:
Solution:
See the endnote to check your answer. 15
MAT107 Chapter 3, Lawrence Morales, 2001; Page 13
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460
Going the Other Direction.
Now suppose we have a decimal number that we want to convert to
sexagesimal notation. You will see that the process is once again
the same that we used to convert from base ten to base twenty (or
any other base). Let’s start with a simple example and work up
from there. We will need to know what the powers of 60 are to do
this so here is a quick little table to help us.
Power of 60 Value
601
60
602
3600
603
216,000
604
12,960,000
Example 5
Convert the decimal value of 85 to modern and Babylonian sexagesimal
notation.
Solution:
Since there are more than 59 ones, we must go to the 60’s place.
There is one 60 in 85 since 85 = (60×1) + 25. We see there are 25 ones left
over.
Therefore, 85 = 1,25.
The Babylonian representation would therefore be:
Alternatively, we could use our methods from Chapter 1. If we divide 85 by
60, we get 1.416666….. So there is one in the sixties place. Subtracting that
one from the quotient gives 0.416666…. which we can then multiply by 60 to
get 25. (Review Chapter 1 for a discussion of this general method.)♦
Example 6
Convert the decimal value of 8,000 to modern sexagesimal notation.
Solution:
For this number, it’s easiest to ask, “What is the highest power of 60 that will
go into 8,000?” From the table we see that 602=3600 is the highest power of
60 that will go into 8,000. 3600 goes into 8000 a total of 2 times, with
remainder 800.
We now move to the 60’s place. 60 goes into 800 a total of 13 times with a
remainder of 20.
Since 20 is less than 59 we have 20 ones.
Therefore, 8,000 = 2,13,20.
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475
476
477
478
479
480
481
482
483
484
485
486
487
Of course, we can check this simply by starting at 2,13,20 and converting
back to decimal to make sure that we get 8,000. (You should do so now on
your own.)
Once again, we could use our alternative method from Chapter 1, in which
case we would dived 8000 by 3600 = 602. Continuing the process will
produce the same result as the one obtained above.♦
Check Point E
Convert the decimal value of 300,000 to modern sexagesimal notation.
Solution:
See the endnotes to check your answer.16
We have one more conversion issue to deal with, and that is decimals. How do we convert the
decimal value of 1/3 into Babylonian notation? How about 0.24? Well, the value 1/3 is not too
hard. We can see that 1/3 is the same as 20/60, so in modern sexagesimal notation, 1/3 = 0;20.
But what about 0.24? In order to see how to convert this, let’s convert 1/3 in a slightly different
way:
1
1 3 60 20
= × =
= 0; 20
3 1 60 60
With this method, we see that multiplying 1/3 by 60 gives us the desired result. So this is the
technique we’ll use, only we’ll streamline it a bit.
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Example 7
Convert 0.24 to modern sexagesimal notation.
Solution:
We’ll display our steps in a table with comments to help explain the process:
60×0.24=14.4
Therefore 0.24 =
14
?
+ 2
60 60
0.4×60=24
24
602
Therefore we have
Finally, 0.24 =
14 24
+
= 0;14; 24
60 602
Check:
14 24
+
60 602
6
=
25
= 0.24
0;14, 24 =
494
495
496
497
498
499
500
501
502
503
504
505
506
507
Comments
Since we get 14.4, this means we
14.4
. But we only want whole values
have
60
14
in the sixtieth’s place. So we take
and
60
then have some undetermined number left
1
over in the 2 place.
60
Since we have 0.4 left over (after
removing the 14), we multiply that by 60.
This tells us how many we have in the
1
place.
602
We combine our results to get the final
answer.
By carefully adding fractions we can get
to 6/25, which when converted to a
decimal is 0.24, confirming our work.
♦
As a result of this example, we now have a method for converting a fraction decimal to
sexagesimal notation.
(1) Multiply the decimal by 60. The whole−number part of the product is the number of 1/60’s
in the number.
(2) Multiply the left over, fractional part of the product in (1) to get the number of 1/602’s.
(3) Continue this process until multiplying by 60 gives you a whole number.
MAT107 Chapter 3, Lawrence Morales, 2001; Page 16
507
508
509
510
511
512
513
Example 8
Convert 5.266 to modern sexagesimal notation.
Solution:
We can ignore the 5 since it is the whole number part. We only want to start
multiplying 60 by the fractional part of this number.
Ones
1
60
1
602
1
603
5+
0.266×60 = 15.96
0.96×60 = 57.6
0.6×60 = 36
514
515
516
517
518
519
520
521
522
5
57
36
We can now assemble all these pieces to see that 5.266 = 5;15,57,36. If you
don’t believe it, here’s a check:
15 57 36 2633
5+ + 2 + 3 =
= 5.266 ♦
60 60 60
500
523
524
525
526
527
528
529
530
531
532
533
534
535
536
537
538
539
540
15
Check Point F
Convert 75.102 to modern sexagesimal notation.
Solution:
See the endnotes to check your answer.17
Check Point G
Convert 13.525 to modern sexagesimal notation.
Solution:
See the endnotes to check your answer.18
MAT107 Chapter 3, Lawrence Morales, 2001; Page 17
540
541
542
543
544
545
546
547
548
549
550
551
552
553
554
555
556
557
558
559
560
Why Base−60?
It is uncertain why the Babylonians chose 60 as their base. Theon of
Think About It
Alexandria (600 C.E.) made a comment that 60 was one of the smaller
numbers that has a large number of divisors and that 60 was chosen for
How many divisors
this reason. Others who have studied the Babylonians think there was a
does 60 have and what
more “natural” origin to the system. For example, the Babylonian took the
are they? Why is the
year to have 360 days. Since 360 is a multiple of 60, they chose this as
concept of having
their base. (But why not some other multiple such as 20?) Dr. Robson has
more divisors helpful?
suggested that it was invented to “shortcut the complicated procedures
needed for multiplication and division in particular.” 19Others have
suggested that the base evolved from two or more groups. Perhaps one set of people had base 10
and another had base 6 and they merged the two. None of these are definitive and are all
theoretical. It has yet to be determined for certain why 60 was chosen as their base.
We saw earlier that their writing system had some disadvantages. The base system also has some
disadvantages. For example, even certain small numbers can require numerous marks in the clay.
For example, 999 only requires three symbols in our decimal system. How many does it require
in the sexagesimal system?20
MAT107 Chapter 3, Lawrence Morales, 2001; Page 18
560
561
562
563
564
565
566
567
568
569
570
571
572
573
574
575
576
577
Part 3: Babylonian Arithmetic
Addition
We now return to the questions of Babylonian mathematics and examine their basic arithmetic
techniques. We will want to look at addition, multiplication, division, and computing the values
of square roots.
Recall from our studies of the Egyptians that addition was as straightforward as gathering like
symbols and then carrying when we had more than 10 in a certain place value. In Babylonian
addition, we will use the same idea…we combine like symbols and when we have more than 59
in a place, we carry a ones symbol up to the next place. One example will hopefully be enough to
illustrate this. Just keep in mind that the idea is exactly the same as what we did with Egyptian
math.
Example 9
Add the following two Babylonian numbers:
578
579
580
581
582
583
584
585
Solution:
We’ll assume that the right columns are the ones. When we gather up all the
symbols in each place (like we did with the Mayans), we get the following.
586
587
588
We need to keep in mind that 10 ’s make one
.
589
590
Also, six ‘s make 60 since each one of them represents 10. When we look at
the ones place, we see that we have a total of 73. So we take 60 of them and
591
592
593
carry to the 60’s place. Note that six ’s make one
(Why?)
MAT107 Chapter 3, Lawrence Morales, 2001; Page 19
in the 60’s place.
593
594
595
596
597
That leaves one
and three ’s in the ones place (for a total of 13). Now we
598
look at the 60’s place in the middle. There are more than 10 ’s here so we
599
need to convert ten of them to a
600
601
602
it get converted to a
to the next place?)
since every 10 ’s makes a
. (Why does
which stays in its current place rather than moving up
603
604
605
606
This leaves 2 ’s and 6 ’s in the 60’s place, for total of 26 60’s. There is
nothing to carry from the 60’s to the 602’s since 26 is less than 60. The 602’s
607
608
place has five ’s in it for a total of 50, so it’s okay as well. So we have our
final result.
609
610
611
612
613
614
615
616
617
Thus, 30,16,40 + 20,9,23 = 50,26,1360
You can check this by converting the original numbers into base 10 and then
adding and comparing to the answer above. ♦
MAT107 Chapter 3, Lawrence Morales, 2001; Page 20
617
618
619
Check Point H
Add the following Babylonian numbers.
620
621
622
623
624
625
626
627
628
629
630
631
632
633
634
635
636
637
Solution:
See the endnotes to check your answer.21
A Primer on Multiplication
When we get to this topic, the method of the Babylonians changes from the method of the
Egyptians. With the Egyptians, we saw that they used the method of doubling to do their
multiplication and that this was a great illustration of multiplication as repeated addition. We do
not know any specific multiplication process that the Babylonians used. Instead, they appeared to
use multiplication tables instead. These tables had the values for numerous products that students
and scribes could look up if they had not already memorized them. Here is a picture of a typical
multiplication table.22 While it is hard to read, it gives you a sense of what they looked like.
638
639
640
MAT107 Chapter 3, Lawrence Morales, 2001; Page 21
640
641
642
643
644
645
646
647
648
649
650
651
652
653
654
655
656
657
658
659
It would be very difficult to try to reproduce these tables and then actually use them, but there is
still some value in trying to learn how to multiply in a base other than 10, a task that we have yet
to undertake. So let’s do so now.
We start by examining our own (U.S.) multiplication algorithm. (If you did not learn arithmetic
in the U.S., you may have a different method of multiplication.) Let’s start with a simple
example: 56×38. To do this problem, we would proceed as follows
1
2
×
4
6
1
5
3
4
8
2
6
8
8
0
8
← This line represents 8×56
← This line represents 30×56
← This line is the sum of the products
This process is a very clever one indeed, when you examine it in more detail. (I’ll bet very few
of you have ever thought about how or why this method works. “It just does.”) We’ll do that on
the next page, where we can see all the steps and reasoning together.
MAT107 Chapter 3, Lawrence Morales, 2001; Page 22
659
660
Multiplying the 6 and 8 is the
equivalent of multiplying 6 ones
by 8 ones, giving 48.
56×38 = (50 + 6)×(30 + 8)
5 6
× 3 8
4 8
Multiplying the 5 and 8 is the
same as multiplying 5 tens by 8
ones, which is 400. Note that
when you up the first two lines,
48+400=448, which is what you
get if you carry the 4 in the first
step and then add it to the 40
you get from 8 times 5.
56×38 = (50 + 6)×(30 + 8)
6
8
8
0
5
× 3
4
4 0
5 6
× 3 8
4 4 8
56×38 = (50 + 6)×(30 + 8)
5
3
4
8
0
6
8
8
0
0
5
× 3
4 4
6 8
6
8
8
0
5
3
4
8
2
6
8
8
0
8
×
4
1
1 5
56×38 = (50 + 6)×(30 + 8)
1
56×38 = (50 + 6)×(30 + 8)
=2128
×
4
1 6
2 1
Multiplying the 3 and 6 is the
same as multiplying 3 tens and
6 ones, giving 180. Multiplying
the 3 and 5 is the same as
multiplying 3 tens and 5 tens,
giving 50×30=1500.
Adding the last two
computations gives 1680.
Add everything up and get
2,128.
661
662
663
MAT107 Chapter 3, Lawrence Morales, 2001; Page 23
663
664
665
666
667
668
What is important to note in the previous page is that the algorithm that we use is essentially
based on multiplying each digit in the first number by every digit in the second number AND
making sure that when doing so, the values of the digits based on their places are properly
accounted for. In our modern algorithm, this is accomplished by placing zeros in the appropriate
places to shift numbers to the left.
5
3
4
8
2
×
4
1 6
2 1
669
670
671
672
673
674
675
676
677
678
679
680
681
682
683
684
685
686
687
688
689
690
691
692
693
694
695
696
697
698
699
700
701
702
6
8
8
0
8
This zero serves to make
sure that we take into
account that the 5 and 3
are in the tens place
Trying to use this same algorithm in base 60 would be possible, but because the base system is
different, it becomes rather confusing. Instead, we’ll use a different method that is equivalent
but will help us to keep track of the place values. It’s a method that actually comes from the
Middle Ages, but we’ll adapt it here for our own purposes. It’s called the gelosia method. (In
Chapter 6, Logs and Cubic, we’ll give more background information on this method.) Before we
start multiplying in base 60 with this new method, we’ll first show how it works in base 10.
Let’s go back to the problem of multiplying 56×38.
In the gelosia method, we begin by placing the numbers to be multiplied on the outer edges of a
grid, as shown below:
5
6
3
8
Note that there are essentially four blocks, each of which has a diagonal line running through it.
The four blocks correspond to the four different multiplications that need to be done: 5×3, 6×3,
5×8, and 6×8. Where two numbers intersect, we place their product, using the diagonal line to
separate the ones place from the tens place. Filling in this grid gives the following:
5
1
6
1
5
4
8
3
8
8
4
0
MAT107 Chapter 3, Lawrence Morales, 2001; Page 24
703
704
705
706
707
708
709
710
711
712
713
714
715
716
717
718
719
720
721
722
723
724
725
726
727
728
729
730
731
732
733
734
735
736
737
738
739
740
741
742
743
744
745
746
747
748
Notice that there is no need to carry anything, yet. We are simply doing all the individual
multiplications and recording the results. Now the diagonal lines play their most important role.
We are going to use them as addition guides. We start at the lower right hand corner of the grid.
All digits that lie within the same diagonal “trough” are added together. If the result is ten or
more, we carry up to the next diagonal.
5
1
6
1
5
4
8
3
8
8
4
0
Carried up
from the
previous
diagonal
For this grid, we would get the following:
5
1
2
1
6
1
5
4
1
8
3
8
8
4
1
0
2
8
In the first diagonal we have an 8, so it gets copied down. The first diagonal represents the ones
place.
In the second diagonal we have 8 + 4 + 0 = 12, so we carry 1 to the next diagonal up and write a
2 in the diagonal’s total box. (Note the one has been carried up and is written in a smaller font
size above for distinction.) The second diagonal represents the tens place.
In the third diagonal we have 4 + 5 + 1 + 1 = 11, so we carry 1 to the next diagonal up and write
a 1 in the diagonal’s total box. The third diagonal represents the hundreds place.
In the fourth diagonal we have 1 + 1 = 2, which gives us a 2, which we write in the diagonal’s
total box. The fourth diagonal represents the thousands
Thousands
Hundreds
place.
Tens
5
1
Now, the answer is easily read from the upper left corner
around the edge to the lower right corner as 2128.
2
Why does this work? Well, note that when we multiply
6×8 and get 48 (lower right corner), this represents 4 tens
and 8 ones. The 8 goes in the first diagonal and the 4
goes up one diagonal into the tens diagonal. When we do
6
1
5
4
Ones
8
3
8
8
4
1
0
2
MAT107 Chapter 3, Lawrence Morales, 2001; Page 25
8
749
750
751
752
753
754
755
756
757
758
759
760
761
762
763
764
765
766
767
768
769
770
771
772
773
774
775
776
777
778
779
780
781
782
783
784
785
786
787
788
789
790
791
792
793
6×3 (upper right corner), we’re really doing 6×30 = 180, which is 1 hundreds and 8 tens. So the 8
goes in the tens diagonal (with the 8 from the previous step), and the 1 goes up into the hundreds
diagonal. Note that the diagonal lines are doing the same thing that we do when we add that extra
zero in our algorithm to make sure that all the places line up.
Let’s look one more example in base 10 before we move to base 60.
Example 10
Use the gelosia method to multiply 392×52.
Solution:
We begin by building a grid of the correct size
3
9
2
5
2
We now fill in the six multiplication blocks. Note that we can place a zero in
the upper part of a block if the product of two digits is less than ten:
3
1
9
4
5
2
1
5
0
1
6
0
5
4
2
0
8
Finally, we add up the diagonal place values, carrying where appropriate:
3
1
2
9
4
5
0
2
1
5
1
0
5
4
2
0
6
3
0
8
8
4
MAT107 Chapter 3, Lawrence Morales, 2001; Page 26
794
795
796
797
798
799
800
801
802
803
804
805
806
807
808
809
810
811
812
813
814
815
816
817
818
819
820
821
822
823
824
825
826
827
828
829
830
831
832
833
834
835
836
837
838
839
We have an answer of 20,384, reading around the edge of the grid. It is not
necessary to draw three separate grids to do this problem. One will suffice.
But three are given here to show you the progression of steps. ♦
There are various gelosia grids in the Appendix at the end of the chapter that you can photocopy
or trace so that you don’t have to draw them by hand.
Check Point I
Multiply 475×879 using the gelosia grid. Work in base 10.
Solution:
See the endnotes to check your answer.23
Think About It
Suppose you had a child who had
already learned their multiplication
tables and you had to teach them
either the modern multiplication
algorithm or the gelosia grid system
for multiplication. Which one do
you think would be more effective?
Why?
Example 11
Use the gelosia grid method to multiply 78.5×3.6
Solution:
What do we do about decimals? It’s simple. Simply do the calculations as
before and then place the decimal point at the appropriate place in the final
answer.
7
2
2
8
8.
2
1
5
1
5 3
4
4
3
.
2
8
0 6
2
4
6
0
MAT107 Chapter 3, Lawrence Morales, 2001; Page 27
Think About It
The first diagonal (at
the lower right corner)
in the grid above
corresponds to the
hundredths place.
Why?
840
841
842
843
844
845
846
847
848
849
850
851
852
853
854
855
856
857
858
859
860
861
862
863
864
865
866
867
868
869
870
871
872
873
874
875
876
877
878
879
880
881
882
883
884
885
We get a result of 28260, but we still need to place our decimal point. Since
there were a total of two digits to the right of the decimals (the 5 in 78.5 and
the 6 in 3.6), we move the decimal place in a total of two places from the
right. So our result is 282.60, which is, of course, correct. ♦
Check Point J
Use the table method to multiply 615×48.2
Solution:
Check the endnotes to check if your answer is correct. You can use the
following grid to do your work.24
The Babylonian Multiplication Table
We are going to extend the table method of multiplication to the sexagesimal system shortly. But
we have one more thing to take care of first. We said earlier that the Babylonians used
multiplication tables to do complicated multiplication problems. So I have created a modern
version of a sexagesimal multiplication table. It is titled “Babylonian Multiplication Table in
Base 60.” You should have it in front of you for the following explanations and examples.
As you look at the table, you will notice that there is a series of rows and columns labeled on the
top and side of the table. To multiply two sexagesimal numbers together we move along the row
containing the first number until intersects the column with the second number. For example, if
we want to multiply the sexagesimal numbers 13×30, we do the following:
1.
2.
3.
On the left side of the column, find the row labeled 13. (It has a 0 26) to its right.
Move along this row until you are in the column that is labeled 30 at the very top. (It will
probably help for you to take a piece of paper and place in under row 13 to help you follow
where you’re at…the table is also alternately shaded to help in this process.)
Where row 13 and column 30 intersect is the “cell” that gives the result. You should see a 6
30 in that cell.
MAT107 Chapter 3, Lawrence Morales, 2001; Page 28
886
887
888
889
890
891
892
893
894
895
896
897
898
899
900
901
902
903
904
905
906
907
908
909
910
911
912
913
914
915
Thus, in the sexagesimal system, 13×30 = 6,30
Let’s check this by converting the result to base 10.
6,30 = 6×60+30×1 = 360+30 = 390
It’s not hard to check that 13×30, in base 10, is 390. Hence, all is well.
Example 12
Multiply 24×13 in sexagesimal.
Solution:
Row 24 intersects with column 13 at the cell 5,12. So 24×13 = 5,12
(Check: 24×13=312. But 5,12=5×60+12=300+12=312) ♦
Check Point K
Multiply 20×40 in sexagesimal.
Solution:
See the endnotes to check your answer.25
Example 13
Multiply 8×23 in sexagesimal.
Solution:
There is a row 8, but there is no column 23. So we think of the problem
differently.
8×23 = 8×20 + 8×3
There is both a column 20 and a column 3, so we do it in two steps:
8×20= 2,40
8×3= 0,24
Total 2,64 = 3,4
916
917
918
919
920
921
922
923
924
Why did we convert 2,64 to 3,4? This is because 2,64 is not a valid base−60
number.
2,64 = 2,(60+4) = (2+1),4 = 3,4
We carry 60 ones up to the next place and get 3,4. ♦
MAT107 Chapter 3, Lawrence Morales, 2001; Page 29
925
926
927
928
929
930
931
932
933
934
935
936
937
938
939
940
941
942
943
944
945
946
947
948
949
950
951
952
953
954
955
956
957
958
959
960
961
962
963
964
965
966
967
968
969
970
Check Point L
Multiply 20×42 in sexagesimal.
Solution:
See the endnotes to check your answer.26
Babylonian Multiplication with Tables
Let’s now look at more complicated multiplication problems in the sexagesimal system.
Example 14
Multiply (5,15) × (3,25) in sexagesimal.
Solution:
This problem is slightly different than those we just finished. The major
difference is that the table we have only allows us to multiply single pairs of
numbers at a time. So, we resort to the gelosia grid method that we practiced
earlier.
We must notice, however, that we are in a base 60 system, so instead of
carrying when we have 10, we will carry when we have 60!
Also, the entries that we put into our gelosia grids will come from the
Base−60 multiplication table. We start by creating the appropriate gelosia
grid:
5
15
3
25
The entries into our multiplication blocks come from the following four
products:
5×3 = 0,15
15×3 = 0,45
5×25 = 2,5
15×25 = 6,15
We can now put these into our grid, being certain to place numbers in the
correct places.
MAT107 Chapter 3, Lawrence Morales, 2001; Page 30
971
972
973
974
975
976
977
978
979
980
981
982
983
984
985
986
987
988
989
990
991
992
993
994
995
996
997
998
999
1000
1001
1002
1003
1004
1005
1006
1007
1008
1009
1010
1011
1012
1013
1014
1015
1016
5
0
15
0
15
45
2
3
6
5
15 25
The first diagonal is the ones, the second is 60’s, the third is 602’s, and the last
is 603’s. By adding (and carrying only if we have groups of 60), we get the
following:
5
0
0
15
0
15
2
17
45
3
6
5
56
15 25
15
From what our table tells us, (5,15) × (3,25) = 0,17,56,15. We should
definitely check this by converting to base 10.
5,15 = 5×60+15 = 315
3,25 = 3×60+25 = 205
17,56,15 = 17×602 + 56×60 + 15 = 64,575
To finish the check we simply check that 315×205=64,575 in base 10, which
it is. ♦
Example 15
Multiply (2,40,50) × (35,18)
Solution:
2
40
50
35
18
MAT107 Chapter 3, Lawrence Morales, 2001; Page 31
1017
1018
1019
1020
1021
1022
1023
1024
1025
1026
1027
1028
1029
1030
1031
1032
1033
1034
1035
1036
1037
1038
1039
1040
1041
1042
1043
1044
1045
1046
1047
1048
1049
1050
1051
1052
1053
1054
1055
1056
1057
1058
1059
1060
1061
1062
Entries into the blocks come from the following products in the Multiplication
Table:
2×35 = 1,10 and 2×18 = 0,36
40×35 = 23,20 and 40×18 = 12,0
50×35 = 29,10 and 50×18 = 15,0
We therefore get the following. Keep in mind that we are carrying ONE up to
the next diagonal when we have 60 in a diagonal:
2
1
1
40
23
10
0
34
50
29
20
12
36
37
10 35
15
0
25
0 18
0
In the third diagonal (602’s), we have 36+12+20+29 = 97 = 60+37 =1,37.
That’s why you see the 37 in the third diagonal’s total box. In the next
diagonal up, we carry one so that we have 0+10+23+1=34.
We can check this pretty easily, again by converting to base 10.
35,18 = (35×601) + (18×600) = 2,118
2,40,50 = (2×602) + (40×601) + (50×600) = 9650
2118×9650 = 20,438,700
Finally we check our total:
1,34,37,25,0 = (1×604)+ (34×60)+ (39×602) + (25×601)+(0×600) = 20,438,700
♦
Check Point M
Multiply (25,50) × (10,15) in sexagesimal.
Solution:
See the endnotes to check your answer.27 Use the grid below to do your work.
MAT107 Chapter 3, Lawrence Morales, 2001; Page 32
1063
1064
1065
1066
1067
1068
1069
The Babylonians also had numbers with fractional parts. When multiplying these numbers
together, the process will be the same.
Example 16
Solution:
16
25
while 40;25 means 40 +
60
60
The method we use is essentially the same as what we did when we multiplied
numbers in base 10 that had fractional parts.
We recall that 24;16 means 24 +
1070
1071
1072
1073
1074
1075
1076
1077
1078
1079
1080
1081
1082
1083
1084
1085
1086
1087
1088
1089
1090
1091
1092
1093
1094
1095
1096
1097
1098
1099
1100
1101
1102
1103
1104
Multiply (24;16) × (40;25)
24 ; 16
1
0
12
48 3
6
6
;
0
40 25
1
18
54
40
Since we need to move two sexagesimal places, we get 1,18;54,40. You
should check this in base 10 to make sure it works. ♦
Example 17
Multiply (30,10;50) × (3,20;30) in sexagesimal.
Solution:
Here is the appropriate table.
30
1
10 ; 50
0
2
1
30
30
30 3
10
3
16
40
0
20
40 20
15
5
25
;
51
0
0
0 30
12
5
0
The result here is 1,40,51,12;5,0 ♦
MAT107 Chapter 3, Lawrence Morales, 2001; Page 33
1105
1106
1107
1108
1109
1110
1111
1112
1113
1114
1115
1116
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1123
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1125
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1128
1129
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1135
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1140
1141
Check Point N
Multiply (18,40;15) × (30,10;35) in sexagesimal.
Solution:
See the endnotes to check your answer.28
Babylonian Multiplication on Tablets
35
The Babylonian tablets that we know about usually
show multiplication by placing the numbers in
particular places relative to each other on the tablet.
For example, let’s look at the (fake) tablet shown
here. (This might be a round “bun” tablet.)
The two numbers to the left of the vertical line are
35 and 13. These are the two numbers being
multiplied. The scribe or student would use the
vertical line (usually) to separate these from the
13
answer, which is on the right of the vertical line.
Here, we see that 35×13=7,35, which you can
easily confirm by looking up 35×13 on the base 60 multiplication table.
The tables we’ve been constructing have been primarily for our own use and do not necessarily
reflect the way the Babylonians would have done multiplication. Remember, our goal was to
learn how to do multiplication in another base.
Babylonian Division
The Babylonians used the fact that division is the opposite of multiplication. That is, they
A
1
recognized that = A × . That is, dividing A by B is the same as multiplying A by the
B
B
reciprocal of B. To suit their division needs, the Babylonian had tables of reciprocals so that they
could do these calculations quickly and efficiently (even though they probably memorized most
of the table entries after a while.) Below you can see a picture of a Babylonian reciprocal
table:29
MAT107 Chapter 3, Lawrence Morales, 2001; Page 34
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1160
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1167
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1169
1170
This is probably going to be next to impossible for you to read, so I’ve placed some of the more
common reciprocals at the bottom of the Base 60 Multiplication Table. Here’s an example: If
you want to know the reciprocal of 32 (which is 1/32) in sexagesimal notation, you simply go to
the column labeled 32 and read the sexagesimal number that is underneath it. From this table you
will see that 1/32 = 0;1,52,30. (This particular reciprocal is highlighted in the boxes on the table
above. Can you read them off the picture? If not, it’s okay…that’s why I put them on the
multiplication table.)
Example 18
Divide (2,40) ÷8 in sexagesimal.
Solution:
We only need to convert this to a multiplication to proceed as before. Since
the reciprocal of 8 is 0;7,30 (from the table) we can rewrite this as
2,40×0;7,30. Here’s the appropriate table of computations.
2
0
0
40
4
14
1
40 ;7
20
20
0
0
0 30
0
Our final answer, after adjusting two sexagesimal places to the left, is 20;0
MAT107 Chapter 3, Lawrence Morales, 2001; Page 35
1171
1172
1173
1174
1175
1176
1177
1178
1179
1180
1181
1182
1183
1184
1185
1186
1187
1188
1189
1190
1191
We can check this by noting that 2,40 = 2×60 + 40 = 160. So, in base 10 we
are just dividing 160 by 8, which is 20, of course. ♦
Example 19
Divide (13,40;30) ÷ 50.
Solution:
The reciprocal of 50 is 0;1,12, so we want (13,40;30) × (0;1,12)
13
0
0
16
40 ; 30
0
0
13
40
30 0;1
2
8
6
36
0
0 12
24
36
0
Let’s check to see if this works:
30
= 820.5.
60
So we are computing 820.5÷50 = 16.41
1192
13,40;30 = (14×601) + (40×600) +
1193
1194
1195
1196
Now checking our Total: 16;24,36 = 16 + 24/60 + 36/602 = 16.41, a match. ♦
MAT107 Chapter 3, Lawrence Morales, 2001; Page 36
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Check Point O
Divide (36,12;17) ÷ 27 in sexagesimal.
Solution:
See the endnotes to check your answer.30
Division on Babylonian Tablets
To show division on clay tablet, the Babylonians would
have done something like what is shown for (1,25) ÷
15.
You can probably see the 1,25 on the top to the left of
the vertical line. The 15 (divisor) is on the left side of
the second row…it’s reciprocal 0;4 (from reciprocal
table) is immediately next to it. Hence, we have to
know (from the context) that these are actually two
different numbers. The answer to the right of the
vertical line is obtained by multiplying 1,25×0;4 to get
5;40. (Check that this is correct.). Note that there is not
a sexagesimal place on the tablet to tell us that 40 is a fractional part of the number. We would
have to be able to discern that from the rest of what’s on the table.
MAT107 Chapter 3, Lawrence Morales, 2001; Page 37
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Part 4: Babylonian Root Approximations
Babylonians and Square Roots
The Babylonians were able to estimate the value of square roots to a reasonable degree of
accuracy with an “iterative” process using basic arithmetic. The best way to see how they did
this is with an example where we try to explain each of the steps carefully.
Example 20
Let’s find an estimate for 11 as the Babylonians might have.
Solution:
The first thing to notice is that 11 is between 3 and 4 (since 32 = 9 and 42 =
16). Since 11 is closer to 32 than it is to 42, this is the Babylonian’s first
estimate for 11 . Note: the first estimate will always come from determining
between what two whole numbers the desired square root lies and then
choosing the whole number that’s closest to the radicand (number inside the
sign) as the first estimate.
We need to keep in mind that the square root of N is found by identifying
some number a such that a×a = N. In other words, two numbers multiplied
together need to give N. In this example, we’re looking for two numbers
(presumably the same) that multiply to give 11. Since our first guess is 3, we
need to find a number x such that 3 × x = 11 . Solving this equation for a gives
11
us x = .
3
2
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1247
1248
1249
1250
1251
1252
 11  121
When we square x we see that   =
≈ 13.444 . This is too big (since
9
3
we want 11 when we square). But 32 = 9 is too small. Hence, our actual
11
square root is somewhere between 3 and . Since we don’t know exactly
3
where in between these two numbers it lies, we will assume that it is half way
between them. To find the point that is half way between two numbers, we
only need to take the average of the two numbers. In this case, we get:
11 9 11
+
3 = 3 3
2
2
 20 
 
3
= 
2
20 10
=
=
6
3
3+
1253
MAT107 Chapter 3, Lawrence Morales, 2001; Page 38
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10
10
is our second estimate for 11 .
≈ 3.333 which, when
3
3
squared, gives 11.11…, which is okay if you don’t mind that much error.
Note: Every estimate after the first is always found by computing the average
of the previous estimate and the number that multiplies by the previous
estimate to give the number under the radical sign.
Therefore,
To compute the third estimate, we need to find the number that, when
10
, gives 11. Thus we need to find x such that:
multiplied by
3
1263
10
x = 11
3
3 33
x = 11 × =
10 10
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1265
2
1266
1267
1268
1269
1270
1271
33
10
 33 
Note that   ≈ 10.89 , so our actual answer has to between
and
10
3
 10 
We now average this with our previous estimate:
10 33 100 99
+
+
3 10 = 30 30
2
2
 199 


30 
=
2
199
=
≈ 3.3166666
60
This is third estimate for 11 . As stated before, it is the average of two
2
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1274
1275
1276
 199 
numbers. When we square this, we find that 
 ≈ 11.0002777 . This is
 60 
accuracy to three decimal places. Not bad. The graphic below attempts to map
out the process for you visually. The first, second and third estimates are in
dotted circles.
MAT107 Chapter 3, Lawrence Morales, 2001; Page 39
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1301
1302
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3 < 11 < 4
3×
11
= 11
3
11
3 = 10
2
3
3+
10 33
× = 11
3 10
10 33
+
3 10 = 199
2
60
♦
Example 21
Estimate
34
Solution:
First estimate: 34 is between 5 and 6, but is closer to 6 (since 34 is closer
to 36 that it is to 25). So our first estimate is 6. Note that this time, the first
guess is closer to the higher of the two integers between which 34 lies.
Second estimate: First find what number times 6 gives 34.
34 17
6 x = 34 ⇒ x =
= . (Always reduce fractions when possible.) Now take
6
3
the average of this with the first estimate.
17 18 17
+
3 = 6 3
2
2
 35 
 
 3
=
2
35
=
≈ 5.8333...
6
6+
1305
2
1306
1307
1308
35
 35 
Thus
is our second estimate. Check:   ≈ 34.02777... , only accurate to
6
 6 
one decimal place.
MAT107 Chapter 3, Lawrence Morales, 2001; Page 40
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Third estimate: First find what number times
35
gives 34.
6
35
204
x = 34 ⇒ x =
. Now we take the average of this with the previous
6
35
estimate.
1310
1311
1312
35 204 1225 1224
+
+
6
35 = 210 210
2
2
 2449 


 210 
=
2
2449
=
≈ 5.830952381
420
1313
1314
2449
. When we square this we see
420
1315
Thus our third estimate is
1316
 2449 

 ≈ 34.00000567 . This is accurate to about five decimal places. ♦
 420 
2
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Check Point P
Find the first three estimates for
29 .
Solution:
See the endnotes to check your answer.31
Of course, we have only described their method here and we’ve used modern notation to do the
work. You can imagine how hard it would be to try to all of this in base 60. We’ll have to leave
that for a more advanced class.
An Alternate Method of Estimating Roots
Other mathematics historians believe that the Babylonians used yet another method to obtain an
estimate for the square root of a given number:
h
,
2a
1334
a2 + h ≈ a +
1335
1336
1337
where a 2 is a square number that is close to (but not bigger than) the number whose square root
is being estimated. Essentially, this method requires that you rewrite the number you want to
1338
1339
take the square root of in the form a 2 + h , where a and h are integers. If you can do that, then
you can use the formula to obtain an estimate.
MAT107 Chapter 3, Lawrence Morales, 2001; Page 41
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Example 22
Estimate 13 .
Solution:
Note that 13 is between the two perfect squares of 9 and 16. With that in mind
we pick a to be 3 since 32 is close to 13 but not bigger than 13. (We would not
want to pick a to be 4 since that would make a2 bigger than 13.) With the
choice of a = 3, then h would be 4 since a 2 + h = 3 2 + 4 = 13 . With the
choices of a and h made, we can use the formula to estimate 13 :
13 = 3 2 + 4
4
≈ 3+
2(3)
2
=3
3
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1358
1359
1360
1361
1362
This gives a rough approximation for the square root of 13. ♦
Example 23
Estimate
Solution:
Note that 47 is between the perfect squares 36 and 49, so we choose a to be 6,
since 62 = 36 is close to 47 but not greater than 47. With a = 6, then h has to
be equal to 11. So we get:
47 = 6 2 + 11
11
≈ 6+
2(6)
11
=6
12
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1367
1368
1369
1370
1371
47
This completes the estimate.♦
Check Point Q
Use the formula to estimate
76 .
Solution:
See endnotes for the answer. 32
MAT107 Chapter 3, Lawrence Morales, 2001; Page 42
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Plimpton 322
One on the most famous Babylonian tablets in existence
is called Plimpton 322 (circa 1700 B.C.E.). Among the
square root problems that the Babylonians undertook was
the question of the relation between the side of a square
and its diagonal. This is just a special case of the
Pythagorean theorem, which gives the relation between
the lengths of the legs of a right triangle and its
hypotenuse.
a 2 + b2 = c2
This is a result that we’ve all used numerous times.
Even though this theorem is named after Pythagoras, who lived in the sixthcentury B.C.E. (well after the Babylonians), the result of the Pythagorean
theorem was known by civilizations well before Pythagoras arrived on the
scene, including the Babylonians and Egyptians.
c
The tablet is made up of four columns of numbers. Upon close
examination, it appears that this table is actually a list of what are called
“Pythagorean Triples.” These are sets of three numbers that satisfy the
Pythagorean Theorem. For example (3,4,5) is a Pythagorean triple because
32 + 42 = 52 . However, the triples it lists are very obscure. For example,
a
one triple it lists is (119,120,169). Others are even more awkward. The
33
why how’s and why’s of this table are hard to unearth. Katz provides an
interesting explanation of how the columns might have been generated, but the details are
probably more than we want to venture into here.
MAT107 Chapter 3, Lawrence Morales, 2001; Page 43
b
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Part 5: Babylonian Algebra
The Babylonians were able to solve systems of equations and this, in turn, allowed them to solve
a variety of quadratic equations. They did not have the quadratic formula, as we do, and they did
not have variables, as we do. Instead, it appears that they had proscribed steps that they would
follow in order to solve these types of problems. We will explore how they solved systems of
equations first and then look at how that allowed them to solve quadratic equations.
Systems of Equations
We start with a simple example34 of how the Babylonians would solve a simple system of linear
equations.
Example 24
 1
l + w = 7
Solve the system  4
l + w = 10
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Solution:
The first thing to note is that the Babylonians did not have variables or
equations like those shown above. Instead, their problems were given and
solved in words. For example, the problem above would most likely be given
in terms of a rectangle. The first equation might have come from a statement
such as “The length plus one fourth of the width is seven.” The second
equation might have come from “Length plus width is ten.” The scribe would
then turn around and ask that the dimensions of the length and width be found.
To do this, the scribe might have written something like what you see in the
left column below. The center column is a more modern algebraic
representation of what is happening. Finally, the right column is one that gives
commentary on what is happening.
Babylonian
7 × 4 = 28
Modern
4l + w = 28

l + w = 10
28 − 10 = 18
3l = 18
1
18 × = 6 (the
3
length)
10 − 6 = 4 (the width)
l =6
w = 10 − 6 = 4
Commentary
Multiply the first equation by 4 to clear
the fractions. The 7 × 4 is computed to
get the new right side of the equation.
Subtract the two equations from each
other and what is left is 3l = 18 .
Multiply both sides of the resulting
equation by 1/3.
Use the value of l to get w
If you were to just look at the left column, it might be difficult to see what they are doing. But
when you place that information alongside a modern version of the problem, you see that they
are essentially doing exactly what we do without the use of variables. This is typical of
MAT107 Chapter 3, Lawrence Morales, 2001; Page 44
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Babylonian “algebra” problems. We call them algebra problems because we can take what they
do and represent their steps in familiar algebraic notation to see that they are doing more than
simple arithmetic. Indeed, they are employing algebraic reasoning to solve simple geometric
problems.
One common type of algebra problem that appears on tablets is the following: Find two numbers
if their sum and product are given. Many of these problems were given in the context of the
dimensions and measurements of a rectangle.
Example 25
Length plus width is 14. Length times width is 45. What is the length and
what is the width?
Solution:
To solve this, we will show the calculations that may have appeared on a
tablet. The Babylonians did not provide a formula or explicit process. They
often just wrote down their calculations, leaving us to figure out what they
were doing, and why they were doing it.
Steps
Step 1
Tablet
Computations
14 ÷ 2 = 7
Step 2
7 × 7 = 49
Square 7 to get 49.
Step 3
49 − 45 = 4
Subtract the area to get 4.
Step 4
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4=2
Translation to Words
Take half the sum of the length and the width,
which is 7.
Take the square root to get 2.
Step 5
7+2=9
The length is half the sum plus the square root.
Length is 9.
Step 6
7−2=5
The width is half the sum minus the square
root. Width is 5.
Woah! What is going on here? First of all, notice that the two dimensions
found satisfy both conditions. 9 + 5 = 14, and 9 × 5 = 45.
To see what is happening, let’s translate what they are doing into modern
algebraic notation.
In Step1, the scribe takes half the sum of the length
and width. This sum is 14, so half of it is 7. That’s easy enough. But why does
he take half?
The answer to that question comes from realizing that if you know that two
numbers add up to 14, then the first guess to take for each of the numbers is
MAT107 Chapter 3, Lawrence Morales, 2001; Page 45
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exactly half the sum. 7 + 7 is certainly 14. However, it is rare that the length
and width will be the same. However, if one number is a more than 7, then
the other number must be 7 minus a. That is, we can write our two numbers
as:
7 + a and 7 − a
Let’s say that the length is 7 + a and the width is 7 − a . Note that if you add
them, you get (7 + a) + (7 − a ) = 14 + a − a = 14 . So, all we need to do is find
the value of a that satisfies the second requirement; namely, the product must
be equal to 45. So, we take the product of these two numbers to get:
( 7 + a )( 7 − a ) = 45
When we multiply the left side out, we get:
49 − a 2 = 45
In this process, notice that we had to square 7 (to get 49)…that’s Step2 above.
Step3 says to subtract the area (from 49) to get 4. Note that if you subtract 45
from both sides of this last equation you get
4 − a2 = 0
which we will rewrite as
4 = a2 .
To solve this for a, we would take the square root of both sides, which is what
Step4 says to do basically. When we do this, we get two answers: a = ±2 .
However, the Babylonians did not recognize negative answers, so the only
answer they would have given was 2.
Now that we know what a is, we can find the length and width, since we
designated the length to be 7 + a and the width to be 7 − a .
Step5 says the length is the half sum (7) plus the square root (a), so length is 7
+ 2 = 9. Likewise, Step6 tells us to get the width by taking 7 – 2 = 5.
While this explanation is a bit long, it does show that the Babylonians are
clearly using algebraic reasoning to find the answer to their question. ♦
MAT107 Chapter 3, Lawrence Morales, 2001; Page 46
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Example 26
Solve the following problem using the Babylonian technique. Show an
algebraic representation of what is going on.
Length plus width is 18. Length times width is 72. Find each.
Solution:
Use this
example as a
guide for
your
homework
problems on
this topic.
Tablet
Computations
18 ÷ 2 = 9
Translation Into Words
Modern Algebra
Half the length plus
width is 9
1
(Length + Width) = 9
2
(9 + a ) + (9 − a ) = 9
9 × 9 = 81
Square 9 to get 81
( 9 + a )( 9 − a ) = 72
81 − 72 = 9
9 =3
Take the square root to
get 3
9 + 3 = 12 (the
length)
9 − 3 = 6 (the
width)
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Subtract the area to get
9
Half the sum plus the
root is length. Length is
12.
Half the sum minus the
root is the width. Width
is 6.
81 − a 2 = 72
81 − 72 = a 2
9 = a2
9 = a2
3=a
Length = 9 + a = 9 + 3 = 12
Width = 9 − a = 9 − 3 = 6
Check Point R
Solve the following problem using the Babylonian technique. Show an
algebraic representation of what is going on.
Length plus width is 30. Length times width is 200. Find each.
Solution:
Check the endnote for the solution.35
Another common type of algebra problem that appears on tablets is the following: Find two
numbers if their difference and product are given. This is very similar to those we’ve just
examined. We will see that the same process, with only a slight modification, will work fine.
MAT107 Chapter 3, Lawrence Morales, 2001; Page 47
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Example 27
The length of a rectangle minus its width is 8. The length times the width is
84. What are the dimensions?
Solution:
We resort back to Babylonian methods.
Steps
Step1
Tablet Computations
8÷2 = 4
Step2
Step3
Step4
4 × 4 = 16
16 + 84 = 100
100 = 10
10 + 4 = 14 . Length is 14 The length is the square root plus half
the difference. Length is 14.
The width is square root minus the half
10 − 4 = 6 . Width is 6
the difference. Width is 6.
Step5
Step6
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Translation to Words
Take half the difference of the length
and the width, which is 4.
Square 4 to get 16.
Add the area to get 100.
Take the square root to get 10.
If you compare this to our previous problems and examples, you see an almost
identical pattern, with only small modifications.
As before, let’s see what is happening algebraically.
In Step 1, we note that we are once again taking half of the given difference
(as opposed to half the given sum.) In this case, half of 8 is 4. We reintroduce
the variable a as before and observe that we can let the length be a + 4 and
the width be a − 4 . (The length is larger since when we take length and
subtract width, we get a positive number.) We can now check that our
condition of l − w = 8 .
l − w = ( a + 4) − ( a − 4)
= a−a+4+4
=8
With expressions for l and w, we can multiply them to satisfy the second
condition.
lw = 84
( a + 4 )( a − 4 ) = 84
When we multiply this out on the left side, we get:
a 2 − 16 = 84
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Step 2 above tells us to square 4, which is 16. The multiplication above does
this.
Step 3 says to add the area to the square. Note that if you add 16 to both sides
of the last equation above, you get a 2 = 100 .
Step 4 says to take the square root, which is 10. Finally, Step5 and Step6 tell
us how to find length and width, just as before. When we put these side by
side, we see this:
Steps
Step 1
Step 2
Tablet Computations
8÷2 = 4
4 × 4 = 16
Translation to Modern Algebra
1
(l − w) = 4
2
( l − w) = ( a + 4) − ( a − 4)
lw = 84
= a−a+4+4
=8
( a + 4 )( a − 4 ) = 84
Step 3
Step 4
Step 5
Step 6
16 + 84 = 100
100 = 10
10 + 4 = 14 . Length is
14.
10 − 4 = 6 . Width is 6.
a 2 − 16 = 86
a 2 = 100
a = 10
Length = a + 4 = 10 + 4 = 14
Width = a − 4 = 10 − 4 = 6
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Example 28
Solve the following problem using the Babylonian Technique. Show an
algebraic representation of what is going on.
The length of a rectangle minus its width is 10. The length times the width is
96. What are the dimensions?
Solution:
Tablet Computations
Translation to Words
10 ÷ 2 = 5
Take half the difference of the
length and the width, which is
5.
Use this
example as a
guide for
your
homework
problems on
this topic.
1
(l − w) = 5
2
l − w = ( a + 5) − ( a − 5)
= a −a +5+5
= 10
5 × 5 = 25
Square 5 to get 25.
lw = 96
( a + 5 )( a − 5) = 96
25 + 96 = 121
121 = 11
11 + 5 = 15 .
Length is 16.
11 − 5 = 6 .
Width is 6.
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Modern Algebra
a 2 − 25 = 96
Add the area to get 121.
a 2 = 121
Take the square root to get 11. a = 11
The length is the square root
Length =
a + 5 = 11 + 5 = 16
plus half the difference.
Length is 16.
The width is square root
Width =
a − 5 = 11 − 5 = 6
minus the half the difference.
Width is 6.
Check Point S
Solve the following problem using the Babylonian Technique. Show an
algebraic representation of what is going on.
The length of a rectangle minus its width is 8. The length times the width is
308. What are the dimensions?
Solution:
See endnote for answer.36
Babylonian Quadratics
The last few examples have been instances where a system of equations was to be solved. We
have primarily shown what we believe to be the procedures that the Babylonians were likely to
use.
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However, there are other ways to solve the last worked−out example. In a more modern algebra
class we might approach the problem a little differently. We will do so here as a transition into
quadratic equations and how the Babylonians solved them.
Example 29
The length plus the width of a rectangle is 18. The length times the width is
72. What are the dimensions of the rectangle?
Solution:
We start by letting the length be l and the width be w. We now have a system
of equations:
l + w = 18

lw = 72
Take the first equation and solve for l to get l = 18 − w . We can now
substitute this into the second equation to get the following:
lw = 72
(18 − w) w = 72
18w − w2 = 72
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This gives a quadratic equation, which, when set equal to 0, can be solved
either by factoring or with the quadratic formula.
w2 − 18w + 72 = 0
( w − 12 )( w − 6 ) = 0
w = 12 or w = 6
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If w = 12, then the length is 18 – 12 = 6, and if w = 6, then the length is 18 – 6 = 12. In either
case, we get the two distinct dimensions, just as the Babylonians did. ♦
The last example gives us hints as to how the Babylonians would solve quadratic equations.
Note that in the last example, solving the given system of equations reduced to solving a
quadratic equation, so if we reverse directions, we see that solving a quadratic should be
equivalent to solving a system of equations. Since the Babylonians definitely knew how to do the
latter, they presumably could do the former. And indeed they could.
The technique involves taking a given quadratic equation and creating from it a system of
equations.
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Example 30
The area of a square plus six times the side gives 16. What is the side?
Solution:
We will resort to using modern notation to help us, but the spirit of the
solution will be consistent with what the Babylonians might have done. In
this problem, if we let the length of the side by x then we can translate it into
an equation. If the side of a square is x units long, then the area of the square
is x 2 . Six times the side would be translated as 6x . Therefore, the equation
we have to solve is:
Area
+ Six times the side
give
16
x 2 + 6 x = 16
The Babylonian solution amounts to the following in modern notation:
x ( x + 6 ) = 16
We’ve simply factored an x out of the left side.
If we let y = x + 6 , then we get x( y ) = 16 . Also, if y = x + 6 , then y − x = 16 ,
we find that we have created a system of equations in two variables. The
system is:
y − x = 6

 xy = 16
Here we have something very similar to length minus width is 6 and length
times width is 16. But this is a typical Babylonian systems-of-equations
problem.
We can set this up as before.
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Tablet Computations
Translation to Words
6÷2 =3
Take half the difference of x
and y, which is 3.
3× 3 = 9
Square 3 to get 9.
Modern Algebra
1
( y − x) = 3
2
y − x = ( a + 3) − ( a − 3)
= a − a +3+3
=6
xy = 16
( a + 3)( a − 3) = 16
9 + 16 = 25
25 = 5
5+3=8.
y=8
5−3 = 2.
x is 2.
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Add the area to get 25.
Take the square root to get 5.
y is the square root plus half
the difference. y = 8..
The width is square root
minus the half the difference.
x is 2.
a 2 − 9 = 16
a 2 = 25
a=5
y=
a + 5 = 11 + 5 = 16
x=
a − 5 = 11 − 5 = 6
We can check this easily. If a square has a side of length 2, then its area is 4.
Six times the side is 12. Adding the area and six times the side gives 4 + 12 =
16, which does satisfy our conditions. ♦
Example 31
Solve x 2 − 8 x = 9 using the Babylonian technique.
Solution:
We start by factoring out on the left:
x ( x − 8) = 9
We let the inside of the parentheses by y so that y = ( x − 8) . We can rewrite
this as x − y = 8 , the difference of two unknowns. This gives us a system of
equations:
x − y = 8

 xy = 9
We introduce our variable a to help us along and build our familiar table. We
can let x = a + 4 and y = a − 4 so that the difference is 8.
MAT107 Chapter 3, Lawrence Morales, 2001; Page 53
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Tablet Computations
Translation to Words
8÷2 = 4
Take half the difference of x
and y, which is 4.
Use this
example as a
guide for
your
homework
problems on
this topic.
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4 × 4 = 16
Square 4 to get 16.
Modern Algebra
1
( x − y) = 4
2
x − y = ( a + 4) − ( a − 4)
=8
xy = 9
( a + 4 )( a − 4 ) = 9
16 + 9 = 25
25 = 5
5 + 4 = 9.
x=9
5 − 4 =1.
y is 1.
Add the product to get 25.
Take the square root to get 5.
a 2 − 16 = 9
a 2 = 25
a=5
x is the square root plus half
the difference. x = 9.
y is square root minus the half
the difference. y is 1
x=
a + 4 = 5+ 4 = 9
y=
a −5 = 5− 4 =1
Note that we don’t really need the last row and the value of y since our goal
was to find x. ♦
Check Point T
Solve x 2 + 4 x = 12 using the Babylonian technique.
Solution:
See Examples above and plug in your result for a check.
Check Point U
Solve x 2 − 4 x = 12 using the Babylonian technique.
Solution:
See Examples above and plug in your result for a check.
General Approaches to Quadratics
To recap, solving Babylonian quadratics boils down to writing the quadratic as a system of
equations. We need to be a little careful when we generalize, however. Note that the quadratic
must be in the form x 2 + bx = c to use this method. Not only that, but the value of c must be
positive since the Babylonians did not recognize negative numbers. In this form, we can factor
out an x on the left.
For example, to solve x 2 + 3 x − 4 = 0 , the problem would first have to be rewritten as
x 2 + 3 x = 4 so that it could be rewritten as x( x + 3) = 4 .
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If the quadratic took on a different form, such as x 2 + c = bx , the Babylonians might have seen
that as a completely different problem and had a different (but perhaps related) technique to
tackle that kind of problem. We should immediately recognize that this is very different than
how we solve quadratics today. In modern algebra, we can solve any quadratic of the form
ax 2 + bx + c = 0 by using the quadratic formula. Thousands of years of “evolution,” including
the recognition and use of negative numbers helps us make the process much more general and
compact. The Babylonians, on the other hand, did not have one general method for solving
quadratics. They had to improvise given the nature of the problem given to them.
Even though they did not have a formula to solve problems like these, that does not stop us from
exploring what kind of formula their process would actually give. In general, suppose that the
Babylonians had a system of equations such as:
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x + y = b

 xy = c
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If we follow the steps of the Babylonians, we should get a formula that will work for any system
of equations that looks like this one.
Tablet
Computations
b÷2 = b/2
b b b2
× =
2 2 4
Translation Into Words
Modern Algebra
Half the length plus
width is b / 2 .
1
b
(l + w) =
2
2
b
 b

 + a +  − a = b
2
 2

b
 b

 + a  − a  = c
2
 2

Square
b2
b
to get
2
4
Subtract the product
b2
−c
4
Take the square root
b2
−c
4
b2
− a2 = c
4
b2
− c = a2
4
b2
− c = a2
4
b2
−c = a
4
b
b2
+
−c
2
4
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Half the sum plus the
root is x
x=
If you look at the final result, you see the value of x is :
MAT107 Chapter 3, Lawrence Morales, 2001; Page 55
b
b2
+
−c
2
4
Think About It
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b
b2
+
−c
2
4
You might call this the “Babylonian Quadratic Formula” for certain
cases. (Of course, they did not have this available to them but their
methods lead to this equation when you translate their steps into
modern algebraic notation.)
How can a more direct
link be made between
the modern quadratic
equation and the
“Babylonian Quadratic
Equation?”
If you look at this closely, it looks very much like what we have in the quadratic formula:
−b ± b 2 − 4ac
2a
With this latest development, we see that the methods of the Babylonians can at least be
indirectly related to the modern quadratic formula. The question at this point, of course, is what
that link really is.
Much more can be said about Babylonian algebra and many more examples could be given, but
we’ll stop here. Hopefully, it is apparent that the Babylonians did engage in algebraic reasoning.
Furthermore, even though their techniques may look different than the algebra that we are used
to using and seeing, much of what the Babylonians did can be demystified by translating their
work into modern notation and techniques.
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Part 6: Homework Problems
Conversions
Convert the following Babylonian numbers to modern sexagesimal notation (i.e. 45,12;30) and
then determine their base-10 value. Please be sure your count the symbols carefully.
2)
1)
Please note that #3 and 4 have dashed lines indicating the separation of whole and fractional
parts.
3)
4)
Convert the following decimal numbers to base 60. Write your final results in both modern
sexagesimal notation (i.e. 45,12;40) and cuneiform notation. Please show your calculations
and/or work.
5)
853
6) 10,000
7)
125
8) 350,000
9)
28.45
10) 253.682
11)
1,453.003
12) 5
1
8
A Babylonian Translation Problem
Suppose a Babylonian teacher/student tablet (perhaps similar to the
one shown)37 contains a calculation that is trying to determine the area
of a rectangular region. The scribe multiplies the length and the width
and writes the following on a tablet:
13) Suppose you know that neither the width nor the length exceed
30 units in length. What can you say about the exact value of the
number shown above? Explain or show how you got your exact
value.
MAT107 Chapter 3, Lawrence Morales, 2001; Page 57
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14) Now assume that you flip the tablet over and discover that the scribe has indicated that the
length of the rectangular region is the following:
Based on your answer to Problem (13), what is the width of the rectangular region? Clearly
explain your reasoning! Write your answer in cuneiform. (Hint: You may want to do
computations in base 10 then convert to base 60.)
Multiplication
Use the Babylonian multiplication and reciprocal table (Table 1 on page 69) to do the following
multiplication problems. You should use the gelosia grid method that was described in this
chapter. (See the blank grids at the end of this chapter…you may have to make copies before
beginning your work, unless you create your own grids.) When you are done write the original
multiplication problem and answer as a scribal student would do so on a tablet.
15)
(4,20,15) × (3,20)
16)
(15,20,10) × (8,7,24)
17)
(20,33) × (13,40)
18)
(12;40,20) × (15;18)
19)
(12,25;35) × (40,20)
20)
(20,14,18) × (7,16,15)
Division
Use the Babylonian multiplication and reciprocal table (Table 1 on page 69) to do the following
division problems. When you are done write the original division problem and answer as a
scribal student would do so on a tablet.
21)
(11,8,2;30) ÷ (15)
22)
(17,5,35) ÷ (50)
23)
(40,30,20,10) ÷ (8)
24)
(16,0;30) ÷ (48)
Tablet Problems
25) The picture to the right represents a problem given to a
student on a clay tablet.
a.
What problem is being described? Justify your answer.
b.
Do the problem using the table method and express your
answer in Babylonian notation.
One of the standard units of length in the Babylonian
measurement system was the nindan (about 6 meters). One of
the standard units of area in the Babylonian measurement
system was the sar (about 36 square meters).
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26) The tablet to the right shows a rectangle with its sides
labeled. Calculate the are area of the rectangle and
write the result (in base 60 cuneiform) inside the
rectable.
27) Suppose you are scribal student and your
scribal teacher tells you to draw a picture of a
rectangle that is 1,25 nindan long and 4,27;18
nindan wide. (These are given in base 60) You
are to label the sides of the rectangle (in base
60 cuneiform notation of course), compute the
area of the rectangle, and write that result in
the interior of the rectangle. He has provided
the student tablet here to the right in which
you can carve your results.
Root Approximations
Use Babylonian methods to do find the first, second, and third approximations to the following
square roots. Show all steps carefully. Leave all your work in terms of fractions….do not
convert anything to decimals.
28)
7
29)
52
30)
32
31)
50
32)
90
33)
79
34)
66
35)
107
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The Alternate Method of Estimating Roots
Estimate the following square roots using the following equation (as described in this chapter):
a2 + h ≈ a +
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1932
h
2a
36)
10
37)
105
38)
230
39)
500
40)
85
41)
555
42)
2000
43)
529
The Babylonians and Pythagorean Triples
As we know, the Babylonians were well aware of Pythagorean triples and the relationship that
holds between the legs of a right triangle and its hypotenuse ( a 2 + b 2 = c 2 ).
One natural geometric question that seems to have arisen in ancient
civilizations is calculating the length of a diagonal of a square if you know
c=?
the length of one side. For example, given a square with sides of length 3,
3
what is the length of the diagonal of the square? (See picture.) Note that
the diagonal of the square corresponds to the hypotenuse of the right
triangle in the lower half corner of the square.
3
44) Use the Pythagorean theorem to find the diagonal of square with
given side length. Keep your answer in radical notation…do not convert to decimals.(For
example, if you were to get 8 for an answer, you would want to completely simplify this
to 2 2 using rules of radicals.)
a)
Length = 3
b) Length = 5
c)
Length = 7
d) Length = 10
e)
Writing: Look at your answers to the previous four parts. What patterns do you see? Write
about them and then use them to describe the length of a diagonal in a square where each
side is x units long.
f)
Now take a square with a side of Length = x. What is the length of the diagonal, in terms x?
Simplify completely and compare this to your observation(s) in part (e). They should be
reconcilable.
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A Famous Babylonian Tablet
45) (Make sure you have done Problem (44) before you
attempt this problem.) The tablet shown is a Babylonian
clay tablet inscribed in about 1600 B.C.E. It is a part of
the Yale University. It shows a square drawn on a tablet.
To determine what this tablet is about, first note the
number that labels the side of the square. This is
obviously 30. There are two numbers in the middle of
the square. The one on the bottom line is the following
(with the “sexagesimal line” added for you):
a.
What is this number in base ten? Round to four decimal places.
b.
The number that is written along the diagonal of the square is the following (with the
“sexagesimal line” added for you):
What is this number, rounded to 4 decimal places?
c.
Find the geometric relationship between these three numbers and then clearly explain what
the scribe was trying to point out on this tablet.
46) Take your result from the previous problem and draw your own tablet similar to the one
pictured above, except have the side of the square be 50 instead of 30. Your drawing
should use cuneiform (not Hindu-Arabic) numbers. Below your drawing, clearly
show/explain why your drawing is accurate (you may use Hindu-Arabic numbers and
modern sexagesimal notation in your explanation).
Babylonian Algebra − Systems of Equations
Solve the following problems using the systems of equations techniques discussed in this chapter.
Show an algebraic representation of what is going on. See Example 26 for examples of how to
write these problems up.
47)
48)
49)
50)
Length plus width is 14. Length times width is 33. Find the dimensions.
Length plus width is 34. Length times width is 280. Find the dimensions.
Length plus width is 32. Length times width is 231. Find the dimensions.
Length plus width is 23. Length times width is 120. Find the dimensions.
MAT107 Chapter 3, Lawrence Morales, 2001; Page 61
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2014
Solve the following problems using the systems of equations techniques discussed in this chapter.
Show an algebraic representation of what is going on. See Example 28 for examples of how to
write these problems up.
51) The length of a rectangle minus its width is 12. The length times the width is 253. What are
the dimensions?
52) The length of a rectangle minus its width is 14. The length times the width is 51. What are
the dimensions?
53) The length of a rectangle minus its width is 8. The length times the width is 660. What are
the dimensions?
54) The length of a rectangle minus its width is 9. The length times the width is 190. What are
the dimensions?
Babylonian Algebra − Solving Quadratic Equations
Use the Babylonian Technique of solving quadratic equations to solve the following. See
Example 31 for a model on how to write these up. You may not use the modern quadratic
formula to do these.
55) Solve x 2 + 5 x = 24
56) Solve x 2 − x = 42
57) Solve x 2 + 2 x = 63
58) Solve x 2 + 4 x = 165
59) Translate the following into a quadratic equation and then solve it using the Babylonian
Technique, like in Problem (55) to (58): The area of a square plus five times its side gives
24. What is the length of the side?
60) Translate the following into a quadratic equation and then solve it using the Babylonian
Technique, like in Problem (55) to (58): The area of a square plus six times its side gives
72. What is the length of the side?
61) Translate the following into a quadratic equation and then solve it using the Babylonian
Technique, like in Problem (55) to (58): The area of a square plus two times its side gives
15. What is the length of the side?
MAT107 Chapter 3, Lawrence Morales, 2001; Page 62
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Practice and Application of the Quadratic Equation
Recall from your previous math classes that an equation of the form ax 2 + bx + c = 0 can be
solved using the “modern” quadratic formula:
−b ± b 2 − 4ac
x=
(Equation 1)
2a
Solve the following equations using the quadratic formula, if possible. Write your answers in
simplified, exact form. That is, do not use decimals; instead, keep your answers in radical form.
62)
x 2 + 3x − 5 = 0
63) 2 x 2 + 9 x = 7
64)
−3x 2 − 8 x + 2 = 0
65) (x + 1) = 10
2
66) The YBC13901 tablet gives a problem that may be translated as “Area [of the square] and
1;20 of a side added give 0;55. What is the side?”
a.
Write a quadratic equation in modern base−ten notation using x as a variable that would
solve this problem. (Hint…let x be the side of the square. If x is the side, what is the “area
of the square”…what is 1;20th of a side?)
b.
Solve the equation using the modern quadratic formula. To do this problem, you will need
to convert to decimal (base−ten) before proceeding. It will help greatly if you clear
fractions before starting to use the quadratic formula. Express your answer in exact
form…that is, no decimal conversion. Keep everything in terms of square roots and
fractions.
c.
Challenge: Take the problem from this tablet and solve it using the Babylonian Technique
(see Example 31.)
New Quadratic Equations
When you translate the problem from YBC13901, you should get an equation with a basic form
of x 2 + bx = c . Note that this is slightly different than the “standard form” of ax 2 + bx + c = 0
commonly referred to today. The Babylonians would have viewed these two equations as
different kinds of problems and hence had a different approach depending on the form of the
equation.
In the case of an equation of the form x 2 + bx = c , the Babylonians appear to have used a
process to get a solution that is equivalent to using the formula:
b
b2
x=− +
+c
2
4
(Equation 2)
Note that this is slightly different from the equation you are used to, but looks similar. Use this
formula to find a solution to the following problems. In order to use this equation, the
coefficient of the x2 term has to stay equal to 1…don’t change it in any way. Also, this equation
MAT107 Chapter 3, Lawrence Morales, 2001; Page 63
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does not require that you move c to the left…it stays on the right since their basic form starts
with c on the right.
67)
68) x 2 + 7 x = 10
69) The YBC13901 tablet gives a problem that may be translated as “Area [of the square] and
1;20 of a side added give 0;55. What is the side?” When you translate the problem from
YBC13901, you should get an equation with a basic form of x 2 + bx = c . Solve this
b
b2
problem using x = − +
+c
2
4
found in #40 as a check
2063
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2065
2066
2067
2068
2069
2070
x2 + 4 x = 5
Yet Another Quadratic Formula
As stated earlier, if the equation took on a different form, the Babylonians usually had a different
approach to solve the equation. If an equation took on the form of x 2 + c = bx , they would use
yet another equation:
x=
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(Equation 2) above. Compare your answer to that
b
b2
+
−c
2
4
(Equation 3)
The equation x 2 + c = bx may not look much different to you than x 2 + bx = c , but to the
Babylonians, they were different kinds of problems. Keep in mind that we are more comfortable
with moving things around within an equation and we have the benefit of the use of negative
numbers, which the Babylonians did not.
b
b2
+
− c (Equation 3), to solve each of the following
2
4
equations that are of the form x 2 + c = bx :
Use this newest equation x =
70)
x2 + 5 = 9 x
71) x 2 + 7 = 8 x
72) x 2 + 10 = 12 x
73) It is well known among mathematics historians that if a quadratic had the form
x 2 + c = bx , then the Babylonians would use a process equivalent to the formula
b
b2
b
b2
− c . See x = +
− c (Equation 3) above. Prove/show that this
x= +
2
4
2
4
formula is equivalent to the modern quadratic formula for a quadratic in the form given.
To do this, you should not use specific numbers. Stick with the variables given. To start
this problem, you may want to set x 2 + c = bx equal to zero, like you would any modern
quadratic equation, and then apply the quadratic formula to the result. You will have to be
careful with the letters b and c and the negative signs.
MAT107 Chapter 3, Lawrence Morales, 2001; Page 64
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Real Babylonian Algebra Problems From Tablets
The following are problems from Babylonian tablets. Solve each one using modern algebraic
methods. Your do NOT have to use Babylonian methods to do these problems. (However, guess
and check will not get full credit.) Some of them are linear, some are quadratic, and some are
systems of linear and/or quadratic equations, so sharpen your algebraic skills. Answer in
complete sentences and include units with your answers. Make sure all variables that you use are
clearly defined at the beginning of your problem.
74) I have added the area and two-thirds the side of my square, and it is 35/60. What is the side
of my square?
75) I have a reed. I know not its dimension. I broke off from it 1 cubit and walked 60 times
along the length [of the remaining reed]. I restored to it what I had broken off, then walked
30 times along its length. The area is 375 square cubits. What was the original length of the
reed?
76) I found a stone but did not weight it; after I added to it 1/7 of its weight and then 1/11 of
this new weight, I weighed the total at 1 mina. What was the original weight of the stone?
Give your answer in minas.
(b.) The scribe gives his answer as 2/3 mina, 8 sheqels, and 22 ½ se. If 1 mina = 60
sheqels, and 1 sheqel = 180 se, verify your answer by checking it against that of the scribe.
(To do this, convert your answers in minas to minas, sheqels, and se’s, or vice versa, to
make sure that they match.)
77) I found a stone but did not weight it; after I subtracted one seventh and then subtracted one
thirteenth [of the remainder], I weighed it at 1 mina. What was the original weight of the
stone?
78) I have multiplied length and width, obtaining area. Then I added to the area the excess of
length over width and obtained the result 183. Further, when I added the length and width,
I obtained 27. Find the length, width, and area.
79) There are two silver rings; 1/7 of the first and 1/11 of the second are broken off, so that
what is broken off weighs 1 sheqel. The first diminished by its 1/7 weighs as much as the
second diminished by its 1/11. What did the silver rings originally weigh?
80) I found a stone but did not weigh it; after I weighed out 6 times its weight and added 2
sheqels, I then added one third of one seventh of this amount multiplied by 24; the weight
was then 1 mina. What was the original weight? (I mina = 60 sheqels)
MAT107 Chapter 3, Lawrence Morales, 2001; Page 65
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Writing
Write a short essay on the given topic. It should not be more than one page and if you can type it
(double−spaced), I would appreciate it. If you cannot type it, your writing must be legible.
Attention to grammar is important, although it does not have to be perfect grammatically…I just
want to be able to understand it.
81) Compare the two Babylonian methods of approximating square roots. Which method is
more accurate? Which on is more efficient? Use at least two specific examples to
demonstrate your conclusions.
82) Use the library or internet to research an aspect of the ancient Babylonian/Mesopotamian
civilization that was not covered in this chapter or in class and give a description of what
you find. Cite all of your sources.
83) If you were going to teach complex multiplication to a child who only knows their
multiplication tables, would you teach the modern method first or the gelosia method first?
Explain your reasoning for your answer.
MAT107 Chapter 3, Lawrence Morales, 2001; Page 66
2155
Appendix: Blank Gelosia Grids
2156
2157
MAT107 Chapter 3, Lawrence Morales, 2001; Page 67
2157
2158
MAT107 Chapter 3, Lawrence Morales, 2001; Page 68
0 16 0 24 0 32 0 40 0 48 0 56 1
0 18 0 27 0 36 0 45 0 54 1
9
5
13 0 26 0 39 0 52 1
1 12 1 21 1 30
4
8
16 0 32 0 48 1
17 0 34 0 51 1
0
8
0
2 20
2 24 2 40
2 15 2 30
1
0
2 10 2
2 12 2 23 2
0
1 30 2
0
0
0
0
0
5
4 30 5
0
0
0
6 40
0
5 50 6 40 7 30 8 20
0
3
7
3 24 3 41 3
4 20 4
4
0
0
0
7 35 8
6 30 7
5 25 5
0
0
20 10
9
8
7
6
6
5
4
4
4
4
8
8
6
6
6
5
5
5
4
4
4
3
3
3
2
2
2
1
1
1
0
9
9
8
8
7
7
6
6
5
5
4
4
3
3
2
2
1
1
0
12
0
0
0
2
3
4
5
6
8
9
16
0
12 30 15
0
0
0
0
15
0
18
16
0
0
20
24
25
27
11
30
0
5
0
13 20 14 40
5
13 30 15
0
0
16 30
16
0
18
0
0
20
0
22
0
16 40 18 20
12 40 14 15 15 50 17 25
0
55 11 20 12 45 14 10 15 35
0
17 30 20
0
0
0
22 30 25
32
0
24
20
0
24
0
0
36
40
0
27
0
0
32
0
0
MAT107 Chapter 3, Lawrence Morales, 2001; Page 69
33
0
33 20 36 40
30
36
0
40
0
44
0
33 45 37 30 41 15
0
0
5
27 30
45
5
48
50
54
36 40 41 15 45 50 50 25
29 10 33 20 37 30 41 40 45 50
28
0
23 20 26 40 30
21
18 45 22 30 26 15 30
13 20 16 40 20
12
9 10 10 0 10 50 11 40 12 30 16 40 20 50 25
18
0
10
15
20
25
30
35
40
45
50
55
11 15 12 30 13 45
20 10 40 12
0
11 40 13 20 15
14
9
8
7
6
5
4
3
2
1
10 10
20
30
40
50
0
10
20
30
40
50
10
9
8
7
6
5
5
4
3
2
1
45 10 50 11 55
0
15
30
45
0
15
30
45
0
15
30
45
45 11 40 14 35 17 30 20 25 23 20 26 15 29 10 32
A Table of Reciprocals in Base−60 Notation
10 12 15
9
9
8
7
6
6
5
4
3
3
2
1
20 10 30 11 40 12 50
40
0
20
40
0
20
40
0
20
40
0
20
40
1
;30 ;20 ;15 ;12 ;10 ;7,30 ;6,40 ;6 ;5 ;4 ;3,45 ;3,20 ;3 ;2,30 ;2,24 ;2,13,20 ;2 ;1,52,30 ;1,40 ;1,30 ;1,20 ;1,15 ;1,12 ;1,06,40
n
n
9
8
8
7
6
6
5
4
4
3
2
2
1
45 10
10
35
0
25
50
15
40
5
30
55
20
45
10
35
10 30 12
9
9
8
8
7
7
6
5
5
4
4
3
2
2
1
1
30 11
0
30
0
30
0
30
0
30
0
30
0
30
0
30
0
30
30
20 10
55
30
5
40
15
50
25
0
35
10
45
20
55
30
5
40
15
50
25
10
8
7
7
7
6
6
5
5
5
4
4
3
3
2
2
2
1
1
0
20 10 25 12 30 14 35 16 40 18 45 20 50 22 55
0
40
20
0
40
20
0
40
20
0
40
20
0
40
20
0
40
20
0
40
20
30 10
15
0
0
45
30
15
0
45
30
15
0
45
30
15
0
45
30
15
0
45
30
15
9 45 10 30 11 15 15
8 40 9
24
10
0
50
36
40
26
12
58
44
3
3
3
3
2
2
2
2
1
1
1
1
0
0
8 48 9 36 10 24 11 12 12
8 15 9
7 20 8
6 36 7 12 7 48 8
6 25 7
5 30 6
4 35 5
4 24 4 48 5 12 5
0
7
3 18 3 36 3 54 4
3
30
16
2
48
34
20
6
52
38
24
10
56
42
28
14
55 1 50 2 45 3 40 4 35 5 30 6 25 7 20 8 15 9 10 10 5 11 0 11 55 12 50 13 45 18 20 22 55 27 30 32
0
0
6 45 7 30
4 40 5 20 6
4 48 5 36 6 24 7 12 8
50 1 40 2 30 3 20 4 10 5
0
0
4 40 5 15 5 50
3 30 4
3 45 4 30 5 15 6
48 1 36 2 24 3 12 4
0
3 20
3 36 4 12 4 48 5 24 6
2 40 3 20 4
45 1 30 2 15 3
40 1 20 2
36 1 12 1 48 2 24 3
0
2 30 2 55 3 20 3 45 4 10
2 30 3
35 1 10 1 45 2 20 2 55 3 30 4
0
5
30 1
25 0 50 1 15 1 40 2
2 20 2 40 3
2
3 15 3
2 56 3 12 3 28 3
2 45 3
2 34 2 48 3
2 23 2 36 2 49 3
2 12 2 24 2 36 2
2
1 50 2
1 39 1 48 1 57 2
3 40 4
0
2 24 2 48 3 12 3 36 4
1 20 1 40 2
0
0
24 0 48 1 12 1 36 2
1
1 28 1 36 1 44 1
20 0 40 1
0
5
1 12 1 18 1
1
1 17 1 24 1 31 1
6
0
3 29 3 48 4
2 24 2 42 3
13
0 44 0 48 0 52 0
19 0 38 0 57 1 16 1 35 1 54 2 13 2 32 2 51 3 10
6
1 25 1 42 1 59 2 16 2 33 2 50
1 20 1 36 1 52 2
1 15 1 30 1 45 2
6
1 18 1 31 1 44 1 57 2 10
1 12 1 24 1 36 1 48 2
1 17 1 28 1 39 1 50
1 10 1 20 1 30 1 40
18 0 36 0 54 1 12 1 30 1 48 2
0
15 0 30 0 45 1
1 10
1 12 1 20
14 0 28 0 42 0 56 1 10 1 24 1 38 1 52 2
0
6
11 0 22 0 33 0 44 0 55 1
12 0 24 0 36 0 48 1
0
10 0 20 0 30 0 40 0 50 1
3
4
3
0
12
0 33 0 36 0 39 0
8
0 12 0 15 0 18 0 21 0 24 0 27 0 30
0 14 0 21 0 28 0 35 0 42 0 49 0 56 1
9
0 12 0 16 0 20 0 24 0 28 0 32 0 36 0 40
0
7
11
0 22 0 24 0 26 0
1
10
0 12 0 18 0 24 0 30 0 36 0 42 0 48 0 54 1
9
6
8
0 55 1
7
8
6
0 10 0 15 0 20 0 25 0 30 0 35 0 40 0 45 0 50
5
0 10 0 12 0 14 0 16 0 18 0 20
0
8
5
4
4
0
6
6
0
3
3
0
4
0
2
2
Table 1 - Babylonian Multiplication Table in Base 60
MAT107 Chapter 3, Lawrence Morales, 2001; Page 70
Specific Gelosia Grids
2by2
4by4
3by3
3by2
2by3
3by4
4by3
5by5
MAT107 Chapter 3, Lawrence Morales, 2001; Page 71
2by2
4by4
3by3
3by2
2by3
3by4
4by3
5by5
MAT107 Chapter 3, Lawrence Morales, 2001; Page 72
Part 6: Endnotes
1
Robson, Dr. Eleanor, Oriental Institute, University of Oxford, “Mesopotamian Mathematics” handout at the 1999 Institute
in the History of Mathematics and its Uses in Teaching.
2
http://library.thinkquest.org/22584/temh2100.htm
3
Calinger, page 20.
4
Calinger, page 20.
5
Calinger, page 21.
6
Calinger, page 22.
7
Calinger, page 22.
8
Robson, Dr. Eleanor, “Counting in Cuneiform”, in Mathematics in School, September, 1998. page 4.
9
Robson
10
See http://it.stlawu.edu/~dmelvill/mesomath/Numbers.html for list of their symbols.
11
http://it.stlawu.edu/~dmelvill/mesomath/Numbers.html
12
Solution to Check Point A
Its decimal value is 2,305,382. In modern sexagesimal notation, it would look like 10,40,23,2
13
Solution to Check Point B
Its decimal value is 263,111. In Babylonian cuneiform, the number would look like:
14
Solution to Check Point C
In modern notation, the number would look like 23,5;21. The decimal value of the number is 1,385.35
15
Solution to Check Point D
In modern notation, the number would look like 59,21;03;12. The decimal value of the number is 3561.0533333
16
Solution to Check Point E
In modern sexagesimal notation, 300,000 = 1,23,20,0. Note that there are no ones left over when you’re done.
17
Solution to Check Point F
In modern sexagesimal notation, 75.102 = 1,15;06,07,12. Don’t forget to convert the whole number part.
18
Solution to Check Point G
In modern sexagesimal notation, 13;31,30. Don’t forget to convert the whole number part.
19
“Counting in Cuneiform”, in Mathematics in School, September, 1998. page 4.
I believe it requires 19 marks…check me to see if I’m wrong.
21
Solution to Check Point H
20
22
23
“Counting in Cuneiform”, page 5
Solution to
MAT107 Chapter 3, Lawrence Morales, 2001; Page 73
Check Point I
4
7
5
3
5
4
4
2
2
4
1
9
6
7
57
4
6
5
08
3
8
3
24
6
3
2
59
5
Solution to Check Point J
The answer is 29643.0
6
1
2
5
0
2
2
4
4
4
0 4
0
9
4
8
1
8
0
6
0
1
2
.
2
4
8
0 2
3
0
25
Solution to Check Point K
You should get 20×40=13,20 in sexagesimal.
26
Solution to Check Point L
You should get 20×42=14,0 in sexagesimal. You’ll need to do 40×60 + 2×60 to get the result.
27
Solution to Check Point M
You should get the following result. 25,50×10,15 = 4,24,47,30. There’s no need to carry in this case.
25
4
4
50
8
10
6
24
20 10
12
15
47
30 15
30
MAT107 Chapter 3, Lawrence Morales, 2001; Page 74
28
Solution to Check Point N
The final answer ends up being 9,23,25,5;58,45
18
40 ;
15
9
20
7
9
0
0
3
6
23
40
23
25
30
30
8
;
45 35
58
45
“Counting in Cuneiform”, page 5
Solution to Check Point O
The reciprocal of 27 is 0;2;13,20, according to the reciprocal table. The final answer ends up being 1,20;27,17,46,40
36
12 ;
17
1
0
0
1
12
7
24
2
20
27
34 2
36
4
41 13
5
0
0
17
;
3
48
12
31
30 10
20
5
29
2
0
10
30 30
46
40 20
40
Solution to Check Point P
First note that 5 <
29 < 6 , and since it’s closer to 5 than 6, our first guess is 5.
29
5+
29
5 = 27 . Thus 27 is the second guess.
5×
= 29 so our second guess is
5
2
5
5
27 145
+
27 145
727
×
= 29 so the third guess is 5 27 =
5 27
2
135
32
Solution t o Check Point Q:
8
3
4
MAT107 Chapter 3, Lawrence Morales, 2001; Page 75
33
Katz, Victor, A History of Mathematics, p.31
Other resources:
http://it.stlawu.edu/~dmelvill/mesomath/index.html for multiplication tables and symbols
http://it.stlawu.edu/~dmelvill/mesomath/calculator/scalc.html for a cuneiform calculator
34
Bunt, page 51
35
Solution to Check Point R
Tablet Computations
30 ÷ 2 = 15
Translation Into Words
Half the length plus width
is 15
Modern Algebra
1
(l + w) = 15
2
(15 + a ) + (15 − a ) = 15
15 × 15 = 225
Square 15 to get 225
(15 + a )(15 − a ) = 200
Subtract the area to get 25
225 − a 2 = 200
225 − 200 = a 2
225 − 200 = 25
25 = a 2
25 = 5
Take the square root to get
5
15 + 5 = 20 (the length)
15 − 5 = 10 (the width)
36
Half the sum plus the root
is length. Length is 20.
Half the sum minus the
root is the width. Width is
10.
25 = a 2
5=a
Length =
15 + a = 15 + 5 = 20
Width = 15 − a = 15 − 5 = 10
Solution to Check Point S
Tablet Computations
Translation to Words
8÷2 = 4
Take half the difference of the
length and the width, which is
4.
Modern Algebra
1
(l − w) = 4
2
l − w = ( a + 4) − ( a − 4)
= a−a+4+4
=8
4 × 4 = 16
Square 4 to get 16.
lw = 308
( a + 4 )( a − 4 ) = 308
16 + 308 = 324
324 = 18
18 + 4 = 22.
Length is 22.
18 − 4 = 14.
Width is 14.
a 2 − 16 = 308
Add the area to get 324.
a 2 = 324
Take the square root to get 18. a = 18
The length is the square root
plus half the difference.
Length is 22.
The width is square root
minus the half the difference.
Width is 14.
37
Length =
a + 4 = 18 + 4 = 22
Width =
a − 4 = 18 − 4 = 14
The tablet shown was found at http://www.royal-athena.com/PAGES/cheappages/CLR105.html and is a tablet “stating
that Amat-Ningirsu and Taribum received a silver object from Shamash-arki and date to be returned; witnessed by four
MAT107 Chapter 3, Lawrence Morales, 2001; Page 76
individuals.” The tablet is for sale on this web site for $400. For more on “Why Base 60?” see http://omega.cohums.ohiostate.edu:8080/hyper-lists/classics-l/listserve_archives/log97/9708b/9708b.66.html.
MAT107 Chapter 3, Lawrence Morales, 2001; Page 77
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MAT107 Chapter 3, Lawrence Morales, 2001; Page 78