The Rhind papyrus; the first handbook of mathematics

Item Type text; Thesis-Reproduction (electronic)

Authors Cobb, Sumner Chase, 1895-

Publisher The University of Arizona.

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The Rhlnd Papyrus

The F i r s t Handbook o f M athematics

by

Sumner 0 . Cobb

Subm itted in p a r t i a l f u l f i l lm e n t o f the

requ irem en ts f o r the degree of

M aster o f Soienoee

in the G raduate C ollege

U n iv e rs ity o f A rizona

1 9 3 6

Approved: U eM ajor p ro fe s s o r

<w)/ * (&x— -----------LDate V 8 6

Table of C o n ten ts .

£ 9 ? f ;/ 9 3? 3

2-

S ec tio n Page

I n t r o d u c t io n ........................................................................... ...............

I The U nit F ra c t io n ............... ................... ........................................... 1

I I The Rhind P a p y ru s ............................... 3

I I I The Egyptian Number System .................... , f

IV The Four Fundamental O p e ra tio n s . . . ........................................ .9

E gyptian U nit F ra c t io n s .............. . . . . . . . . . . . . . . . . 1 3

VI The Table o f 2 over (2n f l ) . ................................................ .20

VII The Table o f 10th e ............................................................... 31

V III Problems 7 to 20. . . . . . . . . . 3 5

IX Com pletion Problem s. . . . . . . . . . . . . ... .......................... .39

X "Aha* P roblem s. ............. . . . . . . . . . . . . . . . . . . . . 4 2

XI Problem 3 3 . ................................. 48

XII The A rith m e tic P ro g re ss io n ........................... .......................... .54

B ib lio g rap h y . . ................ .. . . . . . . . . ........... .............................. 57

I l l u s t r a t i o n s .

E gyptian System o f Number Symbolism. ............................... .. ........... 5

E gyptian Numbers and F r a c t io n s . ..........................................................8

E gyptian A d d itio n , S u b tra c tio n , and M u l t ip l ic a t io n . . . . . . . . 12

Table of 2 D ivided by the Odd N u m b ers .... . . . . . . . . . . . . . 1 7 - 19

H ie ra t ic Symbols and Problem 3 3 . . . . . . . . . . . 4 7

105591

a

I n t ro d u c t io n .

D uring my te a c h in g e x p e r ie n c e , I have o f te n n o tic e d

th a t work in f r a c t io n s i s d i f f i c u l t f o r many s tu d e n ts in

e lem en tary a l g e b r a . . There seems to he an in b o rn f e a r o f

th e word " f r a c t io n " and a g e n e ra l d e s i r e to av o id work in ­

v o lv in g th e u se o f f r a c t i o n s . I have u s u a l ly l a id th e blame

f o r t h i s s t a t e o f a f f a i r s on e a r ly and f a u l ty t r a in in g in

th e s u b je c t . But my c u r io s i ty was a ro u se d . Do p eo p le o f te n

f in d work in f r a c t io n s d i f f i c u l t ?

In an O f f ic e r s * ,T ra in in g Gamp in 1918, I was a ss ig n ed

to te ach a gruop o f 50 men such elem ents o f m athem atics as

th ey would need b e fo re they began an in te n s iv e co u rse in

F ie ld A r t i l l e r y . Our f i r s t le s so n was on f r a c t i o n s . The

group in c lu d ed men from th e p ro fe s s io n s a s w e ll a s th o se

whose sc h o o lin g was o n ly th rough the g ra d e s . These men n o t

o n ly had l i t t l e r e a l co n cep tio n o f th e meaning o f a f r a c t io n

b u t a ls o had l i t t l e a b i l t y to u se a f r a c t io n in th e s im p le s t

o p e ra t io n s . T h is f u r th e r added to my i n t e r e s t in f r a c t io n s *

There seemed to be a rea so n back o f a l l t h i s d i f f i c u l t y . I

d ec id ed to f in d o u t what i t was i f I ev e r had an o p p o r tu n ity .

The h i s to r y o f m athem atics b eg in s w ith th e E gyp tians

and t h e i r e a r l i e s t known w ork, th e Rhind P ap y ru s . S ince a

la rg e p a r t o f t h i s a n c ie n t work i s on a r i th m e t i c , and t h i s

b .

a r i th m e tic * 1b tu rn* i s la r g e ly on u n i t f r a c t i o n s , I dec ided

to s tu d y the Rhimd Papyrus* Here we f in d th e e a r l i e s t con­

c e p t o f a f r a c t i o n and i t s f i r s t sy s te m a tic u s e . The s tu d y

h as been f a s c in a t i n g . There was d i f f i c u l t y w ith f r a c t io n s

3500 y e a rs ago - p le n ty o f i t .

The two volumes on the Rhind Papyrus by D r. A. B. Chace

a re the b a s is o f t h i s t h e s i s . Hie work i s a d e ta i l e d s tu d y

o f th e Papyrus to w hich a s tu d e n t could hope to add l i t t l e *

i f a n y th in g . In th e fo llo w in g pages I hope to p re s e n t a c a re ­

f u l s tu d y o f th e u n i t f r a c t io n and i t s u s e . There w i l l be a

s e c t io n devoted to the ta b le o f rt2 -4- 2 a + 1" which may w e ll be

c a l le d " T h e F i r s t Handbook o f M athem atics % I have con­

f in e d oy I n v e s t ig a t io n to th a t p a r t o f the, Papyips w hich

d e a ls w ith a r i th m e t ic and a lg e b r a .

S ince a f r a c t i o n i s such a w e ll known concept* and

s in c e no v e ry advanced m athem atics i s involved* I have s e t

m yse lf the a d d i t io n a l aim o f making th i s th e s is e n t i r e ly

w ith in the g ra sp o f th e non-m athem atioal mind• So much o f

I n te r e s t in g m athem atics i s n e c e s s a r i ly a c lo sed book to the

laym an. The Rhind Papyrus need n e t b e .

1.

I . The U n it F ra c t io n .

B efo re b eg in n in g our s tu d y o f th e Rhind Papyrus i t

m ight be h e lp fu l to u n d ers tan d what a u n i t f r a c t io n i s and

what p la c e i t o ccu p ies in ou r own a r i th m e t i c .

Any s ta n d a rd a r i th m e t ic c l a s s i f i e s f r a c t io n s a s o f two

k in d s , common and d e c im a l. We f in d th a t ooBnson f r a c t io n s may

be s im p le , compound, o r com plex. Of th e s e , a sim p le f r a c ­

t io n i s one whose num era to r and denom inator a r e s in g le i n t e ­

g e r s , such as 6 /2 9 . A sim ple f r a c t io n i s p ro p e r o r im proper

a c c o rd in g a s the num era to r i s l e s s th a n , o r g r e a te r than th e

d en o m in ato r. Suppose we c o n s id e r on ly th o se sim ple f r a c t io n s

whose num era to rs a r e u n i ty , o r 1 . We m ight c a l l th e se f r a c ­

t io n s the r e c ip ro c a ls o f the I n te g e r s r b u t a s im p le r name

f o r them i s u n i t f r a c t i o n s .

T h is , a s we s h a l l s e e , was th e meaning o f th e word

f r a c t io n in e a r ly E gy p tian m athem atics and la r g e ly th rough­

o u t th e a n c ie n t w o rld . O n e -h a lf , o n e - th i r d , o r o n e - f i f t h ,

f o r exam ple, could be g rasped a s a c o n c e p t, b u t r e a l d i f f i ­

c u l ty was encoun tered when i t became n e c e ssa ry to ex tend the

co n cep t to in c lu d e tw o - f i f th s o r th re e f o u r th s . Much o f the

2*

t ro u b le la y In the in a b l l ty to ex p ress th e se co n cep ts in

sym bolst o r even v e r b a l ly . The B abylon ians were a b le to

ex p ress some o f th e sim ple f r a c t io n s a s p a r t o f t h e i r u n i t

60; th u s , l / 2 was 30 , l / 3 was 2 0 , and so on* But th i s

method f a i l e d where 60 f a i l e d to d iv id e the d enom inato r.

Even a f t e r th e Greeks had developed a q u i te s a t i s f a c to r y

system o f f r a c t i o n s , t r a d i t i o n le d them back o f te n to the

u se o f the u n i t f r a c t i o n . Theodorus (c .4 0 0 B .C .) used u n i t ■ ■ - - . y ' :

f r a c t io n s In ap p ro x im atin g th e s q u a re - ro o t o f 3 .

1 jr-frit 3i tv? > YTy> 1 i: Vt i l

Where no symbol i s used betw een th e f r a c t i o n s , a d d i t io n i s

u n d e rs to o d . T his was th e custom ary p r a c t ic e w herever u n i t, . . . . ■ '

f r a c t io n s were added . In 50 B.C . we f in d Heron u s in g them

and as l a t e a s the 10 th cen tu ry A.D. they ap p ea r in a Hebrew

w r i t in g . D uring th e R ennaisance Buteo favo red them . I t i s

i n t e r e s t i n g to n o te th a t con tinued f r a c t io n s o f the formI 1 - ___ ■

w hich seem to have a co n n e c tio n w ith u n i t f r a c t io n s were n o t

su g g ested u n t i l 1613 in a work by O a ta ld i .

Mow th a t we u n d ers tan d what i s meant by a u n i t f r a c ­

t io n , we a re ready to c o n s id e r what the E g y p tian s could do

w ith th e se "most e lem en ta ry , p ro p e r , s im p le , common f r a c t io n s ."

L D .E .Sm ith : H is to ry o f M athem atics, Vol I I . page 212.

3 .

I I . The Rhlnd P apyrus•

The Ahmes Papyrus sms found a t Thebes in the ru in s o f a

sm a ll b u ild in g n e a r th e Harnesseura* Purchased in 1858 by

A. Henry Rhlnd, i t became known as th e Rhlnd P apyrus« I t

i s now in p o sse ss io n o f the B r i t i s h Museum. I t i s a copy

o f an e a r l i e r work, o r w orks, made by a a o t ib e A,Hoos6, o r

Ahmes. The d a te i s ap p ro x im ate ly 1650 B.C. I t i s prob­

a b le th a t i t r e p re s e n ts the knowledge o f E gyptian a r i t h -

m etio c e n tu r ie s e a r l i e r than 1650 B.C . O ther E gyptian

p ap y ri and fragm en ts g iv e ev idence th a t th e work o f Ahmes

on f r a c t io n s was n o t u n iq u e .

W ritten in h i e r a t i c ( a c u rs iv e form and n o t a s fo rm al

a s th e h ie ro g ly p h ic ) , i t was o r ig in a l ly on a s in g le r o l l of

papyrus n e a r ly 18 f e e t long and 13 inches h ig h . I t s t i t l e :

" D ire c tio n fo r O b ta in in g the Knowledge o f A ll Dark T h ings",

has a c e r t a in ap p ea l to th e average s tu d e n t o f m athem atics.

I t was t r u ly the " F i r s t Handbook o f M athem atics". I t gave

answ ers to type problem s, showed no s o lu t io n s excep t one,

and even co n ta in ed a ta b le o f v a lu e s to h e lp in w orking

p rob lem s. What more do we a sk o f a good handbook today? No­

th in g , excep t th a t i t be a l i t t l e le s s awkward to c a rry

around w ith u s .

4.A fte r a s h o r t In tro d u c to ry p a rag rap h , the work i s d iv id ed

In to th re e g e n e ra l g ro u p s ;

1 . Problem s in a r i th m e tic * la rg e ly devoted to u n i t f r a c t io n s *

2 . P r a c t i c a l problem s in geometry*

3* Problems o f measure®, the d iv is io n o f p ro p e r ty , and even

a fe n sim ple p ro g re ss io n s* :

v.e s h a l l co n fin e ou r in v e s t ig a t io n la rg e ly to th e f i r s t

p o r tio n on u n i t f r a c t io n s * To u n d erstan d th e d i f f i c u l t i e s

th e se f r a c t io n s p re se n ted to the S o rtb e Ahoes we must lo o k

f o r a moment a t th e E gyptian number sy stem .

I x 3 H ______ s b 7 ? 9

U n i t S l i i i n l l t ln ri i

//////

1 1 1 1I I I

m i

u u

I I I I I I 1 1 1

T e n s n n n nnn nnnn nnnnn non000

nnnnnnn

nnnnnnnnnnnnnnnnn

Hundreds e e e. cec eeeccceec

eeecce

ceeccec

ccccccec

ccccccccc

■ :

Thousands I II n? WISIIII m s

I f fI f f

T ens

T h o u s a n d s 1 )) r )))) 1 1 ) m 1111

6 .

I l l • The E gyp tian Humber System*

Of th e two forma o f E gyp tian n o ta t io n , th e h ie ro g ly p h ic

i s much e a s ie r to read than th e more r a p id ly w r i t te n h i e r ­

a t i c , o r c u r s iv e . B efo re 2000 B.C . th e num eral system was

l a r g e ly a d d it iv e * By a method o f r e v e r s in g th e sym bols,

numbers could be w r i t te n from r i g h t to l e f t o r from l e f t to

r i g h t a t w i l l . The l a t t e r method i s " e a s i e r f o r us to read as

i t co rresp o n d s to o u r own o rd e r o f w ritin g * The l&ok o f a

symbol f o r ze ro h in d e red the developm ent o f a p la c e -v a lu e

system s im i la r to ou r A rab ic system . The n e c e ssa ry symbols

w ere j / f o r on®, /^) f o r 10, (3» f o r 100,

f o r 1000,

f o r 1 ,000 ,000 and _ P — f o r 10,000,000*

The symbol f o r one m il l io n ap p ea rs to be a k n e e lin g man

h o ld in g up h is hands in asto n ish m en t* There i s no symbol fo r

one b i l l io n * The a d d i t iv e n a tu re o f E gyptian numbers i s

shown by a c h a r t on page [ S ) • Examples of s e v e r a l numbers

w r i t t e n in th e two methods ap p ea r on page ( ? ) . A l l th e se

exam ples a r e in h ie ro g ly p h ic . A few numbers w r i t te n in h i e r ­

a t i c a r e shown on page { HJ )•

Save f o r 2 /3 , a l l f r a c t io n s were w r i t te n a s u n i t f r a c ­

t io n s and th e s e , by u sa g e , were alw ays w r i t te n w ith no denom­

in a to r r e p e a te d . The f r a c t i o n 2 /7 could n o t , in E g y p tia n ,

be w r i t te n 1 /7 1/7 b u t became l / 4 1 /2 8 . When two f r a c t io n s

f o r 10,000 f o r 100#000

7 .

were w r i t te n s id e by s i d e , a d d i t io n was u n d e rs to o d . The

symbol f o r 2 /3 was^j> w h ile a l l o th e r f r a c t io n s were w r i t ­

ten w ith C o v er th e number r e p re s e n t in g th e d en o m in ato r.

O c c a s io n a lly a d o t rep la ced th e f r a c t io n s ig n •

x iIn h e r H is to ry o f M athem atics, H iss V era S anford s a y s :

* I t i s i n t e r e s t i n g to s p e c u la te w hether th e sym bolism p re ­

v en ted th e use o f f r a c t io n s w ith num erato rs o th e r than one o r

' w hether the e x c lu s iv e use o f u n i t num era to rs was th e reaso n

f o r the sym bolism ."

S an fo rd i A S h o rt H is to ry o f M athem atics. Page 103•

A few exam ples o f u n i t f r a c t io n s w i l l be found on page

e ig h t* The f r a c t i o n l / 2 .was o f te n w r i t te n . ;

8 .

S y m hols>

Ohe. h u n d r e d "th < v $ a n

one ml Ilf on

o n e fen million . O d

fy*so- t h i r d s <̂ >

one -ha I f Zi:

E g y p t i i n N u m b e r s * .

'Tfeod Lcf Tfcic/'ffi'lkt "h L e f t ,

n n n 37/Vi n n n

C Hu / 0 ‘j

' 6 § | | r , n n i" / 7 3 6

u

' l l f f i e e . n 1/ ■2 3 , 3 1 a.

i r f y / > t i a n f r a c t r o n s .

nnnnnnnnn

CCnnm a a i

IV . The Four Ifondam ental O pera tions#

A d d itio n : T his p ro c e ss "Kas e s s e n t i a l l y th e same a s o u r

p re s e n t dec im al method# The numbers w ere p laced in colum ns»

u n i t s under u n i t s , te n s under te n s , and so on# Then th e num­

b e r o f sym bols in each column were counted# F or every te n

gyribols in any one colum n, a s in g le symbol o f th e n e x t h ig h e r

column was added# F or exam ple, i f th e re w ere 12 te n s , one C-»

would he added to the hundreds and two te n s would ap p ear in

th e answ er. An i l l u s t r a t i o n o f a d d i t io n may he found on

page ( 72. )# .

S u b tra c tio n : In s u b t r a c t io n the p r in c ip le o f borrow­

in g was used in th e sane way we employ i t to d ay . T his i s

a l s o i l l u s t r a t e d on page ( / SL ) .

M u lt ip l ic a t io n : D ir e c t m u l t ip l ic a t io n o f an in te g e r

by an in te g e r was aosom pllshed by an in g en io u s d ev ice o f

d o u b lin g and red o u b lin g # Suppose we w ish to m u lt ip ly 432

by 19 . Our work c a r r ie d o u t in ou r own decim al n o ta t io n

b u t a rran g ed a c c o rd in g to th e E gyptian method would ap p ea r

a s fo llo w s :

♦I 432

*2 864

4 1728

8 3456

♦16 6912

1 0 .

The m u l t ip l i e r s I , 2 , and 16 a re checked ( ’ )• These

th re e nunfcera add up to o u r m u l t ip l i e r 19. I f we now add

th e numbers 432, 864, 6912, w hich ap p ea r o p p o s ite th o se

we have checked , t h e i r sum 8208 i s our c o r r e c t r e s u l t .

I t I s e v id e n t th a t th i s p ro ce ss would be lo n g i f o u r m u lti­

p l i e r were la rg e and we were l im ite d to d o u b lin g . To speed

up th e work th e E gyp tians used 10 a s a m u l t i p l i e r . L e t us

se e how they would m u ltip ly 569 by 78 .

I 569

*10 5690

*26 II380

*40 22760

2 1138

4 2276

*8 4552

The sum o f th e numbers o p p o s ite th e checked m u l t ip l i e r s w i l l

g iv e us the r e s u l t 4 4 ,3 8 2 . I t w i l l be n o tic e d th a t in th e

work we have taken on ly doub les o r a m u lt ip le o f te n .

D iv is io n : D iv is io n o f an In te g e r by an in te g e r was

perform ed by su c c e s s iv e m u l t ip l i c a t io n o f th e d iv i s o r u n t i l

th e d iv id en d was o b ta in e d , o r u n t i l a number had been reached

w hich was l e s s than th e d iv id en d by an amount n o t g r e a te r

th an the d iv i s o r , thus le a v in g a rem a in d er. T his p ro c e ss

i s based on th e f a c t th a t

D ividend = Q u o tien t X D iv is o r -f- Rem ainder.

1 1 .

L et us c o n s id e r th e d iv is io n o f 923 "by 24 . The work

would be a rran g ed a s fo llo w s :

I 24

.*10 240

*20 460

2 48

4 ::: 96

*8 192

Yfe see th a t th e sum of the numbers o p p o s ite th e m u l t ip l i e r s

th a t a r e checked i s 912 . T his i s l e s s than th e number we a r e

d iv id in g (923) and the d i f f e r e n c e between 923 and 912 i s I I

w hich i s l e s s than the d iv i s o r (2 4 ) . th e r e fo r e ou r r e s u l t

i s 38 , th e sum of th e checked num bers,as a q u o tie n t w ith a

rem ainder o f I I . T his method i s sometimes c a l le d M u lt ip l i ­

c a t io n o f the Second K ind. ■

With t h i s b r i e f c o n s id e ra t io n o f th e fundam ental o p e r-

a t i o n s , we a re now ready to u n d e rtak e the d iv is io n o f 2 by

th e odd in te g e r s . I t i s h e re th a t we w i l l f in d th e work o f

th e S o rib e Ahmes v e ry e le g a n t in d e ed . -

12*

£ x a m H c s .

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1 3 .

V. E gyptian U nit F ra c t io n s .

In a d d i t io n and s u b tr a c t io n o f f r a c t i o n s , we en co u n te r •

a p e c u l ia r d i f f i c u l t y . Suppose we w ish to add 1 /3 to 1 /2 1 .

The r e s u l t i s e a s i ly expressed in u n i t f r a c t io n s a s 1 /3 1 /2 1 .

Ho d i f f i c u l t y a r i s e s s in c e the denom inators a re n o t a l ik e

and th e re fo re w r i t in g them s id e by s id e v io la te s no r u l e . I f

we were to add l/B to l / 8 , we would double the 1 /8 and c a l l

the r e s u l t l / 4 . The w r i t in g o f two l ik e f r a c t io n s s id e by

s id e was s t r i c t l y a g a in s t good u sa g e . The E gyp tians were no

le s s bound to p re fe r re d m athem atica l forms than we a r e today .

We have a lre a d y seen th a t when the denom inator o f the two

f r a c t io n s was the same even number th e re was no d i f f i c u l t y .

R eal t r o u b le , however, a ro se when the denom inator was odd.

I f 1/5 were to be added to l / 5 , the r e s u l t 1 /5 1/5 was n o t

to be used and th e re was no symbol f o r 2 /5 . So h e re we s t r i k e

an am using snag le a d in g to p le n ty o f m ental g y m n astic s .

L et us b e , f o r - the moment, e a r ly E gyptians faced w ith

the problem to add l / 3 1/5 to l / 5 1 /2 1 . We have n ev er heard

o f a consnon denom inato r, l e a s t o r o th e rw ise ; we must n o t. . . . \w r i te l / 3 1/5 l / 5 1 /2 1 ; and we canno t w r i te 1 /3 2 /5 1 /2 1 .

What a re we to do ? L et us do r a th e r a n a tu r a l th in g : l e t us

th in k o f a number o r a group o f th in g s to which we may r e f e r

14«,

o u r f r a c t io n s , such th a t each f r a c t io n w i l l be a v,hole num­

b e r o f ou r g roup . Suppose v/e tak e 105 lo av es o f bread as

o u r g roup . T his id e a o f r e f e r r in g our f r a c t io n s to a group

would ap p ear to be the same th in g a s ta k in g a common denom­

i n a to r . The p ro cess has been much d iscu ssed by commentators

o f th e P apy rus. W hile R odet, H u ltsh , and P o st see i t a s

common denom inato r. D r. Chace has th i s to say o f i t : "The

id e a o f ta k in g a number, so lv in g the problem f o r th i s num­

b e r , and assum ing the r e s u l t so o b ta in ed h o ld s tru e f o r any

number, i s e x a c tly what th e boy In sch o o l i s in c lin e d to do

w ith a l l p rob lem s, and what th e a u th o r o f our handbook

does in ranch o f h i s w ork."

Chaoes The Rhlnd M athem atical P apyrus, page ( 10 )

We had chosen 105 le av es as our re fe re n c e number. How our

1 /3 1/5 has th e meaning 35 lo a v e s , 21 loaves w h ile 1 /5 I / 2 I

means 21 lo a v e s , 5 lo av es making a t o t a l o f 82 lo a v e s . V/e

oannot say we w i l l have 82/105 o f the whole a s v/o cannot

w r i te th i s f r a c t i o n . The b e s t we oan do i s to ex p ress the 82

a s an ag g reg a te o f d i f f e r e n t p a r ts o f ou r w hole, 105.

v/e must now ta k e f r a c t io n a l m u l t ip l ie r s o f 105 and seek

to m u ltip ly 105 so a s to g e t our 8 2 . We w i l l have to u se a

l i t t l e I n tu i t i o n u n le s s we want to be a lo n g tim e ab o u t i t .

As 2 /3 i s u s u a l ly the f i r s t m u l t i p l i e r , we w i l l s t a r t

w ith t h a t . We remember th a t 2 /3 was the on ly f r a c t io n f o r

which the E gyp tians had a s e p a ra te sym bol. T his may acco u n t

f o r the f a c t th a t Ahmes u ses i t so f r e q u e n t ly .

1 105

*2/3 70

Ve n o t ic e th a t 70 i s somewhere n e a r ou r 82 . L et u s now tak e

h a l f o f our l a s t l i n e above.

1 /3 35

But 35 i s more than we need to add to the 70 to g ive us 8 2 .

S ince 1/10 i s o f te n used to reduce numbers q u ic k ly , we w i l l

tak e 1/10 o f our 1 /3 .

1/30 0 1/2

Suppose we double t h i s l a s t r e s u l t .

' 1/15 7

T his i s u s e f u l . Even i f we add th i s 7 to our 70 above we w i l l

s t i l l la ck 5 to com plete our 82 . Vr’e may have to s t r u g g le a

b i t to g e t i t . L e t us go hack to th e 105 and t r y a new s t a r t

by f i r s t ta k in g 1 /1 0 .

1/10 16 1/2

(doub le) 1 /5 21

H ere i s a 5 b u t i t i s in the wrong lo c a t io n . V.'e want 5 in th e

n u m era to r, n o t in th e denom inato r. L u c k ily , a s E gyp tians we

1 6 .

u n d ers tan d r e c ip ro c a l r e l a t io n s and see th a t i f 1/5 g ives

2 1 , then 1 /21 w i l l g iv e 5• That v/as a l l we needed .

' 1 /21 5

We check th e term s needed to make up our 82 (2 /3 , 1 /1 5 , 1 /21)

and we have o u r answ er.

1 /3 1/5 -f 1/5 1/Sn = 2 /3 1/15 1 /2 1 .

We r e c a l l th a t we r e f e r r e d our o r ig in a l f r a c t io n s to 105

lo av es o f b re a d . As th i s was sim ply an a b s t r a c t id e a , we

may conclude th a t th e above r e s u l t alw ays i s t r u e .

I f we had had a c ce ss to Ahmes Handbook, l e t us see i f

we could have made our work s h o r t e r . Our tro u b le a ro s e from

2 /5 . v'e lo o k up 2 /5 ( page i j ) and f in d

2 /5 = 1 /3 1 /1 5 .

T h e re fo re ,

1 /3 1/5 + 1/5 1 /21 = -1 /3 1 /3 1/15 1 /2 1 = 2 /3 1/15 1 /2 1 .

I t i s e v id e n t th a t th i s Handbook w i l l save us much

tim e and la b o r . L e t us now see how Ames c o n s tru c te d h is

ta b le o f 2 d iv id ed by the odd num bers.

1 7 .

■ D i v i s i o n of a. by odd Numbers.

o)UL>.V*

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20 e

VI* The Table o f X.0L07 -#■ l

The need f o r the ta b le o f 2 d iv id ed by the odd numbers

was g re a t enough to repay the la b o r which Ahmes» o r some o th ­

e r w r i t e r , must have pu t in to i t s c o m p ila tio n . I t was a s

n e c e ssa ry to speed in E gyptian a r i th m e tic a s a ta b le o f log ­

a r ith m s i s to a modern s tu d e n t in trigonom etry* The ta b le

ta k e s the form o f 2 /(iJn + l),.w here 2n+ 1 r e p re s e n ts an odd

number* We have seen th a t even denom inators gave no d i f f i ­

c u l ty s in c e the form 2/2n can be reduced a t once to l / n o r

a u n i t f r a c t i o n . By th e tim e o f Ahmes, an id ea a k in to th a t

o f r a t i o had d ev e lo p ed . The number 2 was d iv id e d , s a y , in to

43 eq u a l p a r t s and w hat i s e s s e n t i a l l y the r a t i o 2 to 43, o r

tw ice 1 /43 , was ex p ressed u s in g modern sym bols, a s

2 :4 3 = 1/42 1/86 1/129 1 /301 .

In h i e r a t i c Ahmes w ro te 1/42 a s ** * " where the d o t i s the

symbol f o r f r a c t i o n , ”— i s 40, and ** i s 2 . In h ie ro g ly p h ic he

wrote^CZ^ZZ> . I t i s a c u r io u s f a c t th a t the d o t i s used in

an E n g lish copy book o f the 18th cen tu ry in oases o f * ~

meaning 1 /2 , 1 /4 .

A f r a c t io n may be s p l i t up in to u n i t f r a c t io n s in

more th an one way. Among th e form s th a t th e f r a c t io n 2/43

2 1 .

may tak e a re the fo llo w in g :

2 /4 3 =r 1 /24 1/258 1/1032

- 1/30 1/86 1/645

= 1/36 1/86 1/645 1/731 1/774

— 1/40 1/660 1/1720

= 1/42 1/86 1/129 1 /3 0 1 .

Vhy Ahmes, from th i s l i s t , chooses the l a s t form Is

h a rd ly a p p a re n t. In th e f i r s t form th e f r a c t io n 1 /24 i s n e a t­

e r the v a lu e o f 2 /43 than I s th e f i r s t f r a c t io n in any o th e r

fo rm . But t h i s seems to he no recom mendation f o r th a t form .

A lthough th e re a r e numerous ru le s f o r form ing u n i t f r a c t io n s

no one o f them a p p l ie s to a l l the ca ses Ahmes g iv e s where n

ru n s from 1 to 50 in the t a b le . This seems good ev idence

th a t the work o f Ahmes combines the r e s u l t s o f e a r l i e r com­

p u te rs each o f whom worked by a s e c r e t fo rm ula o f h i s own.

Or a g a in , i t i s p o s s ib le th a t each s o lu t io n was worked ou t

la b o r io u s ly by rep ea ted t r i a l s .

There i s a g e n e ra l r u le f o r form ing u n i t f r a c t io n s

i f th e denom inator i s th e p ro d u c t o f two f a c to r s whose sum

i s a m u lt ip le o f th e n u m era to r. That i s , where our g iven

f r a c t io n i s a /b c and b -#• c — k a , thenQ i 4 w />ere K = b+cb e b ( b + eg b K * c K

2 2 .

7u-o exam ples o f the u se o f th e g e n e r a l r u le f o l lo w .2- / 5* = -T v ^ ^ = 7% + sfo

/ 5 •=. b * c — 3*S b v C =■ eb K 3 t- S' = A K

_9l-5.1

_ _L_ y___ !—*- 3 -y 7 -5

— — _JL / y J 5

J / = b - c ~ 3*7 b t c = 5. K 3 t 7 ■=• a.K /V= 5*.

Ahmes g iv e s In h is T able 2 /15 1/10 1/30 and

2 /2 1 1/14 1 /4 2 . We s h a l l see l a t e r how th e se v a lu e s a re

based on o th e rs In th e T ab le .

B efo re we In v e s t ig a te th e p a r t i c u l a r r u le s on which

th e Table I s b u i l t , l e t us a t t a c k one d iv i s io n , say 2 by 7 ,

In a g e n e ra l way. To d e te rm in e th e double o f a r e c ip r o c a l ,

such a s th e double o f 1 /7 , i s th e same a s f in d in g the amount

by w hich th e number i t s e l f must be m u lt ip l ie d to produce 2 .

That i s ; by how auoh must we m u lt ip ly 7 to produce 2 ? How

th i s i s r e a l l y a m u l t lp l io a t lo n problem where 7 i s th e m u lti­

p lic a n d and 2 the p ro d u c t, o r a d iv is io n problem where 2 i s

th e d iv id en d and 7 the d iv i s o r . Ahmes would use f r a c t i o n a l

m u l t i p l i e r s . The common m u l t ip l i e r s , such as l / 2 , 1 /1 0 ,

were supplem ented by th e use o f a r e c ip r o c a l r e l a t i o n when

one was e v id e n t . In our d iv is io n o f 2 by 7 th e re is a good

example o f t i l l s .

2 3 .

Y;e have:

1 /2

1 /4

1 7

3 1 /2

1 1 /2 1 /4

We n o t ic e th a t t h i s l a s t number I s c lo se to 2 and r e a l i z e

t h a t we need on ly a sm all m u l t ip l i e r to com plete our w ork.

Here we s h a l l use a r e c ip ro c a l r e la t io n # taken from th e

We now see th a t i f we add 1 1/2 1 /4 to l / 4 we h a re ou r 2 .

T h e re fo re , 2 /7 = l / 4 1 /2 8 . This i s th e v a lu e Ahmes uses

in th e Table and- i t i s q u ite p o s s ib le he went th rough much

the same la b o r .

L et us tu rn to the T able on page /7 and t r y to under­

s ta n d how i t i s o rg a n iz e d . In th e f i r s t h ead in g we f in d the

word Type. T his r e f e r s to s e v e r a l g e n e ra l ways o f w orking

ou t the r e s u l t s and an example o f each w i l l be taken u p .

The second column shows the number by which 2 i s d iv id e d .

The F i r s t M u l t ip l ie r i s th a t m u l t ip l i e r which we s e l e c t be -

cause i t p roduces the C orrespond ing P roduc t ( n ex t column)

which i s v e ry n e a r ly 2 . In our example above where we found

the u n i t f r a c t io n s f o r 2 /7 , 1 /4 i s th e F i r s t M u l t ip l ie r

and 1 1 /2 1 /4 i s the C orrespond ing P ro d u c t. The l a s t two

f i r s t l in e 1

1/7

1 /14 1/2

1 /4

1

' 1 /28

24.Headings e x p la in them oelves* The Remainder i s th e amount

w hich must be added to th e C orresponding P roduct to g ive 2 ,

The Answer i s th e F i r s t M u l t ip l ie r added to th e Remainder

a f t e r i t has been m u lt ip l ie d by th e r e c ip ro c a l o f th e number

by which we a r e d iv id in g 2 . For exam ple, where 2 i s d iv id ed

by 13, we f in d from th e T able th e answ er 1 /8 2/52 1 /1 0 4 .

The f i r s t f r a c t io n i s th e F i r s t M u lt ip l ie r and the o th e r two

a r e the r e s u l t o f m u lt ip ly in g the Remainder 1 /4 1 /8 by l / l 3 ,

th e r e c ip ro c a l o f 13*

L e t us now see how each Case i s worked out* Suppose

we s e l e c t as a ty p ic a l example o f Case A th e d iv is io n o f 2

by 17* In Case A Ahmos alw ays m u l t ip l ie s h i s d iv i s o r f i r s t

by 2 /3 , Thus:

1 17

2 /3 11 1 /3

1 /3 5 2 /3

1 /6 2 1/2 1 /3

* 1 /12 1 1 /4 1 /6

V-’e have now reached a p la c e where we have reduced 17 by f r a c ­

t io n a l m u l t ip l ie r s to a q u a n t i ty l e s s th an 2 . Y.'e can a ls o

see th a t 1 /4 1 /3 i s a l l th a t i s needed to make up 2 . By how

much must 17 be m u lt ip l ie d to produce 1 /4 l / 6 ? I t i s e v i­

d e n t th a t the m u l t ip l i e r 1 /17 w i l l produce 1 from our o r ig ­

in a l 17. Then l / 3 o f 1 /17 w i l l g ive us 1 /3 and 1 /4 o f 1 /17

w i l l g iv e the 1 /4 th a t we need* So we may now w r i te our r e -

2 5 .

s u i t a s 2 /17 — 1/12 1 /51 l / 6 8 .

Case B i s v e ry s im i la r to Case A. Here Ahmes m u lt i­

p l i e s th e d iv i s o r re p e a te d ly hy th e f r a c t i o n l /2 « The d iv i ­

s io n o f 2 "by 13 v / i l l se rv e a s an i l l u s t r a t i o n .

1 13

1 /2 6 1 /2

1 /4 3 1 /4

* 1 /8 1 1 /2 1 /8

Now, 1 /8 i s the F i r s t M u l t ip l ie r and th e amount heeded to

make up 2 i s l / 4 1 /8 ( th e Remainder )•

1 /13 1 ( r e c ip ro c a l r e la t io n )

1/26 1 /2

1 1/52 1 /4

* 1 /104 1/8

R e s u lt : 2 /13 ~ 1 /8 l /5 2 1 /1 0 4 .

There i s no p la c e in th e Table where Case D i s used

a lo n e . I t i s a s s o c ia te d w ith e i t h e r o f th e f i r s t two oases

when th e m u l t ip l i e r 1 /10 oi: l / 7 le used to speed up th e work

o f red u c in g th e d iv is o r to a q u a n t i ty n e a r 2 . As an example

o f Case AD, c o n s id e r 2 d iv id e d by 25 . We have:

1 25

1/10 2 1 /2

• 1/15 1 2 /3

Here we have m u lt ip l ie d f i r s t by l / lO and then by 2 /3 . The

Remainder i s 1 /3 and th i s may be produced by u s in g the m u lti­

p l i e r 1/75 on our d iv i s o r 25 . R e s u lt: 2 /25 — l / l 5 l / 7 5 .

2 6 .

In Case BD, e i t h e r o f th e m u l t ip l ie r s l / lO o r 1 /7

may he used h u t th e on ly o th e r m u l t ip l i e r used i s 1 /2 a s

in Case B0. Example: 2 d iv id ed by 49*

1 49

1/7 7

1/14 3 1/2

'1 /2 8 1 1 /2 1 /4

The F i r s t M u l t ip l ie r i s th e re fo re 1/28 and the Remainder

i s l / 4 .

1/49 1 ( r e c ip r o c a l r e la t io n )

1/98 1 /2

' 1/196 1 /4

R e s u lt : 2 / 4 9 = 1 /28 1 /1 9 6 .

Case C may be ex p la in ed in two w ays.’

Method I . When in the p ro ce ss o f m u lt ip ly in g Ahmes a r r iv e s

a t an In teg e r* he im m ediately u se s the r e c ip ro c a l r e l a t i o n

to b r in g him to a q u a n t i ty n e a r 2 . Take th e ca se where he

i s d iv id in g 2 by 33.

1 33

2 /3 22

* 1/22 3 /2 o r 1 1 /2 ( r e c ip ro c a l)

T his le a v e s a Remainder o f l / 2 .

1 /33 1

* 1 /6 6

R e s u lt : 2 /3 3 = l /2 2 l / 6 6 .

1 /2

2 7 .

Method I I . I t i s alw ays p o s s ib le in Case C to s p l i t up our

f r a c t io n 2 / (2 n + 1) in to 2 /3 m u lt ip l ie d by a u n i t f r a c t i o n .

Thus: 2 /33 = 2 /3 ( l / l l )« A hras g iv e s a r u le f o r m u lt i­

p ly in g a u n i t f r a c t io n by 2 /3 . He s a y s . " Take thou th e

double and th e s ix t h - f o l d . " T his means to double th e denom­

in a to r and then m u lt ip ly i t by 6 . Our r e s u l t by t h i s r u le

w i l l be 1 9 /33 = 1 /22 1 /6 6 , o r th e same answ er we had by

Method I . The Rule o f 2 /3 , a s we may c a l l i t , i s the on ly

d e f i n i t e r u le o f o p e ra t io n th a t Ahmes g iv e s in th a t p a r t o f

th e papyrus which d e a ls w ith u n i t f r a c t i o n s , l e w i l l see

many in s ta n c e s o f i t s u se in l a t e r p rob lem s.

Case g i s used on ly th re e tim es in th e T ab le . Bach

example may w e ll be exam ined• They a r e ex p la in ed by D r. Chace

in th e fo llo w in g m anner.

A .B .Chace; The Rbind P ap y ru s» page ( J iS ) .

1 ) . 2 d iv id e d by 3 5 .

1 35

1/10 3 1 /2

’ 1/30 1 1 /6

1 /5 7

1 /7 5 ( r e c ip r o c a l r e la t io n )

1 /21 1 2 /3

' 1 /42 1 /2 1/3

S ince 1 1 /6 + 1 /2 1 /3 = 2 , then 2 /35 = 1/30 1 /4 2 .

2 8 .

2 .) 2 d iv id e d by 9 1 .

1 91

1/7 13

' 1/70 1 1 /3 1 /10

1 /13 7 ( r e c ip r o c a l r e la t io n )

11/130 2 /3 1/30 (T ab le o f 10th e )

Slnoo 1 1 /5 1 /1 0 1 2 /3 2/30 = 2,, th en 2 /91 = 1/70 1 /1 3 0 .

The Table o f lO th s r e f e r r e d to above w i l l be ex p la in ed in

th e n ex t c h a p te r .

3 .) 2 d iv id e d by 101.

1 101

' 1/101 1 ( r e c ip ro c a l r e la t io n )

The Rem ainder h e re i s 1 , which Ahaes o f te n ta k e s eq u a l to

1 /2 1 /3 1 /6 . Then:

' 1 /202 1/2

1 1/303 1/3

' 1 /606 1 /6

R e su lt! 2 /101 = 1/101 1/202 1/303 1 /6 0 6 .

I t ie i n t e r e s t i n g to n o t ic e th&t th e f i r s t two ex­

am ples coming under Case S may a l s o he so lv ed by a p p ly in g

the g e n e ra l r u le s ta te d above on page (a.» ) .

9 l%—r -f- -L_ - -L . < S t 7 ~ 2. JV ~ 72.S-7 • & 3 c ^ V i /X = 6

^ " 7 ^ 3 7 / 0 + _L 7® /do 7+12 * =

2S ,

Some in t e r e s t i n g f a c t s to n o te from the T able a re

th e fo llo w in g :

1 , Case A lo used 10 tim es a lo n e and w ith D 11 tim e s . Case A

i s used most f r e q u e n t ly when n i s p rim e , y e t i t i s used when

n i s n o t p rim e , f o r example when n i s 9 , A f te r 41 , AD i s

used m ostly f o r prim es and y e t i t i s used f o r 55 and 95 ,

2 , Case B is .u s e d tw ice a lo n e and w ith D 7 tim e s . A l l the

prim es n o t hand led by A o r AD o r B (e x c e p t 101) a r e handled

by BDb

3 , Case C o ccu rs 17 tim es and n i s n ev e r p rim e• T his ca se

alw ays has a s in g le f r a c t i o n a s a rem ainder and thus g iv e s

alw ays a double u n i t f r a c t i o n . Bo r e s u l t has more than fo u r

u n i t f r a c t i o n s ,

I f we lo o k a t th e l a s t column o f the T ab le , we f in d

th a t among th e 128 u n i t f r a c t io n s in th e r e s u l t s th e re a re

on ly 25 h av in g odd d en o m in a to rs . S in ce i t i s th e odd denom­

in a to r s th a t cause tro u b le in E gyp tian a r i th m e t i c , we may

w e ll a s k i f th e se oould n o t have been av o id ed . P ro b ab ly .

Take, f o r In s ta n c e , 2 /5 f o r which Ahmes g iv e s the v a lu e

l / 3 1/15 ( two odd d en o m in a to rs) . In d o u b lin g th i s r e s u l t ,

th e 2 /3 would g iv e no tro u b le a s i t was in c o n s ta n t u s e , b u t

2 /15 would tak e us a t once in to th e T able ag a in to f in d i t s

v a lu e . Bow 2 /5 can be ex p ressed as 1 /4 1 /10 1 /2 0 , a l l w ith

even d en o m in a to rs . But th re e f r a c t io n s cause more la b o r

30 o

than two, e s p e c ia l ly where they a r e used over and ov er again*

\7e m ust a l s o remember th a t eontpactness o f work was a d e s i r ­

a b le th in g on a papyrus • I t i s l i k e l y th a t Ahmes p re fe r re d

to save space r a th e r than to avo id d i f f i c u l t y in h i s opera­

tio n s* I t i s q u ite rem arkab le th a t in n o t exceed ing fo u r

u n i t f r a c t io n s in any o f h i s r e s u l t s th a t he has a s few ■odd

denom inators a s 25*

v=e have seen how th e u se o f the Table speeds up th e

work in a d d itio n * S tra n g e ly enough, m u l t ip l ic a t io n o f one

f r a c t io n by a n o th e r caused l i t t l e d i f f i c u l ty * C o n sid er th e

fo llo w in g example in w hich we a re to m u ltip ly 1 1 /2 1 /7 by

2 1 /4 .

1 1 1 /2 1 /7

*2 2 1 (2 /7 )

We re p la c e 2 /7 by i t s v a lu e in th e T ab le , 1 /4 1 /2 8 , w r i t in g

*2 3 1 /4 1 /28

We now ta k e h a l f o f th e o r ig in a l number;

1 /2 1 /2 1 /4 1 /14

*1/4 1 /4 1 /8 1 /28

Combining th e numbers o p p o s ite th e checked m u l t i p l i e r s , we

have our r e s u l t : 3 l / 4 l /2 8 1 /4 1 /8 1 /28 where we must

w r i te our answ er a s 3 1 /2 1 /8 3 /14 s in o e no f r a c t i o n may

be rep ea ted *

' S u b tra c tio n and d iv is io n o f f r a c t io n s w i l l be taken up

l a t e r a s they ap p ea r in c e r t a in p ro b lem s. Y/e w i l l now look

a t a second ta b le in ou r Handbook.

3 1 .

V II . Table o f 10th e .

Ahmes In c lu d e s a second ta b le in th e Papyrus showing

th e r e s u l t s o f the d iv is io n o f th e f i r s t n in e in te g e r s by

10 • These a re a s fo l lo w s :

1 d iv id ed by 10 g iv es 1/10

2 f f 10 i t 1/5

3 W 10 W 1/5 1/10

4 M 10 * 1 /3 1/15

5 W 10 M 1 /2

f. 11 10 ## 1 /2 1/10

7 W 10 * 2 /3 1/30

8 I f 10 11 2 /3 1/10 1/30

9 * 10 f f 2 /3 1/5 1 /30

Some of th e se v a lu e s a r e o f s p e c ia l i n t e r e s t . As v/e

would expect* 4/10 produces th e same r e s u l t a s 2 /5 w hich we

found in th e T ab le . I t would seem n a tu r a l to ex p ec t 7 /10 to

have the s im p le r form l / 2 l / 5 b u t h e re Ahmes g iv es the v a lu e

2 /3 l / 3 0 • T his ean be ex p la in ed o n ly as a " p re fe r re d fo rm "•

The E gyp tians were v e ry p a r t i a l to th e f r a c t io n 2 /3 and used

i t in most c a se s where they had more than 1 /2 . Vve must re ­

member th a t i t was th e only f r a c t io n they oould w r i te whose

3 2 .

num erato r was more than ! • The f a c t th a t th e re was a symbol

f o r 2 /3 ( ) v e ry l i k e ly inere& sed i t s use* Had 7 /10 been

1/5 1 /2 , 8/10 oould have been l / 2 l / 5 l / lO and 9 /10 could

have been 1 /2 1 /3 1 /1 5 , b o th s im p le r form s than th o se we f in d

aboveo T his ta b le was u s e fu l to Ahmss and we w i l l have occa­

s io n to r e f e r to i t in l a t e r problem so On page (2.1?) we used

th i s ta b le in th e d iv is io n o f 2 by 9 1 . Where 13 was d iv id e d

by 10 in th e th i r d l i n e , we had 1 and 3/10 b u t f o r the f r a c ­

t io n we took the e q u iv a le n t v a lu e 1 /5 l / lO from th i s ta b le

o f 10 the •

D ir e c t ly fo llo w in g the above t a b le , Ahmes g iv e s s e v e ra l

examples to show th e u se o f h is t a b l e . For example $

Problem 3 . D iv ide 6 lo a v e s among 10 men.

Answer: Baoh man r e c e iv e s 1 /2 l / l O . T his v a lu e i s taken d i­

r e c t l y from th e ta b le and i s then proved to be c o r r e c t .

P roofs M u ltip ly l / 2 l / lO by 10 .

1 1/2 1/10

'2 1 1 / 5

4 2 1 / 3 1 /15 ( 2 /5 from Table )

*8 4 2 /3 1/10 1 /30 ( 2 /15 from Table)

Adding th e numbers o p p o s ite the v a lu e s w hich a r e checked ,

1 l / 5 -r 4 2 /3 l / lO l /3 0 — 6 , the c o r r e c t answ er.

I t can be seen th a t were the p ro o f tu rned in to th e work

n eo esea ry to com plete t h i s example muoh la b o r would r e s u l t .

I t j u s t i f i e s th e Table o f 10th s a s a tim e—s a v e r .

33*

S im ila r examples showing th e u se o f t h i s ta b le o f 10the

would add l i t t l e o f i n t e r e s t were i t n o t f o r th e f a c t th a t

th ey p o in t to ev idence th a t th e re w ere perhaps o th e r ta b le s

w hich a re n o t in c o rp o ra te d in th e Ahmes P ap y ru s. L e t us lo o k

a t Problem 4 .

D iv ide 7 lo a v es among 10 men.

Answer: ( from th e ta b le o f 10th e) 2 /3 l / 3 0 .

P ro o f: M u ltip ly 2 /3 1/30 by 10, th e r e s u l t i s 7 .

1 2 /3 1/30

' 2 1 1 /3 1/15

4 2 2 /3 1 /10 1 /3 0 (2 /1 5 Table)

*8 5 1 /3 1 /5 1/15

But in th e Papyrus o p p o s ite *8 vre f in d the v a lu e 5 l / 2 2 /1 0 .

Why ? The same r e s u l t i s g iven in Problem s 5 and 6 when

2 /3 1 /10 1 /30 i s m u lt ip l ie d by 2 g iv in g 1 l / 2 l / l O . Our

f r a c t io n s 1 /3 l /S 1/15 a r e eq u a l to 1 /2 l / l O , so b o th a r e

c o r r e c t . Where th e same r e s u l t i s used th re e tim es , Dr.Chaoe

b e l ie v e s i t shows re fe re n c e to some o th e r m u l t ip l ic a t io n t a ­

b le which Ahmes d id n o t in c lu d e in h is P ap y ru s .

A .B.Chaoe; The Rhind P ap y ru s, page ( 2. ) .

P o s s ib ly i t i s on ly a c a se o f r e f e r r in g th e f r a c t io n s to 30

and n o tin g th a t 104 6 + 2 ™ 15f 3 . T his shows th a t th e f r a c ­

t io n s 15/30 o r l / 2 and 3/30 o r l / lO a re e q u iv a le n t to th e

34*

th re e f r a c t io n s we r e f e r r e d to 30* The r e s u l t g iven by Ahaes

a ls o shows a g a in h ie p re fe re n c e f o r even denom inators as w e ll

a s th e f a c t t h a t to an E gyptian two f r a c t io n s were alw ays b e t ­

t e r than th ree*

l e t u s see how much work th e u se o f even denom inators

av o id s in Problem 5 w here 8 lo av es a r e d iv id e d among 10 men*

Answer* 2 /3 l / lQ 1/30*

Check: (by A haes)

1 2 /3 1 /10 1/30

*2 1 1 /2 l / lO ( same a s Prob* 4)

4 3 l A

’8 6 1 /3 1/15 ( 2 /5 from T able)

1 1/2 1/10 + 6 1 /3 2/15 = 8

Check: ( w ith o u t u s in g even denom inators in l i n e 2)

1 2 /3 1 /10 1 /30

*2 1 1 /3 1 /5 1/15

4 2 2 /3 1 /3 1/15 1 /10 1/30

(2 /5 and 2 / l5 from Table)

*8 6 1/10 1/30 1 /5 1/15

1 1 /3 1 /5 1 /1 5 + 6 2/10 1/30 2/5 2 /1 5 = 8 .

A lthough t h i s checks* to th e E g y p tian s th e r e p e t i t i o n o f 1/5

and 1/15 was a r e a l co m p lica tio n * I t i s a p p a re n t from the

above exam ples why Ahmes p r e f e r s even denom inators *

We w i l l now tu rn to Problem s 7 th rough 15 w hich form

an i n t e r e s t i n g group in them selves *

3 5 .

V II I . Problem s 7 to 20 .

Problem s 7 to 1 5 , i f vze om it problem 8 v;hloh w i l l be

tak en in a l a t e r g roup , seem to form a group in th em se lv es .

Here Ahmes a p p a re n tly h as a p rooese in mind a lth o u g h he does

n o t make any comment on i t . The r e s u l t s o f th e se e x a m p le s

p u t in ta b u la r form i m e d ia t e ly e a to h our a t t e n t i o n . Each

example i s th e m u l t ip l ic a t io n o f c e r t a in f r a c t io n s by th e

same m u l t i p l i e r , nam ely 1 1 /2 1 /4 . L e t us work o u t P ro b . 7

a s i t ap p ea rs in Ahmes P ap y ru s .

Problem 7 . . M u ltip ly 1 /4 1/28 by 1 1 /2 1 /4 .

1 1 /4 1 /28 (a s p a r t s o f 28) th e se a re 7 1

1 /2 1 /8 1/56 " " " 3 1 /2 1 /2

1 /4 1 /16 1 /112 w " " 1 1 /2 1 /4 1 /4

sum: 14

Answer: 1 /2 , s in c e 14 i s 1 /2 o f 2 8 .

We shou ld n o t ic e th a t 1 /4 l /2 8 ~ 2 /7 (T ab le o f 2 )

and th a t 1 1 /2 1 /4 = 1 3 /4 = 7 /4 . T h ere fo re 7 /4 • 2 /7 = l / 2 .

The number 28 i s chosen a s a number o f re fe re n c e p ro b ab ly

b ecau se 28 i s th e l a r g e s t denom inator in th e problem .

The o th e r problem s in t h i s group a re a l l worked in a

s im i la r manner so we w i l l sim ply s t a t e each in tu rn and g iv e

th e answ er.

Problem 9 . M u ltip ly l / 2 l / l 4 by 1 l / 2 1 /4 .

Anawer: 4 /7 • 7 /4 = 1 .

3 6 ,

Problem 10 . ( s t r a n g e ly enough th i s i s an ex a c t d u p l ic a te

o f problem 7 ab o v e )•

Problem 1 1 . M u ltip ly 1 /7 by 1 1 /2 l / 4 .

Answers 1 /7 • 7 /4 % 1 /4 .

Problem 12. m u lt ip ly 1 /14 by 1 1 /2 1 /4 .

Answer: 1 /1 4 • 7 /4 = 1 /8 .

Problem 1 3 . A r e p e t i t i o n o f problem 12 ex cep t th a t 1 /14 i s

h e re exp ressed by th e two e q u iv a le n t f r a c t io n s 1 /16 1/112*

Problem 14 . M u ltip ly 1 /2 8 by 1 1 /2 1 /4 .

Answers 1 /28 • 7 /4 = 1 /1 6 .

Problem 1 5 . A r e p e t i t i o n o f problem 14 ex cep t th a t 1 /28 i s

h e re ex p ressed a s 1 /32 1 /2 2 4 . .

There seems no p o in t to th i s r e p e t i t i o n u n le s s Ahaes

w ished to c a l l ou r a t t e n t io n to th e se problem s p a r t i c u l a r l y .

L et us re a r ra n g e th e se r e s u l t s and ta b u la te them.

Problem . R e s u l t . M u lt ip lic a n d . M u l t ip l ie r

#9 , #10. 1 / . ... 1 /2 1 /14 o r 4 /7 7 /4

#7 . 1 /2 1 /4 1 /28 o r 2 /7 7 /4

#11 . 1 /4 (1 /8 1/56) 3 /7 7 /4

#12, #13. 1 /8 1 /16 1/112 o r 1 /14 7 /4

#14, #15. 1 /16 1 /3 2 1/224 o r 1 /28 7 /4

H ere we see a ta b le w hich i s r e a l l y com plete and which

cou ld have been u s e fu l in s e v e r a l ways i f our S c r ib e had seen

f i t to use i t . We n o t ic e th a t th e r e s u l t s form a geom etric

p ro g re s s io n . Ahmes l a t e r in the Papyrus c a l l s th i s a " la d d e r " .

37»

The m lt ip l lo a o d B a ls o f o r a a " la d d e r " <. We s h a l l see th a t

most o f th e work o f Ahmeo i s o f a p r a c t i c a l n a tu r e , h u t h e re

we have som eth ing v e ry n e a r to pure number th e o ry . I t i s

p o s s ib le th a t Ahmes was n o t e n t i r e ly in te r e s te d In p r a c t i c a l

mathematics®

In problem s 8 and 16 to 2 0 , Ahmes does a second m u lt i­

p l i c a t i o n g roup , m u lt ip ly in g th e u n i t f r a c t io n s l / 4 , 1 /2 , 1 /3 ,

1 /6 , 1 /1 2 , 1 /24 by one q u a n t i ty 1 2 /3 l / 3 o r , a s we s e e , by 2 .

These r e s u l t s , re a rran g ed and grouped , g iv e us th e two fo llo w ­

in g tables®

Problem® M u lt ip l ic a n d . M u l t ip l i e r . R e s u lt

#20 . 1/24 1 2 /3 3 /3 1 /12

#19, 1 /12 1 2 /3 1 /3 1 /6

#18 . 1 /6 1 2 /3 1 /3 1 /3

#17. 1 /3 1 2 /3 1 /3 2 /3

#8® 1 /4 1 2 /3 1 /3 1 /2

#16. 1 /2 1 2 /3 1 /3 1

Here a g a in we have " la d d e rs " a lth o u g h s h o r te r than in

th e f i r s t group® The q u a n t i ty 1 2 /3 l / 3 was a " p re fe r re d "

form f o r 2 and we f in d Ahmes u s in g i t o f te n . Thus such em­

p h a s is on what i s no more than "doub ling" i s somewhat a c ­

counted f o r . A gain , i t seems to show th a t Ahmes was som ething

more than p ra c tic a l®

3 8 .

B efore we go on to th e n ex t c h a p te r , i t w i l l be I n s t r u o - •.

t iv e to lo o k th rough Problem 17 a s i t shows a new use o f th e

T able o f 2 /{ 2 n + l) • The problem i s t h i s :

To m u lt ip ly l / 3 by 1 2 /3 1 /3 .

1 1 /3

2 /3 1 /6 1 /18

( th i s i s by ttie Rule o f 2 /3 , page 27 ) .

1 /3 1/9. /

(because 1/6 1 /18 eq u a ls 2 /9 in the T ab le , and 3 /2 of.

2 /8 eq u a ls 1 / 9 ) .

T o ta l : 2 /3 ( denom inators a p p lie d to 1 8 ) .

T his i s a r e v e r s a l o f th e u s u a l u se o f the T able and

i s w orth n o t in g . The T able o f 2 /(2 n + l ) was indeed a r e a l

Handbook o f M athem atics.

IX. Completion Problems.

Problem s 21 to 23 may be p u t u n d er a g e n e ra l head ing

" Com pletion Problem s'*. T his p ro ee ss which i s used ex ten ­

s iv e ly in l a t e r problem s i s r e a l l y an E gyp tian method o f

s u b t r a c t io n . Problem 21 w i l l se rv e a s a sim ple i l l u s t r a ­

t io n .

Problem 21 $ " I t i s s a id to th e e : Complete 2 /3 1/15 to 1 " .

Do i t thus : Apply 2 /3 1 /15 to the r e fe re n c e number 15.

2 /3 o f 15 eq u a ls 10

1 /15 o f 15 eq u a ls 1

The sum o f 10 and 1 I s 11 , which le a v e s 4 to com plete th e

u n i t o f 15 . How v/e must m u lt ip ly 15 so a s to g e t 4 .

1 15

1/10 1 1/2( Table o f 10the)

'1 /5 3

'1 /1 5 1 ( r e c ip ro c a l r e la t io n )

S in ce 3 and 1 eq u a ls 4 , l / 5 l / l 5 must be added to 2 /3 1 /15

to com plete i t to 1 .

P ro o f : Apply the f r a c t io n s 2 /3 1/15 1 /15 l / 5 to the number

30 , g iv in g 2 0 , 2 , 2 , 6 , w hich add to g ive 30. T h ere fo re our

answer ch eck s . In good E gy p tian p ro ced u re , we should have

su p p lie d th e v a lu e 1/10 1/30 from the T able f o r 2 /1 5 . The

oheolc would he han d led in th e same m anner.

4 0 .

Problem 23 In c lu d e s s e v e ra l I n te r e s t in g v a r ia t io n s

o f co m p le tio n . We a r e asked to com plete th e q u a n t i ty

1 /4 1 /8 1 /10 1 /30 1 /45 to 2 /3 . T his would n o t he too easy

in a r i th m e t ic o f to d a y . I t was more d i f f i c u l t In 1650 B .C .

Here Ahaes s e l e c t s 45 a s a number to w hich he a p p l ie s h i s

f r a c t io n s o The ch o ice o f 45 i s u n f o r tu n a te ,b u t Ahaes o f te n

chooses h is l a r g e s t denom inato r a s h is r e fe re n c e num ber. T his

seems good ev idence th a t th e E gyp tians had no n o tio n o f L .C .D .

when ta k in g a number f o r r e f e r e n c e . L e t us fo llo w Ahmes a s

he a p p l ie s h is f r a c t io n s to 4 5 .

1 /4 o f 45 g iv es 11 1 /4

1 /8 " 5 1 /S 1 /8

1/10 ' ' 4 1 /2

1/30 " 1 1 /2

1/45 " 1

S ince th e sum i s 23 1 /2 l / 4 1 /8 , 6 1 /8 i s needed to make up

30 which i s 2 /3 o f 4 5 . Here Ahaes sim ply says : 6 1 /8 i s

eq u a l to 1 /9 l /4 0 o f 45 . He p roves h ie answ er by a p p ly in g

a l l h i s f r a c t io n s to 45 and ad d in g 1 /3 to show the whole i s

o n e . He may have o b ta in ed h is 1 /9 1/40 by a method s im i la r

to th e one he used in P rob . 21 .

1 45

1/10 4 1 /2

1 /5 9

*1/9 5 ( r e c ip r o c a l r e la t io n )

1/20 2 1 /4

'1 /4 0 1 1 /8

41=

The sum o f th e numbers o p p o s ite th e eheoked item s g iv e s

6 1 /8 , and th i s i s 1 /9 l /4 0 o f 45,

v.e a r e now f a m i l ia r w ith the E gyptian methods of

a d d i t io n , m u l t ip l i c a t io n , and s u b t r a c t io n . D iv is io n d id

n o t s tan d o u t as a s e p a ra te p ro c e ss w ith th e E g y p tia n s .

I t was , in a s e n s e , a k ind o f m u l t ip l ic a t io n and u s u a l ly

invo lved com pletion a s w e l l . We w i l l need to u n d e rs tan d

d iv is io n a s i t i s used in th e s o lu t io n o f eq u a tio n s in

th e nex t c h a p te r .

X . " Aha " Problei

B efo re p ro ceed in g to th e s o lu t io n o f " Aha " o r

" C u a n tity M Problem s, i t T rill be w e ll to lo o k a t an ex­

ample in v o lv in g the d iv is io n by a f r a c t i o n a l q u a n t i ty .

Suppose o u r problem i s t h i s : A number to g e th e r w ith i t s

f i f t h eq u a ls 2 1 . F ind th e number*

In a lg e b ra we would l e t x equal th e number and form

the eq u a tio n :

x -t* l/S X = 2 1 .

6 /5 x - 2 1 .

x = 17 1 /2 .

The E g y p tia n s , s t a r t i n g w ith th e same s ta te m e n t, would

a sk : Y/hat number m u lt ip l ie d by 1 1/5 w i l l g iv e 21 ? So they

would n ex t m u lt ip ly 1 1 /5 by t h e i r p ro cess o f d o u b lin g to

produce a r e s u l t a s n e a r 21 a s p o s s ib le . Thus *

*1 1 1 /5

2 2 1 /3 1/15 ( Table v a lu e o f 2 /5 )

4 4 2 /3 1/10 1 /30 ( " 2 /15)

8 9 1 /3 1 /5 1 /15

*16 18 2 /3 1 /3 1/15 1/10 1/30 ( " 2 /5 , 2 /15)

The sum o f th e numbers o p p o s ite th e checked item s g iv e s us

20 l / 5 1 /15 l / lO 1 /3 0 . How much i s needed to make up 21 ?

T his i s now a com pletion problem .

We vrish to com plete 20 1 /5 1/15 2/10 1 /30 to 21o L et us ap­

p ly our f r a c t io n s to 3 0 .

1/5 o f 30 g ivee 6

1 /15 . w- " 2

1 /10 " " 3

1 /30 " w 1

S in ce th e eua i s 12 , 30 minus 12 , o r 18 i s n eeded . What f r a c ­

t io n a l p a r t s o f 30 w i l l g iv e 18 ? By t r i a l , wo f in d th a t

1 /2 o f 30 (15) and 1/10 o f 30 (3) w i l l g iv e us 18 . So we

see th a t 1 /2 1 /10 added to 20 l / 5 l / l 5 1 /10 l /3 0 w i l l g iv e

us th e 21 we want* But we m ast f in d what m u l t ip l i e r o f 1 1 /5

w i l l produce l / 2 1 /1 0 0

1 1 1 /5

' 1 /2 1/2 2/10

By com bining th e th r e e checked m u l t i p l i e r s , 1 , 16 , 1 /2 , we

have e u r c o r r e c t r e s u l t 17 1/2* T his i s a co m p ara tiv e ly

sim p le problem in d iv i s io n . O ften Ahaes has to employ much

in g e n u ity to c a r ry him th rough the d i f f i c u l t i e s he runs in to

in d iv i s io n .

The Aha Problem s a r e good exam ples o f eq u a tio n s so lv ed

by t r i a l . Abates u se s the p ro ce ss which was l a t e r known as

F a lse P o s i t io n h u t in a v e ry sim p le form . T h is p ro cess p layed

an im p o rtan t r o le in th e s o lu t io n o f eq u a tio n s b e fo re the

id e a o f t r a n s p o s i t io n was f irm ly e s ta b l i s h e d w hich was n o t

u n t i l the 16 th c e n tu ry A.D.

4 4 .

Let us look a t Problem 24 .

A q u a n t i ty and i t s 1 /7 added togeth er beeome 1 9 . What i s

th e q u a n t i ty ?

A lg e b ra !o a l ly , t h i s seems sim p le e n o u ^ i.

x + 1 /7 x - 19

8 /7 x — 19

x = 16 5 /8 o r 16 1 /2 1 /8 in u n i t f r a c t i o n s »

Ahmes so lv e s t h i s e q u a tio n by f a l s e p o s i t io n , assum ing 7 a s

th e answ er. We s h a l l f in d th a t o u r S c rib e u s u a l ly assumes

f o r h i s answ er th e d enom inato r, o r th e p ro d u c t o f s e v e r a l

denom inators i f he has more th an one f r a c t i o n in h i s p rob lem .

I f 7 i s to be th e q u a n t i ty , then s

once g iv es 7

1 /7 g iv e s 1

T o ta l g iv es 8 .

Then he says : As many tim es a s 8 must be m u lt ip l ie d to g iv e

1 9 , so many tim es m ust 7 be m u lt ip l ie d to g iv e th e re q u ire d

number ( th e q u a n t i ty ) .

T h e re fo re , we m u lt ip ly 8 to g iv e 19.

1 8

*2 16

1 /3 4

'1 /4 2

' 1/8 1

T o ta l : 2 l / 4 l / 8 .

S ince 8 must be m u lt ip l ie d 2 1 /4 1 /8 tim es to produce 19 ,

7 must be m u lt ip l ie d 2 1 /4 1 /8 tim es to g iv e the re q u ire d

q u a n tity * But Alamos knows th a t 2 1 /4 1 /8 tim es 7 i s th e same

r e s u l t a s 7 tim es 2 1 /4 l / 8 , so he chooses to tak e th e l a t t e r

p roduct*

'1 2 1 /4 1 /8

*2 4 1 /2 1 /4

»4 9 1 /2

T o ta l $ 16 1 /2 1 /8 , w hich I s th e anewer*

Check $ q u a n t i ty 16 l / 2 1 /8

1/7 " 2 1 /4 l / 8 (from above)

T o ta l * 1 9 .

By u s in g th i s r u le o f f a l s e p o s i t io n , th e S c r ib e so lv e s

more com plicated problem s w hich a re worded n o t u n l ik e th o se

in modern a lg e b r a .

Problem 28 $ A q u a n t i ty and i t s 2 /3 a r e added to g e th e r and

from the sum 1 /3 o f th e sum i s s u b tra c te d and 10 rem ains *

What i s th e q u a n t i ty ?

E xpressed a s an e q u a tio n , th e problem becomes :

( x + 2 /3 x ) - 1 /3 ( 2 /3 x ) = 10.

o r 2 /3 ( x + 2 /3 x ) = 1 0 .

Ahraes assum es 9 a s h ie answ er ( th e p re d u c t o f 3 x3) and

g e ts a c o r r e c t s o lu t io n in th re e s te p s s in c e 9 happens to he

th e c o r r e c t v a lu e l 2 / 3 ( 9 t 6) = 2 /3 (1 5 ) = 10*

Problem 29 hag even a more f a m i l ia r eenads

A q u a n t i ty and I t s 2 /3 a r e added to g e th e r* and l / 3 e f

th e sun i s added , th e n 2 /3 o f t h i s sum i s taken away and th e

r e s u l t i s 10 . What i s th e q u a n t i ty ?

S ta te d a s an e q u a tio n th i s beeomes %

( x f 2 /3 z ) + 1 /3 ( x > 2 / 3 x ) —(2 /3 ) (4 /3 ) ( X+-2/3 x ) = 10.

Ahmes assum es 27, th e p ro d u c t o f 3x3x3, a s h is answ er.

T h ere fo re : ( x -t 2 /3 x ) — 27-h 18 = 4 5 .

Then he has 4 5 + 1 5 = 60

2 /3 (6 0 ) = 4 0

D iffe re n c e 20*

T h ere fo re $ As m n y tim es a s 20 must be m u lt ip l ie d to g iv e

1 0 , so many tim es must 27 be m u lt ip l ie d to g iv e th e q u a n t i ty .

S in ce 1 /2 o f 20 I s 10 , th en 1 /2 o f 27 , o r 13 1 /2 , i s th e r e ­

q u ired q u a n t i ty .

Check : q u a n t i ty 13 1 /2

2 /3 " 9

sum 22 1 /2

1 /3 sum 7 1 /2

t h e i r sum 30

2 /3 * 20

d i f f e r e n c e 10 ( check)

I t seems q u i te rem arkab le th a t Ahmes i s a b le to so lv e

a problem l i k e t h i s w ith such com parative ease when much le s s

co m p lica ted p ro c e s se s caused him so much la b o r .

F

4 7 .

Hieraf(c Symbols

160 0

48*

X I, Problem 33*

In Problem s 30 to 34* th e "Aha" ta k e s on a form which

m ight be termed "D iv is io n by a F ra c t io n a l Dumber"* We h a v e .

a l re a d y co n s id e red a sim p le example o f t h i s in C h ap te r X,

Dow t/o s h a l l u n d e rta k e a more com plica ted d iv i s io n . W hile

any one o f th e p ro b le m 30 to 34 w i l l se rv e e q u a lly w e ll as

an i l l u s t r a t i o n , l e t us c o n s id e r problem 33, whloh w i l l be

found in h ie ro g ly p h ic symbols on p a g e f7 » The s ta te m e n t i s

t h i s $

A q u a n t i ty , i t s 2 / 3 . i t s 1 /2 , and i t s 1 /7 , added to­

g e th e r , become 37 , What i s th e q u a n t i ty ?

Suppose we f i r s t so lv e i t by algebra* We have $

x f 2 /3 x -f l /2 x -+ l /7 x = 37 .

97/42x - 37*

x cr 16 2 /9 7 .

T his answ er in u n i t f r a c t io n s w i l l be 16 1/56 1/679 1 /7 7 6 ,

L e t us now se e h ow Ahmes f in d s a s o lu t io n .

F i r s t : M u ltip ly l 2 /3 1 /2 1 /7 to g e t 37 .

1 1 2 /3 1 /2 1 /7

2 4 1 /3 1 /4 1 /28 (2 /7 from T able)

4 8 2 /3 1 /2 1 /14

8 18 1 / 3 1 / 7

’16 36 2 /3 1 /4 1 /28 (2 /7 from Table)

4 9 ,

T his v a lu e 56 2 /3 1 /4 1 /28 le very c lo se to th e 5?

we n eed . We now have a com pletion problem s to com plete

2 /3 1 /4 1 /2 8 to 1 , w hich added to th e 36 w i l l make 57 , We

choose 42 a s a re fe re n c e number ( the p ro d u c t o f 2x3x7, the

denom inators o f ou r o r ig in a l f r a c t io n s )»

Apply 2 /3 1 /4 3 /28 to 4 2 ,

1 42

T 2 /3 28

1 /2 21

* 2/4 10 1/2

*1/28 l 2 /2 ( rec ip ro ca l r e l a t i o n l in e two )

The sum o f 28 , 10 1 /2 , and 1 2 /2 i s 40* T h ere fo re th e re r e ­

mains 2 , o r 1 /2 1 o f 42 , needed to make up a u n i t which w i l l

produce our 37 ,

How Ahmes tu rn s back to problem 31 where he had ap p lied

1 2 /3 1 /2 1/7 ( the same m u l t ip l i e r he i s u s in g in problem 33)

to 42 and f in d s t h a t i t g iv e s 9 7 ,

onoe g iven 42

2 /3 o f 42 « 28

1 /2 o f 42 " 21

1 /7 Of 42 * 6

T o ta l s 9 7 .

I f 1 2 /3 1 /2 1 /7 g iv e s 9 7 , then by a r e c ip ro c a l r e l a t i o n ,

1 /97 w i l l g iv e 1 a s a p a r t o f 42 , To g e t 2 as a p a r t o f 42

he h as on ly to double 2 /9 7 , T h ere fo re 2 /97 w il l g iv e 2/42 o r

1/21® T his i s a l l he needed. Answer: 16 1/56 1/679 1 /7 7 6 .

5 0 .

As i s so o f te n the c a s e , the p ro o f ia much h a rd e r

than th e s o lu t io n . I t io w orth go ing th rough a s i t c o n ta in s

some o f th e most e le g a n t use o f u n i t f r a c t io n s to he found

in th e work o f Ahmcs. To prove th a t ou r r e s u l t i s c o r r e c t

we s h a l l have to m u lt ip ly 16 1 /56 1/679 1/776 by 1 2 /3 1 /2 l / 7

and show th a t th e numbers o b ta in ed combine to g iv e 37 . In th e

second l i n e o f the p ro o f w i l l be found a sp le n d id example o f

th e ru le f o r m u lt ip ly in g by 2 /3 . We r e c a l l th a t we take th e

double and th e s ix - f o ld o f th e denom inator in 2 /3 o f 1 /6 7 9 .

P ro o f : '

1 16 1 /5 6 1/679 1/776

2 /3 10 2 /3 1 /84 1/1358 3/4074 1 /1164

1 /2 8 1/112 1/1358 1/1552

1 /7 2 1 /4 1 /28 1 /392 1/4753 1/5432

T o ta l : 36 2 /3 l / 4 1 /28 p lu s a l l th e sm a lle r f r a c t io n s

whose sum must eq u a l 1 /28 1 /84 to make up 37 . We s h a l l ap p ly

a l l th e se s m a lle r f r a c t io n s to the l a r g e s t d enom inato r, a s

Ahaes d o es , to f in d t h e i r s u e . The v a lu e s when a p p lie d to

5432 ap p ea r on th e fo llo w in g page . Of th e se th e on ly one o f

s p e c ia l i n t e r e s t i s 1/392 o f 5432 equa l to 13 6 /7 . To reduce

6 /7 to u n i t f r a c t i o n s we may use th e T ab le .

6 /7 ~ 3 (2 /7 ) %= 3( 1 /4 1/28)

(2 + l ) ( 1 /4 1 /28)

1 /2 1 /4 1 /14 1 /2 8 .

We s h a l l now ap p ly ou r f r a c t io n s to 5432.

5 1 .

1/56 o f 5432 g iv es 97

1/679 M * 8

1/776 M « 7

1 /84 * " 64 2 /3

1/1358 * * 4

1/4074 * • 1 1 / 3

1/1164 # " 4 2 /3

1/112 # * 48 1 /2

1/1SS8 " 4

1/1552 3 1 /2

1/5432 W " 1

1/4753 M " 1 1 /7

1/392 • 13 3 /2 1 /4 1/14 1 /28

These v a lu e s oombla# to g iv e 258 2 /3 which m e t be proved

equa l to 1 /28 1 /8 4 . The s im p le s t way to do th i s would be

to ap p ly l /2 8 1 /84 to 5432 and then cheek the r e s u l t w ith

258 2 /3 . Thus : l /2 8 g iv e s 194

1/84 g iv e s 64 2 /3

T o ta l : 258 2 /3 ( oheek)

The S c r ib e , how ever, makes u n n ecessa ry work In p ro v in g th e(

above by now go ing back and ap p ly in g 36 2 /3 1 /4 1 /28 to 5432.

1 5432

*2/3 3621 1 /3

1 /2 2716

'1 /4 1358

*1/28 194

5 2 .

Combining the f r a c t i o n a l p a r ts o f 36 2 /3 l / 4 l /2 8

a s a p p lie d to 5432, ve g e t 5173 1 /3 . How Ahmee s&ye th a t

258 2 /3 rem ains to make up th e w ho le , o r 5432. He then

d is c o v e rs th a t 258 2 /3 eq u a ls 194 p lu s 64 2 /3 . S eeing

th a t 194 i s 1 /28 o f 5432 and th a t 64 2 /3 i s l /8 4 o f 5432,

he i s s a t i s f i e d t h a t h i s answ er 16 1 /56 2/679 1/776 i s

c o r r e c t . W ith a l i t t l e s tu d y o f the ta b le on th e p reced ­

in g page, t h i s f a c t could have been seen a t once . This

problem and p ro o f a r e a good example o f th e d i f f i c u l t i e s

Ahmee had to surm ount in h a n d lin g f r a c t io n s by the use o f

a re fe re n c e number in s te a d o f ou r more u s e fu l lo w es t com­

mon denom inato r.

A r a th e r am using v a r i a t i o n o f th i s problem in v o lv in g

d iv is io n by a f r a c t io n i s found in Problem 6 7 . Here we have

som ething o f more p r a c t i c a l v a lu e to the E g y p tian . The

problem i s s ta te d thus :

The herdsm an oame to th e s to c k - ta k in g w ith 70 c a t t l e .

The a c c o u n ta n t s a id to th e herdsm an, "Very few t r i b u t e c a t ­

t l e a r t thou b r in g in g ; p ray where a re a l l thy t r i b u t e c a t ­

t l e ? " The herdsm an r e p l ie d to h im , "What I have b ro u g h t i s

2 /3 o f l / 3 o f th e c a t t l e th a t thou h a s t oom aittod to me.

Count and thou w i l t f in d th a t I have b ro u g h t th e f u l l num­

b e r . "

S in ce 2 /3 o f l / 3 i s 2 /9 , and 2 /9 o f the whole herd i s

5 3 .

7 0 , then our herdsman must have had 315 c a t t l e com m itted

to h i s o a r s . T his i s th e r e s u l t th a t Ahmes a r r iv e s a t in

th e fo llo w in g manner.

2 /3 o f l / 3 i s 2/9 o r 1 /6 1 /18 from th e T ab le .

Get 1 by o p e ra t in g on 1 /6 1 /18 thus %

1 1 /6 1 /18

2 1 /3 1 /9

*4 2 /3 1 /6 1/18( 2 /9 from T able)

* l/2 1 /9 ( f o r 1 /6 1 /18 i s 2 /9 )

S in ce 2 /3 l / 6 1 /18 and 1 /9 combine to g iv e 1 , v;e must now

m u ltip ly 70 by the 4 1 /2 w hich produced th e 1 .

1

2

•4

* 1/2

Adding 280 and 35 ,

70

140

280

35

we have the t o t a l herd eq u a l to 315.

54*

X II . The A rith m e tic P ro g ress io n *

The u se o f th e a r i th m e t ic p ro g re ss io n seems to he th e

h ig h e s t p o in t reaohed by Ahmes in h is work on a lg eb ra* Prob­

lem 40 w i l l se rv e a s an i l l u s t r a t i o n *

Problem 40 : D iv id e 100 lo av es among 5 men In suoh a

way th a t th e s h a re s re c e iv e d s h a l l be in a r i th m e tic p ro­

g re s s io n and th a t l / 7 o f th e sum o f th e l a r g e s t 3 sh a re s

s h a l l be eq u a l to th e sum o f th e s m a lle s t two* What i s th e

d i f f e r e n c e in th e s h a re s ?

L et us see how Ahmes h an d le s t h i s . He sa y s :

Do i t th u s $ Lake th e d i f f e r e n c e 5 1 /2 , then th e a -

aoun te th a t th e 5 men re c e iv e w i l l be 1 , 6 1 /2 , 12 , 17 1 /2 ,

and 23 , g iv in g a t o t a l o f 6 0 . As many tim es a s i t i s nec­

e s sa ry to m u lt ip ly 60 to make 100, so many tim es m ust th e se

term s be m u lt ip l ie d to make th e t ru e s e r ie s *

1 60

2 /3 40

T o ta l $ 100 .

T h e re fo re , s in c e 60 i s m u lt ip l ie d by 1 2 /3 to g ive

100, each term o f th e f a l s e s e r i e s must be m u lt ip l ie d by

1 2 /3 to produce th e t ru e s e r i e s .

5 5 ,

1 becomes 1 2 /3

6 1 /2 W 10 2 /3 1 /6

12 * 20

17 1 /2 # 29 1/6

23

T o ta l $

e

100 .

38 1 /3

The s e lu t lo n o f th e problem M agee on the manner in

w hich Ahmes d ieco v ered th e d i f f e r e n c e 5 1 /2 . He may have

found I t by t r i a l and e r r o r , b u t D r. Chaoe i s o f th e op in ­

io n th a t he used th e fo llo w in g m ethod.

A .B.Chaoe; The Rbind P ap y ru s , page ( /Z ) .

Suppose the 5 term s in th e s e r i e s a r e

a b o d e

whose v a lu e s a r e 1 2 3 4 5 ,

How : a-f-b - 3 , and l / 7 ( e + d 4 e ) = 1 /7 (1 2 ) = 1 1 /2 1 /7 1 /1 4 .

C onsider a second s e r i e s w ith a common d i f f e r e n c e o f 2 .

a b c d @

1 3 5 7 9

Here % a r b = 4 , and l/7(o-<- d + e) = 3 .

How : 3 minus 1 1 / 2 1 /7 1 /1 4 = 1 1 /4 1 /2 8 ,

and 4 minus 3 — 1«

T h ere fo re : an in c re a s e o f 1 in the d i f f e r e n c e has produced

a d e c re a se o f 1 /4 1 /2 8 . To make a+ b = l /7 (o V d i- e ) , he must

m u lt ip ly th e i n c r e a s e d ) by a s many tim es a s 1 /4 l /2 8 (2 /7 )

i s co n ta in ed In 1 1 /4 l /2 8 ( 9 /7 ) , which Is 4 l / 2 t im e s .

I f to t h i s 4 1 /2 we add 1 which i s the o r ig in a l d i f ­

f e r e n c e , we g e t Ahmee1 aaeuewd d if fe re n c e * This i s the th eo ry

o f D r. Ghaoe a s to how Ahmes a r r iv e d a t th e 5 l / 2 .

A n o th er, and e n t i r e l y d i f f e r e n t th e o ry , i s g iv en "by

C a jo r i who b a se s h i s work on th a t o f C a n to r .

56,

P ie r ia n C a jo r i ; H is to ry o f t h e m t i e s , page 23 .

M oritz C a s te r ; V orleesngen fther G esohiohte d e r M athem atik,

V o l. I p a g e 31*

C a jo r i b e l ie v e s Ahmes s t a r t s w ith th i s s e r i e s t

a a -d a - 2d a - 3d a -4 d .

Then l /7 ( 3 a -3d ) » 2a - 7d

3a -3d 14a -49d

46d 11a

46d -44d 11a »44d

2d 11a -44d

d — l l / 2 ( a -4 d )

d 5 1 /2 o f th e s m a lle s t term*

I f Ahmee assum es 1 to be h ie s m a lle s t te rm , then h ie d i f ­

fe re n c e i s 5 1 /2 .

While t h i s i s undoub ted ly an in g en io u s s o lu t io n , i t

does n o t seem to be a s t y p ic a l ly E gyp tian a s th a t su g g ested

by D r, Ghaoe.

5 7 .

B ib llo

I The Hhlnd M athem atical P ap y ru s, Volumes I and I I .

M athem atical A s so c ia tio n o f A m erica. 1927 - 1929.

Arnold Buffum Chaco, London.

I I H is to ry o f M athem atics. P ie r ia n C a jo r i . 1929.

The MacMillan Company, Hew Y ork.

I I I H is to ry o f M athem atics. \7. W. R. B a l l 1915.

S ix th E d i t io n . MacMillan Company L im ited . London.

IV H is to ry o f M athem atics, Volumes I and I I . 1923.

D. B. S m ith . Ginn and Company. Hew Y ork.

V S h o rt H is to ry o f M athem atics. V era S an fo rd . 1930.

Houghton M if f l in Company. Hew Y ork.

VI American M athem atical M onthly, Volume 38 (1931)

Page 194. Theorems R e la t in g to th e Hhlnd P ap y ru s .

G. A. M il le r .

V II E noyelopoed ia B rit& n n io a , 14th E d i t io n , Volume I I .

A r i th m e tic .

V II I P h ilo so p h y o f A r i th m e tic . Eduard B rooks. 1901.

The Hormal P u b lish in g Company. P h i la d e lp h ia .

E 1 7 1 1 . I=i3b - 7 3 C3

a 3 9 0 0 i 0 0 1 2 8 0 7 3 7 b

' '9 3 4 ,9 3

£ 9 7 9 /