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\centerline{MONT 106Q -- Mathematical Thinking}
\centerline{Problem Set 2 -- Egyptian Mathematics -- Selected Solutiosn}
\centerline{September 27, 2016}
\bigskip
\noindent
I.  {\it Egyptian multplication and division.}
\item{C)} Divide by the Egyptian method:  $167 \div 18$.
\item{} {\it Solution:}  We want to calculate with $18$ to yield $167$, so we start by doubling:
$$\eqalign{1 \times 18 &= 18\cr
           2 \times 18 &= 36\cr
           4 \times 18 &= 72\cr
           8 \times 18 &= 144\cr}$$
Using these we can make $9 \times 18 = 144 + 18 = 162$.  The left over part (remainder) is
$167 - 162 = 5$.   Now to get this fractional part of the 
quotient in an easy way, recall that the Egyptians did allow the non-unit fraction $2/3$.  This
would be useful here, for instance if we start by multiplying by $2/3$, then multiply by $1/2$ once
and $2/3$ again:
$$\eqalign{2/3 \times 18 &= 12\cr
            1/3 \times 18 &= 6\cr
            1/6 \times 18 &= 3\cr
            1/9 \times 18 &= 2\cr}$$
So we get $167 \div 18 = 9 + {1\over 6} + {1\over 9}$.  (This was not really an exact science
for the Egyptians -- it seems as though they were happy if they just found something that worked!)
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\noindent
II.  {\it Egyptian fractions.} A part of the table of conversions of $2/n$ 
into unit fractions from the Rhind papyrus is given in 
the handout from class on Wednesday, September 21.
Each row gives an expansion
$${2\over n} = {1\over a} + {1\over b} + {1\over c} + {1\over d}$$
for one of the odd integers $3 \le n \le 101$.  
(if there is no entry for $c$ and/or $d$, then those terms 
are not present).
\item{A)}  Verify that the rows for $n = 13$ and $n = 29$ are
\item{} {\it Solution:}  What I meant for you to do on this one was something like this:
From the table ${2\over 13} = {1\over 8} + {1\over 52} + {1\over 104}$.  But putting the fractions
over a common denominator and adding,
$${1\over 8} + {1\over 52} + {1\over 104} = {13 + 2 + 1\over 104} = {2\cdot 8\over 13\cdot 8} = {2\over 13}.$$
This verifies that the entry is {\it exactly} correct.  With a finite decimal approximation, the most
you can say is that the results are close.  (The decimal expansion for $2/13$ is infinite and repeating.)
The other one is similar.
\item{B)}  Use the table to find
a decomposition of ${4\over 27}$ into unit fractions. 
\item{} {\it Solution:}  From the table
$${2\over 27} = {1\over 18} + {1\over 54}$$
so doubling and canceling factors of $2$:
$${4\over 27} = {2\over 18} + {2 \over 54} = {1\over 9} + {1\over 27}.$$ 
(There are other correct ways to do this also.)
\item{C)}  Show that for any positive integer $n$
$${2\over n} = {1\over n} + {1\over 2n} + {1\over 3n} + {1 \over 6n}.$$
Yet this expansion appears only once in the table from the Rhind
papyrus.  For which $n$ did the compiler of the table fall back
on this form?  Can you see why they might have preferred to avoid using this if
possible?
\item{} {\it Solution:}  The formula is valid for general $n$ because if we find the common denominator
$6n$, then 
$${1\over n} + {1\over 2n} + {1\over 3n} + {1 \over 6n} = {6 + 3 + 2 + 1\over 6n} = {12 \over 6n} = {2\over n}.$$
It's not enough to check that this works for one (or any finite number of) specific value(s) of $n$ because
the question is asking for a general proof that the formula {\it always works}.  This form appears in the 
Rhind papyrus $2/n$ table only for $n = 101$.  It is likely that the Egyptians prefered to use sums with
2 or 3 terms when they could find them to simplify calculations.   
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\noindent
IV. {\it An Egyptian approximation to $\pi$.}  We take $\pi$
to be the ratio between the area of a circle and the square
of the radius:  $\pi = {A\over r^2}$.
\item{A)}  In the Rhind papyrus, the area of a circle with
diameter $d$ is given as 
$$A_{circ} = \left({8\over 9} d\right)^2.$$
(This is not exactly correct, but yields decent approximate results.)  
What is the implied approximation to the value of $\pi$ here?
\item{} {\it Solution:}  The Egyptian formula would be correct if
$$\left({8d\over 9}\right)^2 = \pi \left({d\over 2}\right)^2$$
which would say $\pi = {4 \cdot 64\over 81} \doteq 3.16$.  That is quite
close to the exact value of $3.14159\cdots$.  
\item{B)}  Here's one way this may have been derived. Take
a square of side $9$ units, divide the edges into 
three equal pieces, and cut off four corner triangles 
with side length 3 to make an octagon.  The octagon 
is pretty close in area to the circle inscribed in
the original square.  Explain why the octagon has area
$63 \doteq 8^2$, and then explain how the approximation
$A_{circ} = \left({8\over 9} d\right)^2$ would result.  
\item{} {\it Solution/Comment:}  If we inscribe a circle of radius $d$ in a square of
side $d$, the area is very close to the area of the octagon obtained by cutting off
the four corners of the square as described.  This octagon has exact area  ${7 d^2\over 9}$.
But then 
$${7 d^2 \over 9} = {63 d^2\over 81} \doteq {64 d^2\over 81} = \left({8d \over 9}\right)^2.$$
This is thought to be how the Egyptians derived this approximate area formula for the circle.  
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