Last Time | Next Time |
We began with Egyptian multiplication, which is based on the fact that the Egyptians didn't mind doubling things: that was easy for them; and that's the key to Egyptian multiplication.
Successive doublings means powers of two, so this reduces to the Fraudini trick! That's the good news.
We double the larger of the two, generally, and use the binary factorization of the other to choose which rows to include in the final answer (indicated by the asterix):
23=16+4+2+1 |
|
672+168+84+42 = 966 |
Now add up those rows marked with an asterix (*), and you'll get the answer (966 = 23*42).
This works because of the distributive rule: \[ 23*42 = (16+4+2+1)*42 = 16*42+4*42+2*42+42 = 672+168+84+42 = 966 \]
1 | 23 | |
2 | ||
4 | ||
8 | ||
16 | ||
32 | ||
64 | ||
128 | Too big! |
1 | 321 | |
2 | ||
4 | ||
8 | ||
16 | ||
32 | ||
64 | ||
128 | Too big! |
Add up those rows marked with an asterix (*), and you'll get the answer (35952).
We can think of division as just using the multiplication table "backwards". So if we write the quotient (which is what we're looking for) as \[ \frac{dividend}{divisor}= quotient \] We can think of this as a product instead: \[ divisor*quotient = dividend \] For the product we take one of the parts of the product (the divisor, say), and double it in the center column. Making up the other part of the product (the quotient) with powers of two, we then find the product (the dividend) by adding up the corresponding numbers from the center.
In the division problem we know the dividend, so we reverse the process: we find numbers in the center that sum to the dividend, and then add up the corresponding powers of two on the left to give the quotient, which is what we're after.
Example: Let's try this one from last time (rather than 23*42(=966), suppose you want $\frac{966}{23}$(=42)).
1 | 23 | |
2 | ||
4 | ||
8 | ||
16 | ||
32 | ||
64 | Too big! |
1 | 8 | |
2 | 16 | |
4 | 32 | * |
So the answer is 4 (how do we get 4?)
1 | 8 | * |
2 | 16 | |
4 | 32 | * |
So the answer is 5 (how do we get 5?)
Let's look at an example: divide 35 by 8.
In a way we turn it into a multiplication problem: what times 8 equals 35? So we know the 8, and use it to "double" -- but then to "halve", when 8 won't go evenly into 35:
1 | 8 | |
2 | 16 | |
4 | 32 | * |
1/2 | 4 | |
1/4 | 2 | * |
1/8 | 1 | * |
So the answer is 4+1/4+1/8
But they didn't restrict themselves to "halving", as our next example shows. Divide 6 by 7:
1 | 7 | |
1/2 | 3+1/2 | * |
1/4 | 1+1/2+1/4 | * |
1/7 | 1 | |
1/14 | 1/2 | * |
1/28 | 1/4 | * |
So the answer is 1/2+1/4+1/14+1/28 (we usually order them from largest to smallest).
Notice that the Egyptians didn't use decimals -- you shouldn't either!
Why did Egyptians do things this way? (an example division problem, 3/5)
Dominic Olivastro, "Ancient Puzzles", suggests a third reason why this use of unary fractions is good. Consider the problem Ahmes poses of dividing 3 loaves of bread between 5 people. We would answer "each person gets 3/5-ths of a loaf". If we implemented our solution, we might then cut 2 loaves into 3/5 | 2/5 pieces, with bread for 3 people; then cut one of the smaller pieces in half, giving the other two people 2/5 + 1/5 pieces. Mathematically acceptable, but try this with kids and they will insist that it is not an even division. Some have larger pieces, some have smaller.
Ahmes would calculate 3/5 as : 3/5 = ()3 + ()5 + ()15 [ = 1/3 + 1/5 + 1/15 ] Now cut one loaf into fifths, cut two more into thirds, then take one of the 1/3-rd pieces and cut it into 5-ths (for the 1/15-th pieces), and you can now distribute everyone's 3/5-ths share in a way that _looks_ equal, since they will have exactly the same size pieces. (And no, I don't want to argue about the crust.)
so we look up $\frac{2}{7}$, and find that $\frac{2}{7}=\frac{1}{4}+\frac{1}{28}$. Therefore,
\[ \displaystyle \frac{6}{7}= 2*(\frac{1}{4}+\frac{1}{28})+\frac{1}{4}+\frac{1}{28}= \frac{1}{2}+\frac{1}{14}+\frac{1}{4}+\frac{1}{28} \]
(Start with halves, and then what?)