Egyptian division prolonged---fractions

Last time, we used Egyptian division to find a whole number quotient and a remainder for the division `1543 -: 21`, in the process formed the table

`\qquad\qquad\qquad[(1543, x),(21, 1),(168,8),(1344,64), (1533,73)]`.

To calculate the fractional part in decimal form, we begin by subtracting the last row from the first, and append the result at the bottom:

`\qquad\qquad\qquad[(1543, x),(21, 1),(168,8),(1344,64), (1533,73), (10, 10/21)]`.

In practice, of course, don't actually subtract, but just write the remainder we have already found, and that same remainder divided by the divisor in fractional form in the next column.

Now we take the divisor row, `[21, 1]`, halve it to get `[10.5, 0.5]`, and append at the bottom :

`\qquad\qquad\qquad[(1543, x),(21, 1),(168,8),(1344,64), (1533,73), (10, 10/21),(10.5, 0.5)]`.

We now continue, halving the bottom row and appending, halving the bottom row and appending, until the second number in the bottom row is comparable to the accuracy we require.  Suppose we want the fractional part of the quotient to be accurate to two decimal places, then :

`\qquad\qquad\qquad[(1543, x),(21, 1),(168,8),(1344,64), (1533,73), (10, 10/21),(10.5, 0.5), (5.25, 0.25), (2.625, 0.125), (1.3125,0.0625), (0.65625,0.03125), (0.328125,0.015625), (0.1640625, 0.0078125)]`.

We now make a sum as close as we can to the `10` in the target row `[10, 10/21]`, by adding together some of the rows below it.  `10.5` is larger than `10`, so we ignore that row.  `5.25` is smaller, so that row get used.  `10 - 5.25 = 4.75` left to account for. This is larger than `2.625`, so we use that row too, leaving `4.75 - 2.625 = 2.125` to account for.  Continuing, `2.125 - 1.3125 = 0.8125`; `0.8125 - 0.65625 = 0.15625`; skip `0.328125`.  For the last entry, we perform the subtraction, even though the result is negative, `0.15625 - 0.1640625 = -0.0078125`.  The reason is that it leaves an error `-0.0078125` that is smaller in magnitude than the error we would have without the subtraction, i.e. `0.1640625`. This gives us more accuracy than we expected.

Deleting (or striking out) the rows to be ignored, we get :

`\qquad\qquad\qquad [(1543, x),(21, 1),(168,8),(1344,64), (1533,73), (10, 10/21),(5.25, 0.25), (2.625, 0.125), (1.3125,0.0625), (0.65625,0.03125), (0.1640625, 0.0078125)]`.

Adding the contributing rows, we get

`5.25+2.625+1.3125+0.65625+0.1640625 = 10.0078125`,

(or, easier, `10 - -0.0078125 = 10.0078125`) and

`0.25+0.125+0.0625+0.03125+0.0078125 = 0.4765625`,

so the table finally becomes

`\qquad\qquad\qquad [(1543, x),(21, 1),(168,8),(1344,64), (1533,73), (10, 10/21),(5.25, 0.25), (2.625, 0.125), (1.3125,0.0625), (0.65625,0.03125), (0.1640625, 0.0078125),(10.0078125, 0.4765625)]`.

Since `10.0078125` is within a tenth of a percent of `10`, we must have that `0.4765625` is within a tenth of a percent of `10/21`, i.e. `10/21 \approx 0.477`, this compares well with the accurate answer of `0.476190...`  i.e. we have almost three decimal places of accuracy.

So `1543/71 \approx 73.477.`

Although it makes relatively few demands in terms of times tables, and this part requires only that we know how to halve decimal quantities, one has to put in considerable labor for each decimal place.

The method is much better adapted to binary fractions than to decimals.  These are sometimes useful, e.g. when one is working to sixteenths or thirty-seconds of an inch, say.  In binary fractions, the final table looks like this,

`\qquad\qquad\qquad [(1543, x),(21, 1),(168,8),(1344,64), (1533,73), (10, 10/21),(5 1/4, 1/4), (2 5/8, 1/8), (1 5/16, 1/16), (21/32,1/32), (21/64, 1/64),(21/128, 1/128),(10 1/128, 61/128)]`,

so that `1543/71 \approx 73 61/128.`