A history of Greek mathematics - Wilbourhall.org

•
544 COMMENTATORS AND BYZANTINES
E
X
*
1 1
J^
1^ 3
176 ~" /16 + 3 . 11\ + /16 + 3. 11\ 1~ = 77 + 112'
U (—7^-)
16 (-^)3
3 1 1 11
and again — = H =
6
112 /16 + 2.7\ /16 + 2.7\1 70 80
7 (-3—)
16 (— r~)i
a 1 1
1
/3) ==
.
'
cdf cd + df
f cd + df
a
a
Example.
28 28 1 111
+
1320 10.12.11" 120 + 132 120+132 90 99
'
28~
'
28
The object is, of course, to choose the factors of the denominator,
and the multiplier m in (2), in such a way as to make
the two denominators on the right-hand side integral.
When the fraction has to be decomposed into a sum of three
or more submultiples, we take out an obvious submultiple
first, then if necessary a second, until one of the formulae
will separate what remains into two submultiples. Or we
take out a part which is
not a submultiple but which can be
divided into two submultiples by one of the formulae.
For example, to decompose -^j^. The factors of 61 6 are 8.7 7
or 7 . 88. lake out gg, and ^T e
= gg 6 T6 - = 8 8 7 7 = 88 77 11 5
and T
2
T = eV A by formula (1), so that ^ = -£? TV is A
Take ^V The factors of 6460 are 85.76 or 95.68. Take
2 3 9
out q-
1^, and 6 4 6 o
= st ihtwo
• Again take out 93-, and we have
ws 9*5 eif or •
is 9V is
r^ie actual problem here is to find
which latter expression reduces to
3^3 rd of HJ^to eV,
20 • ^39.
The sort of problems solved in the book are (1) the division
of a number into parts in the proportion of certain given
numbers, (2) the solution of simple equations such as this:
From a certain treasure we take away j^th, then from the
remainder y7 th of that remainder, and we find 150 units left;
what was the treasure? \ =.R.