130 A History ofMathematics
Solutions to exercises
- Clearly ‘Is’ denotes an odd place (counting from the end); and the point is that your starting
point is to look at the number up to the last odd place (e.g. 5 for 576, or 13 for 1369). The
whole part of the root of this number—which is a single figure—gives you the first figure of
your answer.
You now have 2 as the root of 4, the largest square less than 5. You subtract its square (4)
from the 5, and drop down the rest giving 176. You now double the two (4 again) and put it
under the 7, so it is effectively 40. Al-Uql ̄idis ̄i’s expression means that you are looking for anx
such that 40xadded tox^2 ‘exhausts’ the 176 you have left. In other words,( 40 +x)x=176.
In fact this is satisfied byx=4.
The method is simply using the usual formula for(a+b)^2 , witha=20 andb=the
unknownx. If this is slightly confusing, try some other three- and four-figure squares. Then
see how it generalizes to larger ones (it does). - As in Chapter 2, let us use algebra to simplify. Call the length AB ‘a’. Then BC=a,BD=a/2,
and so CD=(a/ 2 )
√
- Hence by construction, DE=CD=(a/ 2 )
√
- So AE=a(( 1 +
√
5 )/ 2 ).
This is the right length for the ‘golden section’ construction of Chapter 2; the triangle ABG
whose sides are in the ratio 1 :( 1 +√
5 )/2:( 1 +
√
5 )/2 has angles 36◦,72◦,72◦, and the
construction proceeds as required.- Al-Khw ̄arizm ̄i gave six equation models, and these were always followed by his successors
through the medieval and early modern period. There are three ‘trivial’ ones: roots equal to
numbers, roots equal to squares, and squares equal to numbers; and three ‘serious’ ones, roots
and squares equal numbers, roots and numbers equal squares, and squares and numbers
equal roots. (The point is that all coefficients must be positive.) Again, because there must be
a positivesolution, the form (which we would think worth including) ‘squares and roots and
numbers equals zero’ (e.g.x^2 + 3 x+ 2 =0) is excluded. - AD is equal to AB+BD, ora+b. ‘The rectangle AD by DB’ in Euclid’s language means the area
of a rectangle whose sides are equal to AD and DB, so it is the product(a+b)b. Since C is the
midpoint of AB, CB=a/2; while CD=CB+BD=(a/ 2 )+b. From this the statement follows. - (a) Al-Khw ̄arizm ̄i’s method starts by halving the roots—result 1. Square this, result 1; add
to 1 (the ‘numbers’), result 2. Now our problem is to take the square root. If we can (call the
result
√
2 as usual), subtract half the roots, that is, 1, and get the answer√
2 −1. (b) The line
BE has length 2; and we must construct AB so that the square on AB and the rectangle AB·BE
are equal to 1. We divide BE in half at F, so BF=1. Euclid II.6 says that EA·AB together
with the square on BF (i.e. 1+ 1 =2) equals the square on AF. So we construct a square of
area 2 (compare theMeno!); its side is AF. Subtract BF (i.e. 1), and you have the result AB. This
depends on the fact that you can construct AF, whose length is√
2, geometrically without
saying what the length is.- Call the ‘amount’x. If 10 is added to an amount (10+x), and the amount (i.e. the sum) is
multiplied by the root of 5, we have( 10 +x)
√
- This is said to be equal to the product of the
amount (this word is being over-used) with itself; that is, tox^2 .So,( 10 +x)
√
5 =x^2 as stated.
By the usual rules for quadratics: write it asx^2 −√
5 x− 10√
5 =0. Then the solution isx=