160 A History ofMathematics
- Let us translate Euclid’s statement into algebra. You start with a quantitya. You subtract from
thisk 1 awherek 1 >^12. Then you subtract againk 2 (a−k 1 a), wherek 2 >^12 , and so on. Euclid
claims that by repeating the subtractionsa−k 1 a−k 2 (a−k 1 a)−···you will arrive at a quantity
less than any assigned quantityb.
In fact, afternsubtractions, you havea( 1 −k 1 )(···)( 1 −kn). If eachkiequals^12 , you get
( 1 / 2 n)a, which tends to 0; even more so if eachki>^12.
However, the condition is actually too restrictive. By a ‘classical’ theorem, the product
a( 1 −k 1 )(···)( 1 −kn)tends to 0 if and only if the sumk 1 +k 2 +···is divergent. - Guess 12 (a multiple of 3 and 4) for the height; then underground it is 4+3, so above ground
it is 5, which is wrong, as we need 30. So multiply the guess by 30/ 5 =6, giving 72. This is
right (24+ 18 =42 underground, leaving 30 above).
Guess 4; then 4+ 4 + 2 + 1 =11. We want it to be 99 (one denaro short of 100). So multiply
the guess (4) by 9, giving 36.
You would need to convert the 2 years 7 months and 15 days to years (assuming the interest
is annual—10 per cent per month seems extortionate, but such things did happen). Suppose a
month is 30 days, and that there are 12 in a year, to simplify. This gives 2^1524 × 10010 =^2180 for
the multiplier; you then have to multiply this by the principal. The calculation looks appalling;
I do not know how a medieval student was supposed to do it.
(b) In a year—12 months—the interest is 12 pence in 1 pound, or 24012 ×100 per cent. This
is clearly 5 per cent. - We haveu−v=4 anduv=1 (divide 3 by 3 and cube). This gives(u−v)^2 =16 and so
(u+v)^2 =20. Hence,u+v= 2
√
5 andu= 2 +√
5,v=− 2 +√
5.
We now take cube roots and subtract:x=^3√
2 +
√
5 −^3
√
2 −
√
5.
(b) The point is of course that the differenceis1; you can check it first of all on a calculator.
Bothuandvare (if you think in terms of algebraic number theory!) exact cubes of expressions
a+b√
5; takea=±^12 ,b=^12. Hence, the difference of the cube roots is( 1 +√
5 )/ 2 −
(− 1 +
√
5 )/ 2 =1.
(c) (1) Suppose the equation isx^3 +cx= d. Letx =^3√
u−^3√
v. Thenx^3 = u−v−
3 (^3√
u·^3√
v)(^3√
u−^3√
v), using the formula for(a−b)^3 and rearranging. Ifuv=(c/ 3 )^3 , then√ (^3) u·√ (^3) v=c/3. Sox (^3) =u−v−cx. Butu−v=d,soxdoes satisfy the equation.
(2) It is enough to construct a geometrical decomposition which verifies the formulaa^3 +
3 ab^2 = 3 a^2 b+b^3 +(a−b)^3. From there you can ‘geometrize’ the argument in part (1). I leave
the details to you.
- This is not so pleasant, if no harder. We haveu−v=500,uv= 271 .So(u+v)^2 =250,000 274 ,
andu=^12 (
√
250,000 274 )+ 500 ),v=^12 (√
250,000 274 )− 500 ). Now writex=^3√
u−^3√
v
again. (This could be worked out with a calculator, if you have the patience; answer about
7.895.)- You are givenb=x−yandd=x^3 −y^3. Viète points out that ifu=x+y, thenu+b= 2 xand
u−b= 2 y(obvious). Now( 2 x)^3 −( 2 y)^3 =(u+b)^3 −(u−b)^3 = 6 u^2 b+ 2 b^3 by expanding
the brackets and subtracting. But we know that( 2 x)^3 −( 2 y)^3 = 8 (x^3 −y^3 )= 8 d.
We can therefore deduceu, since (by the above), 6u^2 b+ 2 b^3 = 8 d,sou^2 =( 4 d−b^3 )/ 3 b
(the displayed equation in the extract). Now we knowx−y=bandx+y=u, and soxandy.