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More Liber Abaci

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A certain man had one pair of rabbits together in a certain enclosed place, and one wishes to know how many are created from the pair in one year when it is the nature of them in a single month to bear another pair, and in the second month those born to bear also.

Because the abovewritten pair in the first month bore, you will double it; there will be two pairs in one month. One of these, namely the first, bears in the second month, and thus there are in the second month 3 pairs; of these in on month two are pregnant, and in the third month 2 pairs of rabbits are born, and thus there are 5 pairs in the month...

You can indeed see in the margin how we operated, namely that we added the first number to the second, namely the 1 to the 2, and the second to the third, and the third to the fourth, and the fourth to the fifth, and thus one after another until we added the tenth to the eleventh, namely the 144 to the 233, and we had the abovewritten sum of rabbits, namely 377, and thus you can in order find it for an unending number of months.

Much of Liber Abaci is incredibly dull, with pages and pages of stuff of the form "Then because three from five is two, you put a two under the three, and then because eight is less than six you take the four that is next to the six and make forty-six, and take eight from forty-six and that is thirty-eight, so you put the eight from the thirty-eight under the six...". And oh, gosh, it's just deadly. I had hoped I might learn something interesting about the way people did arithmetic in 1202, but it turns out it's almost exactly the same as the way we do it today.

But there's some fun stuff too. In the section just before the one about the famous rabbits, he presents (without proof) the formula 2^n-1 · (2ⁿ -1) for perfect numbers. Euler proved in the 18th century that all even perfect numbers have this form. It's still unknown whether there are any odd perfect numbers.

Elsewhere, Leonardo considers a problem in which seven men are going to Rome, and each has seven sacks, each of which contains seven loaves of bread, each of which is pierced with seven knives, each of which has seven scabbards, and asks for the total amount of stuff going to Rome.

Negative numbers weren't widely used in Europe until the 16th century, but Liber Abaci does consider several problems whose solution requires the use of negative numbers, and Leonardo seems to fully appreciate their behavior.

Some sources say that Leonardo was only able to understand negative numbers as a financial loss; for example Dr. Math says:

Fibonacci, about 1200, allowed negative solutions in financial problems where they could be interpreted as a loss rather than a gain.

This, however, is untrue. Understanding a negative number as a loss; that is, as a relative decrease from one value to another over time, is a much less subtle idea than to understand a negative number as an absolute quantity in itself, and it is in the latter way that Leonardo seems to have understood negative numbers.

In Liber Abaci, Leonardo considers the solution of the following system of simultaneous equations:

A + P = 2(B + C)
B + P = 3(C + D)
C + P = 4(D + A)
D + P = 5(A + B)

(Note that although there are only four equations for the five unknowns, the four equations do determine the relative proportions of the five unknowns, and so the problem makes sense because all the solutions are equivalent under a change of units.)

Leonardo presents the problem as follows:

Also there are four men; the first with the purse has double the second and third, the second with the purse has triple the third and fourth; the third with the purse has quadruple the fourth and first. The fourth similarly with the purse has quintuple the first and second;

and then asserts (correctly) that the problem cannot be solved with positive numbers only:

this problem is not solvable unless it is conceded that the first man can have a debit,

and then presents the solution:

and thus in smallest numbers the second has 4, the third 1, the fourth 4, and the purse 11, and the debit of the first man is 1;

That is, the solution has B=4, C=1, D=4, P=11, and A= -1.

Leonardo also demonstrates understanding of how negative numbers participate in arithmetic operations:

and thus the first with the purse has 10, namely double the second and third;

That is, -1 + 11 = 2 · (1 + 4);

also the second with the purse has 15, namely triple the third and fourth; and the third with the purse has quadruple the fourth and the first, because if from the 4 that the fourth man has is subtracted the debit of the first, then there will remain 3, and this many is said to be had between the fourth and first men.

The explanation of the problem goes on at considerable length, at least two full pages in the original, including such observations as:

Therefore the second's denari and the fourth's denari are the sum of the denari of the four men; this is inconsistent unless one of the others, namely the first or third has a debit which will be equal to the capital of the other, because their capital is added to the second and fourth's denari; and from this sum is subtracted the debit of the other, undoubtedly there will remain the sum of the second and fourth's denari, that is the sum of the denari of the four men."

That is, he reasons that A + B + C + D = B + D, and so therefore A = -C.

Quotations are from Fibonacci's Liber Abaci: A Translation into Modern English of Leonardo Pisano's Book of Calculation, L. E. Sigler. Springer, 2002. pp. 484-486.

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