Bombelli: Algebra

In 1572 Rafael Bombelli published the first three volumes of his famous book Algebra (the intended further two volumes were not completed before his death). We quote from the text where Bombelli is using fractions to approximate to square roots. We insert in square brackets the equations that Bombelli is working with in modern notation and some comments. He begins with preamble, then gives his method without justification, finally giving a detailed justification of his method:-

Many methods of forming fractions have been given in the works of other authors; the one attacking and accusing the other without due cause (in my opinion) for they are all looking at the same end. It is indeed true that one method may be briefer than another, but it is enough that all are at hand and the one that is most easy will without doubt be accepted by men and be put in use without casting aspersions on another method. ... In short, I shall set forth the method which is most pleasing to me today and it will rest in men's judgement to appraise what they see; meanwhile I shall continue my discourse going now to the discussion itself.

Let us first assume that if we wish to find the approximate root of 13 that this will be 3 with 4 left over. The remainder should be divided by 6 (double the 3 above) which gives 23\large\frac{2}{3}\normalsize . This is the first fraction to be added to the 3, making 3233\large\frac{2}{3}\normalsize which is the approximate root of 13. Since the square of this number is 134913\large\frac{4}{9}\normalsize , it is 49\large\frac{4}{9}\normalsize too large, and if one wishes a closer approximation, the 6, which is double the 3, should be added to the fraction 23\large\frac{2}{3}\normalsize giving 6236\large\frac{2}{3}\normalsize and this number should be divided into 4 which is the difference between 13 and 9 ...

Bombelli continues to describe his method. However let us turn to his justification. The reader may wish to compare what Bombelli is doing with the calculation of a continued fraction. We return to this point after the quote.

Let us suppose we are required to find the root of 13. The nearest square is 9, which has root 3. I let the approximate root of 13 be 3 plus 1 unknown.
[ 3+x=133 + x = √13]
Its square is 9 plus 6 unknowns plus 1 power. We set this equal to 13.
[ (3+x)2=9+6x+x2=13(3 + x)^{2} = 9 + 6x + x^{2} = 13]
Subtracting 9 from either side of the equation we are left with 4 equal to 6 unknowns plus 1 power.
[ 6x+x2=46x + x^{2} = 4]
Many people have neglected the power and merely set 6 unknowns equal to 4. The unknown then come out to 23\large\frac{2}{3}\normalsize
[ 6x=46x = 4 gives x=23x = \large\frac{2}{3}\normalsize ]
and the approximate value of the root is 3233\large\frac{2}{3}\normalsize since it has been set equal to 3 plus 1 unknown.
[ 13=3+x=323√13 = 3 + x = 3\large\frac{2}{3}\normalsize]
However, taking the power into account, if the unknown is equal to 2 /3 the power will be 23\large\frac{2}{3}\normalsize of an unknown which, added to the 6 unknowns, will give us 6 and 23\large\frac{2}{3}\normalsize unknowns, which equal 4
[ 6x+x2=46x + x^{2} = 4 implies 6x+23x=46x + \large\frac{2}{3}\normalsize x = 4 ]
so the unknown will be equal to 35\large\frac{3}{5}\normalsize, and since the approximate is 3 plus 1 unknown, it comes to 3353\large\frac{3}{5}\normalsize .
[ x=4/(6+23)x = 4/(6 + \large\frac{2}{3}\normalsize ) implies 3+x=3353 + x = 3\large\frac{3}{5}\normalsize ]
But if the unknown is equal to 35\large\frac{3}{5}\normalsize the power will be 35\large\frac{3}{5}\normalsize of an unknown and we obtain 6356\large\frac{3}{5}\normalsize unknowns equal to 4
[ 6x+x2=46x + x^{2} = 4 implies 6x+35x=46x + \large\frac{3}{5}\normalsize x = 4]
Then the unknown comes to 2033\large\frac{20}{33}\normalsize .
[ 6x+35x=46x + \large\frac{3}{5}\normalsize x = 4 implies x=4/(6+35)=2033x = 4/(6 + \large\frac{3}{5}\normalsize ) = \large\frac{20}{33}\normalsize ]
... this process can be carried to within an imperceptible difference.

Notice that Bombelli has set up a recursive procedure x4(6+x)x \mapsto \large\frac{4}{(6+x)}\normalsize which he realises converges to the correct answer.

With some modern tools, let us investigate what Bombelli is doing. Here is Maple code for Bombelli's recursive procedure
bombelli:= proc(n)
option remember;
if n = 0 then 0 else 4/(6+bombelli(n - 1)) end if;
end proc:
This gives the first 10 fractions as:
23,35,2033,66109,109180,7201189,23783927,39276485,2594042837,85674141481\large\frac{2}{3}\normalsize , \large\frac{3}{5}\normalsize , \large\frac{20}{33}\normalsize , \large\frac{66}{109}\normalsize , \large\frac{109}{180}\normalsize , \large\frac{720}{1189}\normalsize , \large\frac{2378}{3927}\normalsize , \large\frac{3927}{6485}\normalsize , \large\frac{25940}{42837}\normalsize , \large\frac{85674}{141481}\normalsize
The last term as a decimal correct to 12 places is .605551275436

Compare this with the fractional part of √13:   .605551275464

Of course we can also see that Bombelli's procedure leads to a continued fraction expansion of a square root. Cataldi's work on continued fractions appeared in 1613, well after Bombelli's work of 1572. Since Bombelli is not thinking in terms of continued fractions it would not be appropriate to claim that he was the inventor of continued fractions because of the above work. However, it is quite possible that Cataldi had the idea of a continued fraction after reading Bombelli's book.

Last Updated August 2006