Ramble, Part 9: A Matter of Magnitude
Dec. 19th, 2006 02:38 pm![[personal profile]](https://www.dreamwidth.org/img/silk/identity/user.png){kind=small}
stoutfellowI'm not going to say very much about mathematics in the Muslim and Hindu civilizations, mostly because I really don't know as much about them as I should. However, there are a few items that I'd like to point out before going on to postclassical Western mathematics; those will be the subjects of the next two or three posts.
As I understand it, one lesson that people in the hard sciences learn early on is the necessity of keeping track of units. You don't add areas to masses, and if you divide a length by a time you get a speed1. I'd like to devote a little time to a related problem in mathematics, which arises when algebra is conceived of in geometric terms.
Geometric algebra, as practiced from Euclid's time on, worked in terms of magnitudes, and the problem with magnitudes is that they do come in different types - lengths, areas, volumes. Where Diophantus could blithely add x3 to 3x2, geometric algebraists could not, unless the "3" was somehow conceived of as the same sort of magnitude (probably a length) as x.
Again, in geometric algebra, the notion of a "square root" was problematic. One could speak of the square root of an area X, meaning the side of a square with the same area as X (and this is what is meant by such a phrase as "squaring the circle"), but the idea was not applicable as such to, for example, lengths. What was available was the notion of a "mean proportional" - i.e., an average ("mean") in the sense of proportionality, as opposed to the usual additive notion of average. (Technically, we're distinguishing "geometric means" - significantly so named - from "arithmetic means".) To be precise, given two magnitudes of the same kind a and b, the mean proportional is a magnitude x of that kind, such that a:x = x:b. That is, x is the square root of ab. Similarly, one could speak of "two mean proportionals", meaning two magnitudes x and y so that a:x = x:y = y:b. (x is the cube root of a2b, and y that of ab2.)
Now, if a and b are lengths, it's possible, using straightedge and compass, to construct the mean proportional. Start with a segment AB having length a; extend it past B to C, so that the segment BC has length b. Now draw a semicircle whose diameter is AC. At B, raise the perpendicular to AC, intersecting the semicircle at D. Then BD is the mean proportional.
Here's an oddity. The best example of that progress is something attributed to Omar Khayyam. Yes, the poet: he was, apparently, quite versatile. (It's questionable whether this was actually his work, but then, it's also questionable to what extent the Rubaiyat is his. Let's be generous, OK?) Omar presented a general solution to cubic equations of the form x3+b2x+a3 = cx2, where a,b,c,x are the lengths of line segments. (Note the care he takes to keep the units right.) The (rather complex) procedure involves the construction of a semicircle and a rectangular hyperbola. It does not locate negative roots of the equation (naturally enough), and, depending on the precise values of a,b,c, it sometimes misses some positive roots as well. For those so inclined, here's Omar's procedure.
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1. I do mean speed - possibly signed - and not velocity, which is, after all, a vector rather than a scalar.
As I understand it, one lesson that people in the hard sciences learn early on is the necessity of keeping track of units. You don't add areas to masses, and if you divide a length by a time you get a speed1. I'd like to devote a little time to a related problem in mathematics, which arises when algebra is conceived of in geometric terms.
Geometric algebra, as practiced from Euclid's time on, worked in terms of magnitudes, and the problem with magnitudes is that they do come in different types - lengths, areas, volumes. Where Diophantus could blithely add x3 to 3x2, geometric algebraists could not, unless the "3" was somehow conceived of as the same sort of magnitude (probably a length) as x.
Again, in geometric algebra, the notion of a "square root" was problematic. One could speak of the square root of an area X, meaning the side of a square with the same area as X (and this is what is meant by such a phrase as "squaring the circle"), but the idea was not applicable as such to, for example, lengths. What was available was the notion of a "mean proportional" - i.e., an average ("mean") in the sense of proportionality, as opposed to the usual additive notion of average. (Technically, we're distinguishing "geometric means" - significantly so named - from "arithmetic means".) To be precise, given two magnitudes of the same kind a and b, the mean proportional is a magnitude x of that kind, such that a:x = x:b. That is, x is the square root of ab. Similarly, one could speak of "two mean proportionals", meaning two magnitudes x and y so that a:x = x:y = y:b. (x is the cube root of a2b, and y that of ab2.)
Now, if a and b are lengths, it's possible, using straightedge and compass, to construct the mean proportional. Start with a segment AB having length a; extend it past B to C, so that the segment BC has length b. Now draw a semicircle whose diameter is AC. At B, raise the perpendicular to AC, intersecting the semicircle at D. Then BD is the mean proportional.
Here's why this works. First off, the triangle ADC is right. (To see this, let O be the center of the semicircle. The triangles AOD and COD are isosceles, so the angles OAD and ODA are equal, as are OCD and ODC. Now use the fact that the sum of the angles of a triangle is 180 degrees, and the fact that the angles AOD and COD also add up to 180.) So are the triangles ABD and DBC, and - check the angles - they are similar. Therefore |AB|/|BD| = |DB|/|BC|, which is what we wanted.Unfortunately, it's not possible to construct two mean proportionals with straightedge and compass. If we extend our range of tools to include general conics, that construction becomes possible.
Just quickly, and using modern notions of geometry, consider the parabola x2=ay and the hyperbola xy=ab. They intersect in a single point, and the coordinates of that point are two mean proportionals between a and b.Using mean proportionals, Euclid was able to - in modern terms - solve quadratic equations, as I mentioned earlier. Using two mean proportionals, it becomes possible (under the right circumstances) to solve cubic equations, and here's where Islamic mathematics comes into the picture: they made considerable progress in doing just that.
Here's an oddity. The best example of that progress is something attributed to Omar Khayyam. Yes, the poet: he was, apparently, quite versatile. (It's questionable whether this was actually his work, but then, it's also questionable to what extent the Rubaiyat is his. Let's be generous, OK?) Omar presented a general solution to cubic equations of the form x3+b2x+a3 = cx2, where a,b,c,x are the lengths of line segments. (Note the care he takes to keep the units right.) The (rather complex) procedure involves the construction of a semicircle and a rectangular hyperbola. It does not locate negative roots of the equation (naturally enough), and, depending on the precise values of a,b,c, it sometimes misses some positive roots as well. For those so inclined, here's Omar's procedure.
First, construct a segment AB with length a3/b2. (This can be done using straightedge and compass, but the details are tedious.) Extend it to a point C so that BC=c. Draw a semicircle with diameter AC; raise the perpendicular to AC at B, cutting the semicircle at D. On BD, mark off E so that BE=b. Draw the line EF parallel to AC. Find G on BC so that (BG)(ED)=(BE)(AB), and complete the rectangle DBGH. Draw a rectangular hyperbola through H with asymptotes EF and ED, cutting the semicircle at J. Drop the perpendicular to AC from J, cutting EF at K and AC at L. Then BL is a root of the cubic equation. (The proof that this works is rather lengthy, and I've already gone on long enough.) [Taken from Eves, An Introduction to the History of Mathematics, 6th ed., p. 247.]The ingenuity involved in this construction is admirable. The constraints of geometric algebra required such ingenuity. What algebra was capable of, once freed from those constraints, was yet to be discovered.
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Ramble Contents
1. I do mean speed - possibly signed - and not velocity, which is, after all, a vector rather than a scalar.
