Consider the modular form of the Stern-Brocot (S-B) tree. It is built starting with the 2×2 matrices a = [[00][10]] and b = [[01][00]]. (check out “Matrix Gateway to Geometric Algebra, Spacetime and Spinors by G. Sobczyk for an interesting connection) That pair of “vectors” when multiplied (it’s non-commutative) gives ab = [[10][00] and ba = [[00][10]]…work it out (using the usual rule for matrix multiplication). The sum of the two, ab+ba, give the identity matrix for 2×2 matrices: ab+ba = [[10][01]] = I.
The two matrices I+a and I+b are the L = [[10][11]] and R = [[11][01]] “moves” of the modular S-B tree (all of the multiplications that follow are multiplications by L or R on the right of a given matrix as in [A]x[L] = [[ab][cd]]x[[10][11]]). Effectively, the L “adds left”…it adds the column on the right of the matrix to the column on the left while keeping the right column constant. R does the opposite: it “adds right” by adding the column on the left to the column on the right and keeps the left column constant. Try it with the matrix [[ab][cd]] multiplied on the right by L and R to see how that works.
Starting with the identity matrix, I = [[10][01]], multiplication on the right by either of our L or R matrices above gives the corresponding entry in the modular S-B tree. The connection to the fractions of the S-B tree can be seen by multiplying the resulting matrix (that you get on either side of I) times the column vector [1,1]T. The components of the resulting vector [x,y]T give the numerator (x) and the denominator (y) of the corresponding entry in the S-B tree.
Thinking of our “operators” L and R this way gives a kind of “machine language” view of what is happening in the modular S-B tree. If you think about the effect that operators L and R have (which is entirely due to the presence “inside” them of our a = [[00][10]] and b = [[01][00]] matrices since multiplication by I alone changes nothing) you will notice that that effect changes depending on where you are in the tree.
First, think about what is happening all the way on the right of the tree. The first multiplication of I by R (let’s just call it the string R, leaving out the I) changes [1,1]T (the vector associated with the identity matrix…what you get when you multiply I times the vector [1,1]T) into [2,1]T…BIG effect compared to what it does way down on the right when it multiplies the string RRRRRRRRR…R16=R16 which corresponds to the vector [16,1]T. The first product changes the first component of the input vector by a factor of 2/1: [1,1]T-> [2,1]T, while the second product only changes [16,1]T->[17,1]T changing that first component 16 by a factor of 17/16. As you go down the right side the change effected by one additional multiplication by R disappears: the input is hardly changed…the ratio of the entry on the right in row n divided by the entry in row n+1 goes to 1.
It is in this sense that b = [[01][00]] is an infinitesimal
R does what it does wholly due to the fact that it is the sum I+b…to the presence of b within it. The first multiplication of I by R yields an increase of magnitude 1…in the S-B tree we go from 1/1 to 2/1 on the right side. Look at the relative change. It is a change of 100%. In row 16 we get to I x R16 = 16/1. With the next multiplication by R we are still adding 1, but that produces a relative change of only 1/16th or about 6% (we go from 16/1 to 17/1). The relative change of continued multiplication by R is given by its code complement on the far left of the same row…the reciprocal of the value on the far right. So while row after row the fractions on the far right go off to infinity those on the far left approach 0. The relative effect of multiplication by R becomes infinitesimal…it is exactly what is given by the far left fraction in each row. As the relative change goes to 0 we are making the transition from discrete steps to smooth, continuous change.
Thinking of the modular S-B tree as a projective space (which it is…the matrices in it are the matrix form of projective transformations of the projective line over Z, the integers), as n goes to infinity b becomes infinitesimal in its effect on the projective space…it is infinitesimal at infinity: at the point at infinity (which, by the way, is missing from the affine line built on the projective line). The fractions on the right in the S-B tree never actually reach infinity, but they approach infinity. Thinking of the S-B tree as a projective space as well as an affine space they do and at infinity b is an infinitesimal….not equal to 0, but less than any rational number. Likewise for a = [[00][10]].
What happens in between?
Let’s look at what happens in the middle of the tree. Just to the right of I (or just to the right of a line drawn down the middle of the S-B tree) we see the sequence of vertices that grow as RLLLLL… The sequence of fractions associated with those vertices starts at [2,1]T (using our vector form of the fraction 2/1) and goes to 3/2, then 4/3/ then 5/4…etc. With each step down the tree just to the right of the center line we see that L has the effect of driving [2,1]T towards [1,1]T. Again we see the effect L has via continued multiplication …its effect “disappears” as the n in RLn goes to infinity. Again we see the a in L take on the role of an infinitesimal….again, not equal to 0, but less than any rational number. Likewise for the b in R just to the left of the center line in the S-B tree.
Now let’s look at what happens when R and L alternate. What we get is the product RLRLRL… What does this correspond to? It corresponds to the irrational number referred to as the Golden Mean (labeled with the Greek letter Φ). Essentially the effects of R and L “cancel out” and we get a ratio that has the peculiar feature of having a special relationship between complements and reciprocals (1-1/Φ = 1/Φ2 and the modular S-B code for Φ = 101010… is the code complement of that for 1/Φ = 010101…). That special sequence and its complement divide the S-B tree into thirds more and more precisely as we move down the tree along those infinite paths. The number of vertices between and on either side of 101010… and 010101… get closer and closer to an equal 1/3rd share of the total number of vertices in a row. “At infinity” where the a and b in R and L become infinitesimals (one more time! – not equal to 0, but less than any rational number) those shares are exactly 1/3.
Much of what the blog entries “The Real deal” and “The Real deal, too” are showing can be seen as a “machine language level view” of operations on binary strings. The role of matrices R and L is that of basic machine operations in a computer I/O process. Transforming the tree is like reprograming the I/O processing. We get a different output for the same input code with the same R and L machine operations.
The 00000…string (the infinite string of 0’s) is the fixed point of the DD and anti-DD trees. It runs down the left side of all the trees and is never “relocated”.
21st Century Paradox 