Notes on Trinary:

      - The conventional trinary representation of a positive number x is given
        by

	x = SUM(i>=0) 3^(-i) * t_i,  where 0 <= t_i <= 2 for i > 0, and
				     t_0 can be arbitrarily large

	However, to accomodate exact real representation using interval
	approximation, we must increase the range from [0,2] to [-2,2].

      - Addition:

	To add two exact reals with a trinary stream expansion, we are
	working from left to right, so the traditional notion of
	carrying digits is insufficient.  The following outlines an
	algorithm for the addition of two trinary streams:

	Suppose you want to add 0.22-120-2... and 0.-120102-2... The
	first step is just a straightforward addition of each place:

	  0. 2 2-1 2 0-2 ...
	+ 0.-1 2 0 1 2-2 ...
	--------------------
	  0. 1 4-2 2 2-4 ...

	Now, this result is correct, but it is no longer in our modified
	trinary form, as it has digits in the range [-4,4].  However, it
	is clear that all digits will remain in this range, so we need
	only worry about converting this result to proper trinary.
	However, we are processing from left to right, so we have to be
	a little clever.  We implement a sort of "inverse carrying"
	procedure as follows: We allow for one unit of "elbow room" in
	the i_th trit (i.e. ensure it is in range [-1,1]) so that we can
	always put (i+1)st trit into range [-1,1].  Thus we step a two
	digit window down the line, ensuring that we can always
	accomdate the move of the next trit into the range [-1,1].

	0. 1 4-2 2 2-4...       start
	0. 1			1 in [-1,1], no adj. necessary
	   2 1			4 not in [-1,1], sub 3, incr. 1 to 2
	     0 1		-2 not in [-1,1], add 3, decr. 1 to 0
	       2-1		2 not in [-1,1], sub 3, incr. 1 to 2
	         0-1		2 not in [-1,1], sub 3, incr. -1 to 0
		  -2-1...       -4 not in [-1,1], add 3, decr. -1 to -2
        ---------------------  
        0. 2 0 2 0-2 ...

	This method works ad infinitum, since any move from the [-4,4]
	range to the closest edge of the [-1,1] range is at most 3,
	which means at most a change of 1 in the previous digit.  But
	since the previous digit was ensure to be in [-1,1], adding or
	subtracting one will always result in a value in [-2,2], the
	desired range.

	Note that this method fails for binary (-1 <= t_i <= 1) since
	there doesn't exist a successful elbow room range.  The method
	does work for all bases >= 3.


	- Multiplication:

	  Multiplication of two exact reals with a trinary stream
	  expansion is considerably more complicated.  Consider the
	  multiplication of a = 0.1-202-1... and b = 0.2-1-210.... First
	  we construct a multiplication table using the digits:

	    | 0. 1-2 0 2-1 ...
	  ----------------------
	   0| 0. 0 0 0 0 0 ...
	   .| 0. . . . . . ...
	   2| 0. 2-4 0 4-2 ...
	  -1| 0.-1 2 0-2 1 ...
	  -2| 0.-2 4 0-4 2 ...
	   1| 0. 1-2 0 2-1 ...
	   0| 0. 0 0 0 0 0 ...
	   .|  . . . . . . ...
	   .|  . . . . . . ...
	   .|  . . . . . . ...


	  We can write the product a*b as SUM(i>=0) 3^(-i)*C_i, where
	  C_i is the ith column of the multiplication.  Also, as in the
	  addition case, we notice that the range of the digits in the
	  table is [-4,4], since x*y, where x,y is in [-2,2], must lie
	  in [-4,4].  So we are again left with the problem of ensuring
	  that the result of this addition is in fact in trinary as we
	  have defined it (that is, all digits are in [-2,2].  To do
	  this, we take as t_i of a*b the closest integer in [-2,2] to
	  each C_i, and add t_i - C_i to C_(i+1), and do the same for
	  t_(i+1), and so on.