[John W. Burgeson]
 I pulled the following from Pickover's LISTSERV  now I am really over
 my head!
 
 What information can't pi contain in its digit stream?

 Say that for example you have 3 atoms arranged in an equilateral triangle
 formation. That means that at least one of the coordinates is equal to
 the square root of 3, or 1.732etc. The three coordinates of the atoms
 maybe could be (0,0), (1,0), and (1, 1.732..).

 Because the square root of three is irrational, its expansion as a real
 number will continue forever without a pattern. Thus, if pi is to
 "contain" the sqrt(3) that means that at one point, the digits of pi
 would have to revert to the digits of sqrt(3), and continue that way for
 infinity. Which would mean that pi is not transcendental. Pi has been
 proven to be transcendental, so an infinite expansiion of sqrt(3), or any
 other root for that matter, is impossible.
 
 Ignoring the question of the proper definition of "transcendental,"
 I still don't see this follows. The infinite expansion of pi ought
 to be able to contain a finite number of infinite expansions. Maybe.
It all depends on how you interpret the word "contain".
For any increasing sequence of natural numbers 0 < n_1 < n_2 <n_3 < ...,
there is a real number A with decimal expansion given by the rule
ith decimal of A = n_ith decimal of pi.
There are uncountably many such sequences, in fact, even uncountably
many such sequences with the property that 1 <= n_(i+1)  n_i <= 2 for
all i. The quote from Pickover's listserv deals with a very strict
subclass of sequences, namely those which are of the form n_i = N + i
for some number N.
 Stein Arild Strømme <http://www.mi.uib.no/~stromme>
This archive was generated by hypermail 2b29 : Thu Aug 02 2001  05:38:16 EDT