bts@unc.UUCP (10/19/83)
Here's a question concerning enumerations of the
rational numbers and divergent series. I don't know the
answer, and I've never seen this question asked before. (If
you've seen it elsewhere, please provide references as best
you can.) And, I don't have any applications in mind, so
try solving it for fun only.
It's possible to enumerate the set of rational numbers
(see any textbook on analysis or set theory) as a sequence
q1, q2, q3,... , with or without repetitions. Also, given
any sequence of numbers a1, a2, a3,... it's possible to find a
sequence b1, b2, b3,... so that the a's are the sequence of
partial sums of the infinite series of b's, b1+b2+b3+b4+...,
i.e.
a1 = b1
a2 = b1 + b2
...
an = b1 + b2 + ... + bn
...
Just let b1 = a1, and after that bn = an-a(n-1). If the a's
are rational numbers, clearly the b's are rational, also.
Here's the question: Is it possible to have two
enumerations of the rational numbers q1, q2, q3 ,... and
r1, r2, r3 ,... so that the r's are the sequence of partial
sums of the infinite series of q's, q1+q2+q3+... ? Is it
possible to do this so that there are no repetitions in
either sequence? If so, can you give the enumeration effec-
tively, or does your proof rely on non-constructive methods
(e.g. the Axiom of Choice)?
Bruce Smith, UNC-Chapel Hill
decvax!duke!unc!bts (USENET)
bts.unc@udel-relay (other NETworks)bts@unc.UUCP (Bruce Smith) (10/19/83)
In my query on enumerating the rational numbers
(unc.6029, "A question on enumerating the rationals"), feel
free to have the sequences enumerate the non-zero rationals,
instead of the whole set. Having 0 as a term in a series
means there'll be at least one repeat in its sequence of
partial sums.
Bruce Smith, UNC-Chapel Hillleichter@yale-com.UUCP (Jerry Leichter) (10/20/83)
There is a trivial answer of "no" to the existence of an enumeration of the rationals, without repetition, whose partial sums are also an EWOR: 0 must occur in the enumeration, forcing a repetition in the partial sum at that point. As to the general question: Looks much harder. I'd guess "yes", based on gut feel only. I'd like to see a proof one way or the other, though. -- Jerry decvax!yale-comix!leichter leichter@yale