11 Feb
2015
11 Feb
'15
8:50 p.m.
That is, using the greedy representation.
On Feb 11, 2015, at 7:47 PM, Daniel Asimov <asimov@msri.org> wrote:
On Feb 11, 2015, at 5:14 PM, Keith F. Lynch <kfl@KeithLynch.net> wrote:
rcs@xmission.com wrote:
The only theorem I know of is that a rational number always has a finite representation in factorial base; therefore e is irrational.
How does that follow? The existence of an infinite representation for e in factorial base doesn't prove there isn't also a finite representation.
It follows from the (easy) theorem that for all fractions f in [0,1] the factorial fraction representation is unique.