Good point. I left out some important nits: Irrationals have a unique representation in factorial base. Any rational has exactly two representations: one that's finite, and one analogous to the xxx99999... glitch with finite decimal representations. In factorial base, 1 can be represented as 1.000000... or as 0.123456... . (Zero is an exception here, although I suppose you could argue that 0 and -0 are two different representations.) Then you need to check that e-2 = .11111... isn't finite, and also doesn't fit the ....n(n+1)(n+2)... pattern. After adding the proofs for these nits, this proof that e is irrational doesn't look particularly nicer than the standard proof. In fact, the two proof themes look pretty similar. Rich ------- Quoting "Keith F. Lynch" <kfl@KeithLynch.net>:
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.
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