Re: Is continuum completely filled up?

On Feb 14, 4:16 pm, "David R Tribble" <d...@xxxxxxxxxxx> wrote:
cbrown wrote:
For example, we could represent the real numbers "base pi" if we chose
[*]. In that case, all rational numbers would require an infinite non-
repeating sequence of coefficients in their representation; but we
would represent the number "pi" as "1.000...".

However, if we chose to represent the reals that way, we would have to
change the usual algorithm for multiplying; because clearly pi*pi is
not pi! (i.e., 1.0000...*1.000... should not equal 1.00...!).

No, that's incorrect. In base pi, the number 1.000(pi) is 1. The
number 10(pi) is pi. So 1(pi) x 1(pi) = 1(pi) = 1 x pi^0 = 1, as
expected. 10(pi) x 10(pi) = 100(pi) = pi^2, also as expected.

Is "I was channeling Homer Simpson" a valid excuse?

You are quite correct. b^0 = 1 for /any/ base.

Cheers - Chas

.

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