Re: Does the number of nines increase?

On 6/30/2024 2:05 PM, Ross Finlayson wrote:

On 06/30/2024 08:44 AM, Jim Burns wrote:

On 6/30/2024 10:48 AM, Ross Finlayson wrote:

On 06/30/2024 02:55 AM, FromTheRafters wrote:

>

This *complete* ordered field of reals guarantees
cauchy sequence convergence.
See how real numbers are defined.

>
It's axiomatic,
and about the usual open topology.
>
There are others, ....

>
...but not in that discussion.
>
Yes,
we CAN discuss things other than the real numbers.
>
However,
if we ARE discussing the real numbers,
then we AREN'T doing that.
>
Define the Dedekind.complete real numbers.
Prove the Intermediate Value Theorem.
>
Counter.propose(?) the rational numbers,
for which the Intermediate Value Theorem is false.
>
So what?
The rational numbers aren't the real numbers.
We can still apply the Intermediate Value Theorem
to the real numbers,
which is all anyone has claimed.
>
>

 Well, iota-values are defined and satisfy making
for the IVT which results the FTC's, Fundamental
Theorems of Calculus.
 Then, with regards to their having representations
as infinite sequences, which are the same sort of
usual way in which it's demonstrated that series
with the property of being Cauchy or having a limit
are being Eudoxus and all, that iota-values start
on the one end like .000... and on the other like
.111..., yet differ, goes to that you've forgotten
or chosen to ignore that the representation, of
the numbers in usual positional notation, is shared
by both of these models of continuous domains.

[...]
r[0] = .01
r[1] = .0011
r[2] = .000111
r[3] = .00001111
r[4] = .0000011111
...
Now, think of a possible formula. Something like this shit:
;^) lol.
r[0] = .01
r[1] = .01 * 10^(-1) + .1 * 10^(-3) = .0011
r[2] = .0011 * 10^(-1) + .1 * 10^(-5) = .000111
r[3] = .000111 * 10^(-1) + .1 * 10^(-7) = .00001111
r[4] = .00001111 * 10^(-1) + .1 * 10^(-9) = .0000011111
r[5] = .0000011111 * 10^(-1) + .1 * 10^(-11) = .000000111111
...
Taken to infinity, it would be: .(0)(1) or something? How to properly write it:
.0...1...
?