Re: Calculus XOR Probability

W. Dale Hall said:
Tony Orlow wrote:
W. Dale Hall said:

Tony Orlow wrote:

... stuff deleted ...

I am sure I don't get it.

You had this remark:


1:000...000/7 is one of the following:

0:142857....142857.142857...142857 with remainder of 1
0:142857....1428571.428571...428571 with remainder of 3
0:142857....14285714.285714...285714 with remainder of 2
0:142857....142857142.857142...857142 with remainder of 6
0:142857....1428571428.571428...571428 with remainder of 4
0:142857....14285714285.714285...714285 with remainder of 5

Since the repeating pattern continues forever, it does not terminate
at the zero point, and therefore does not evenly divide any power of
10. Where the zero point falls in the repeating pattern determines
the finite remainder, but that can't be determined in such an
infinite division.


Earlier, you wrote this:


The T-riffics require a repeating pattern extending from the
countable neighborhood of one limit point to the countable
neighborhood of the next, the transition from one to the next within
each countable neighborhood being explicitly defined. You have not
described where or how the repeating 4 becomes the repeating six
digit sequence. Sorry, that doesn't make sense, and it's not my
system.

I've decided that I can't figure out why the "numbers" you wrote

0:142857....142857.142857...142857
0:142857....1428571.428571...428571
0:142857....14285714.285714...285714
0:142857....142857142.857142...857142
0:142857....1428571428.571428...571428
0:142857....14285714285.714285...714285

are valid members of your system. If they were, then the
"integral" parts would be these values:

0:142857....42857142857
0:142857....28571428571
0:142857....85714285714
0:142857....57142857142
0:142857....71428571428
0:142857....14285714285

and the difference of any two would be of the form

0:0000...????

which is the sort of sequence that you seem to
have forbidden.

For instance, the first minus the second:

0:000000...14285714286

where I'm sure you recognize that the "right sequence"
doesn't terminate after a finite number of digits.

Just how does one perform such a subtraction and
still have a valid sequence? From what I gather,
the only valid sequences beginning with the sequence
of zeros we see on the left must have a finite
sequence of non-zero digits on the right, terminated
by zeros, like this:

0:0000...00012345

I have decided that I really don't understand your
rules for formation of numbers. All I can gather is
that the sequence on the left

0:142857142587....

has a *unique* extension to a valid number, and it
is impossible to establish what that unique extension
is. Is that what you have in mind?


I am not sure what you mean by "unique extension", but you make some good
points. I have already admitted that division by a number coprime with the base
creates an ambiguous number at the finite end of the string, and can have any
remainder but 0. So that results in the problem you describe. I believe this
can be resolved by always using a remainder of 1, since that leaves us with an
integral number of repetitions of the digit pattern. If the pattern begins with
142857... it should end with 142857 at the other end. This assumes an integral
number of 6 digit intervals in the infinite string of digits, but since the
string is logx(N) long for number base x, and it can't be determined whether
that is divisible by any number, and since divisibility of a finite into an
infinite is already problematic, it's worth just assuming we can divide the
infinite expanse of digits into an integral number of six-digit intervals and
see how that works.


So, in that case, the only valid answer to N/7 would be
0:142857....142587.142857...142857. That should clear up that problem, don't
you think? If not, why not?



For this particular question, I don't care about the fraction N/7.

What is problematic (and what you forgot to address) was the fact that
the number

0:000...14285714286

is a difference of your numbers. You explicitly forbid this sort of
number. How can you do arithmetic if numbers can't even be subtracted?

THAT'S the question I'm asking here.

Dale


I corrected that problem by requiring that the repeating pattern be in integral
numbers of repetitions. Then, when you want to do arithmetic with a m-period
string and a n-period string, you end up with a n*m-period string. See?
--
Smiles,

Tony
.

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