Re: Well Ordering the Reals

David R Tribble said:
> Tony Orlow wrote:
> Tony Orlow wrote:
> >> define x>y if the most significant digit where they differ is a 1 in x and a 0
> >> in y.
> >
>
> David R Tribble said:
> >> What if my infinite naturals have no most significant digit?
> >>
> >> Suppose
> >> x = ...101010101010,
> >> i.e., an infinite sequence of alternating 1,0 digits, and that
> >> y = ...110110110110,
> >> i.e., an infinite sequence of repeating 1,1,0 digits.
> >>
> >> I don't see how I can use your definition to determine whether x
> >> or y is larger. It's true that x and y are both even, and that they
> >> differ in half their bits, but I can't say much about their "most
> >> significant" digits.
> >
>
> Tony Orlow wrote:
> > Perhaps now you see why I express the T-riffics in terms like 1:000...001,
> > where there IS a most significant digit. What you have here is one sequence
> > that goes 101 repeatedly and one that alternates between 101 and 010, and the
> > last two bits are the same. The only way the second can be greater than the
> > first is if the most significant bit falls on one of its 1's, which is a 1/3
> > chance. It can't be determined without knowing a most significant bit.
>
> But x and y have an infinite number of digits; they have no leftmost
> (most significant) digit. They are infinite numbers. Surely that
> means that they are composed of an infinite number of nonzero
> digits, right?
>
> Are you saying that the infinite numbers x and y above cannot be
> represented in your notation?
>
>
Yes, they are not proper T-riffic numbers, because they have no infinite
digital point or most significant digit. That is the problem with infinite
numbers which, so far, has prevented them from working. Someone had challenged
the notion of infinite wholes, asking why .....111111,...11110, ...11112, etc,
were all divisible, apparently, by 7. It's a bit of a parlor trick, and it
still doesn't work for ....11115, but it is explainable and does illustrate a
problem with this kind of notation.

I pointed out that this ....11111 was not my N. N is 1:000...000. When I
divided this by 7 I got the normal 6 repeating digits found in x/7, for x in N.
Now, since I cannot say whether the number of 0's in 1:000...000 is divisible
by 6, I cannot say where this repeating string of digits is going to land with
respect to the digital point, and so I cannot tell the remainder. But, what it
showed was that 1:000...000/7 can have 6 of the 7 possible remainders, 1
through 6, but cannot have a remainder of 0. This makes sense if we have a
number which is some power of 10, which shares no prime factors with 7. Seeing
this work, where the other system really failed, is encouraging. At some point
I'd like to play with this a little more with other numbers, and see how
general the rule is regarding remainders of such divisions.
--
Smiles,

Tony
http://www.people.cornell.edu/pages/aeo6/WellOrder/
.

Relevant Pages

  • Re: Well Ordering the Reals
    ... > Tony Orlow wrote: ... >>> As in all regular digital systems, the leftmost nonzero digit is most ... those infinite numbers that are infinitely ...
    (sci.math)
  • Re: The Modified Halting Problem, Take ??? .
    ... Turing tapes" will not be a set or even a proper class in NAFL, ... each tape itself is an infinite entity. ... They're usually described as potentially infinite or finitely unbounded. ... and there is no last digit of Pi ever computed which one would think ...
    (sci.logic)
  • Re: An uncountable countable set
    ... >>> string up to digit number n. ... You state that an infinite string could be ... but contains only finite digit positions. ...
    (sci.math)
  • Re: abundance of irrationals!)
    ... >> WE have a number of axiom systems, ... such set meets Cantor's definition of an infinite set. ... digit position in the nth listed number. ... >> with internal contradictions, but what we are talking about has been ...
    (sci.math)
  • Re: abundance of irrationals!)
    ... >> It is not a matter of understanding, but a matter of contradiction. ... > in which infinite sets are REQUIRED by the axioms. ... The axiom requires n+1 for given n. ... > last digit of one of your listed numbers, ...
    (sci.math)
Loading