Re: Well Ordering the Reals

David R Tribble said:
> Tony Orlow wrote:
> >> As in all regular digital systems, the leftmost nonzero digit is most
> >> significant. You knew that.
> >
>
> David R Tribble said:
> >> But what about those infinite numbers that don't have a leftmost
> >> nonzero digit? You know, those infinite numbers that are infinitely
> >> long sequences of digits?
> >>
> >> Two examples (which I gave in another post):
> >> a = ...101010101010,
> >> which is just '01' repeated forever, and
> >> b = ...110110110110,
> >> which is '110' repeated forever. Neither of these infinite numbers has
> >> a most-significant digit, so we can't tell which one is larger by
> >> comparing their "leftmost" digits.
> >>
> >> Then there is the next problem of defining what it means to add
> >> infinite numbers. What is a+b?
> >
>
> Tony Orlow wrote:
> > Let's see if these can be added. Not all adic numbers will work, but some do.
> > Let's start with the smallest common repetition, six digits, and see if it
> > involves a carry bit:
> >
> > 101010
> > +110110
> > ---------
> > 1100000
> >
> > Well, it does involve a carry bit to the left, but the rightmost bit is a 0, so
> > we can just replace it with a 1 for all but the first bit, and say, going left,
> > it is the string 100000, with 10001 repeated forever to the left.
>
> You seem to be saying that
> a+b = ...1000110001100000
> but that's obviously wrong.
>
> Wouldn't it be more logical to say that
> ...101010101010
> + ...110110110110
> ------------------
> = ...000000000000
> with a carry of 1 that goes, um, where?

If you are going to correct someone, I suggest that next time you do it
correctly. Do you see where my rightmost 1 is? That is the result of adding 1
and 1 from the two strings, plus the carry from the previous bit, which gives a
1, with a carry. If you want two strings that add up to 1:000...000 (an
infinite numbers of zeroes and a 1) I recommend you take two strings which are
complementary, such that each has a 1 where the other has a 0, and add 1 to one
of them. Then you will get the trick you desire. Still, that doesn't reflect on
the T-riffics, because those aren't porper T-riffic numbers.

>
> Of course, you object to the lack of a most significant digit, even
> though that's pretty much the definition of an infinite number:
> an infinite number of nonzero digits, so no "last" digit.

That's the set-theoretical concept, anyway.

>
> At any rate, we could try a similar (but different) addition using
> your notation (where c and d are not the same as a and b above):
> c = 0:101010...101010
> d = 0:110110...110110
> + ------------------
> s = 0:1000000...000000
>
> Hmm. It looks like s is one bit longer than c and d, even though
> they are all (supposedly) infinitely long numbers. Or perhaps
> we've simply just run out of bits, even though each number has an
> infinite supply of them?

So what? There are larger infinities and smaller infinities. Haven't I been
consistent in saying even one bit makes a difference to an infinity, and even
one additional element? When you add two digital numbers of a given length,
what are the chances that the sum will be one bit longer? about 50/50. What's
wrong with that? Does it work TOO much like finite math for you? Is it not
mysterious enough to actually represent anything infinite?

>
> Or maybe, just maybe, digital representations just don't work for
> infinite numbers?
Or, maybe they do, and you are grasping for straw men to knock down. When you
find an actual contradiction, let me know.
>
>

--
Smiles,

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

Relevant Pages

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