Re: Well Ordering the Reals

David R Tribble said:
> Tony Orlow wrote:
> >> Do you deny my set of H-riffic numbers are not inductively defined,
> >> given these two rules?
> >>
> >> 1) 0 is a real number
> >> 2) If x is a real number, then -2^x and 2^x are real numbers.
> >
>
> David R Tribble said:
> >> You then have to prove that every real x is of the form 2^(+-p), where
> >> p in turn is either 0 or of the form 2^(+-q), recursively.
> >>
> >> To start with, you could show us what p is for 2^p = 3.
> >>
> >> x = 3
> >> 2^p = 3
> >> log 2^p = log 3
> >> p log 2 = log 3
> >> p = log 3 / log 2 = 1.58496+
> >>
> >> p = 2^q
> >> 2^q = log 3 / log 2
> >> log 2^q = log(log 3 / log 2)
> >> q log 2 = log(log 3 / log 2)
> >> q = log(log 3 / log 2) / log 2
> >> q = log(1.58496+) / log 2 = 0.66444+
> >>
> >> q = 2^r
> >> 2^r = log(log 3 / log 2) / log 2
> >> log 2^r = log(log(log 3 / log 2) / log 2)
> >> r log 2 = log(log(log 3 / log 2) / log 2)
> >> r = log(log(log 3 / log 2) / log 2) / log 2
> >> r = log(0.66444+) / log 2 = -0.58977+
> >>
> >> r = 2^t
> >> 2^t = log(log(log 3 / log 2) / log 2) / log 2
> >> log 2^t = log(log(log(log 3 / log 2) / log 2) / log 2)
> >> log 2^t = log(-0.58977+) = error
> >>
> >> Uh oh. It looks like 3 cannot be generated from your rules.
> >
>
> Tony Orlow wrote:
> > You seem to be forgetting that you can have minus signs in your exponentiation.
> > why don't you try r=-2^t, so 2^t= 0.58977+?
>
> Okay, that might work.
It does.
>
> But your mapping only generates a countably infinite number of
> reals, so it can't possibly denumerate all of the uncountable reals.
That may be true, but that's also true of the digital number systems, if you
only allow finite numbers of digits. That is what Cantor's nested interval
proof of the uncountability of the reals is all about. Will you ever get to
point c? Not in any finite number of steps, but it is there, requiring an
infinite bitstring to represent in that system, like 1/3 in decimal. In an
uncountably infinite set indexed by bit strings, the vast majority of elements
will require bit strings of infinite length. Given bit strings of infinite
length, this set definition generates the complete uncountable set of reals,
including oo, and -oo, the predecessor to 0, coming full circle at oo.
>
> In fact (if I'm not mistaken), your mapping omits all transcendental
> reals, since all the reals it generates are of the form 2^p or 2^-p,
> which are algebraic numbers only. Where is pi in your mapping?
I am not sure where pi is. Maybe I can generate it at some point. But, is 2
^sqrt(1/2) not transcendental? It's certainly irrational.
>
>
> Tony Orlow wrote:
> > Let's go from the other end.
> >
> > 3>0 -> 2^0=1
> > 3>1 -> 2^1=2
> > 3>2 -> 2^2=4
> > 3<4 -> 2^-4=0.0625
> > 3>0.0625 -> 2^0.0625=1.04427
> > 3>1.04427 -> 2^1.04427=2.06232
> > 3>2.06232 -> 2^2.06232=4.17657
> > 3<4.17657 -> 2^-4.17657=0.0553003
> > 3>0.0553003 -> 2^0.0553003=1.03908
> > 3>1.03908 -> 2^1.03908=2.05492
> > 3>2.05492 -> 2^2.05492=4.15521
> > 3<4.15521 -> 2^-4.15521=0.0561251
> > 3>0.0561251 -> 2^0.0561251=1.03967
> > 3>1.03967 -> 2^1.03967=2.05576
> > 3>2.05576 -> 2^2.05576=4.15763
> > 3<4.15763 -> 2^-4.15763=0.056031
> > 3>0.056031 -> 2^0.056031=1.0396
> > 3>1.0396 -> 2^1.0396=2.05566
> > 3>2.05566 -> 2^2.05566=4.15734
> > 3<4.15734 -> 2^-4.15734=0.0560423
> > 3>0.0560423 -> 2^0.0560423=1.03961
> > 3>1.03961 -> 2^1.03961=2.05567
> > 3>2.05567 -> 2^2.05567=4.15737
> > 3<4.15737 ->2^-4.15737=0.0560411
> >
> > Hmmm this could go on a long time, forever in fact, but we can certainly take
> > note of the pattern of bits that emerges. It would appear that 3 is
> > ...1011101110111? That's soemthing like the number that Dave Rusin had
> > confirmed equalled 3, though the 0's and 1's may have been inverted.
>
> And I thought you didn't believe in infinite numbers without a
> leftmost digit.
I "believe" in adic numbers, but they have problems. The T-riffics are meant to
be an extnesion of normal digital number systems which overcomes some of the
difficulties of adics when it comes to normal arithmetic. Here, we are talking
about the H-riffic reals and bit strings which express signs of nested
exponents, so it's a different kind of bit string. It may make sense to express
H-riffics in T-riffic notation, but I have not yet seen any need or reason to
do this yet. So, I am talking about a repeated sequence of operations which
will get us to the value of 3.
>
>

--
Smiles,

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

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