Re: Well Ordering the Reals

Tony Orlow wrote:
>> 1:0101...01010 with N (oo) digits, or 4(2^N)/3=2^(N+2)/3
>

David R Tribble said:
>> The expression 4(2^N)/3 is meaningless, of course, by your own
>> admission, since we don't know what value N really is. If we say
>> x has log2(N) digits, what is its value then?
>

Tony Orlow wrote:
> 4N/3, of course. What you have is 1.01010..., which is 4/3, times 2 to whatever
> value that limit point has. Look, N could be 50, so if you have 50 digits
> between the N point and the 0 point, you are going to get 2^52/3 as a value for
> the string. It could be aleph_0, omega, e_0, or whatever. It's a way of
> expressing a value with a pattern which CAN be infinite. If it is infinite, and
> therefore immeasurable, you can compare such patterns over infinite range
> formulaically. When you have such formulas, it becomes easy to compare them for
> magnitude.

Assuming that log(x) < x for infinite x, or x/2 < x, or x < x+1,
or a plenitude of other false equations about infinite numbers.

You've been assuming all along that infinite arithmetic operates
exactly the same as finite arithmetic, but that is blatantly false.
And you have never demonstrated otherwise. Until you do,
nothing based on such an assumption can be taken seriously.

Take a step towards the horizon, and you're exactly as far from
it as before. That's infinity.

.

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