Re: abundance of irrationals!)


Randy Poe wrote:
> mueck...@xxxxxxxxxxxxxxxxx wrote:
>
> > Then consider this list 2 please:
> >
> > .
> > 0 1
> > 0 1 0 1
> > 0 1 0 1 0 1 0 1
>
> Does this mean 0.01, 0.0101, 0.01010101, ...?
>
> > and read from top to bottom (like the Chinese do or like we do in
> order
> > to read the diagonal number of a list).
>
> I don't understand what you mean. What is the first number
> in this list? What is the second?

There is every real number of[0,1].
>
> > Then you find ALL the bits of 1/3 = 0.010101... in the positions
> > 1,2,3,6,11,22,...
>
> A number x being in your list means that there is m
> such that the m-th number in your list is x.

No, these numbers exist in the same way as Cantor's antidiaonal. For
every digit of number x I can show you the place in my list where it
is. The number 0 = 0.000... for instance has always the first digit (0)
of each line.

But we need no quarrel about this kind of existence. Imporant is only
that the bit-positons of the list are countable, but there are more
bit-positions than reals in [0,1] because every real deviates at leat a
one position from another real.

> > a_1 = 1 (Position of the first 0 after the point)
> > a_n = 2*a_(n-1), if n is even,
> > a_n = -1 + 2*a_(n-1), if n is odd.
>
> Are you saying that a_n is the value of the n-th digit
> for EVERY number in your list?

This sequence gives the bit positions for 1/3. For other nunmbes there
ar other bit positions.
>
> That doesn't make sense. Your recursion rule says:
>
> a_1 = 1
> a_2 = 2
> a_3 = 3
> a_4 = 6
> a_5 = 11
> a_6 = 22
>
Take the first bit of line 1, the second bit of line 2, the third bit
of lne 3, the sixt of line 4, the eleventh of line 5, the 22nd of line
6 and so on. Then you have 0.010101... .

Regards, WM

.

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