Re: abundance of irrationals!)


Virgil wrote:
> In article <1115569851.909098.102670@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
> mueckenh@xxxxxxxxxxxxxxxxx wrote:
>
> > Randy Poe wrote:
> >
> > > > > > > > My list:
> > > > > > > > 0.1;
> > > > > > > > 0.01; 0.11;
> > > > > > > > 0.001; 0.101; 0.011; 0.111;
> > > > > > > > ...
> > > > > > > > contains every positive real number of (0,1).
> >
> > > > Do you agree that all digit-positions of all real numbers of
(0,1)
> > > are
> > > > in my list?
> > >
> > > I don't know what this means. What are the "digit-positions"
> > > of real number?
> >
> > Ask for the k-th digit of any real number of (0,1). You can find it
in
> > the list, for the first time in line k in number l at position m =
i.

Correction: m = k

>
> > Is there any natural
> > number m such that any b_m cannot be found?
>
> Irrelevant!

>
> > Why should you require to
> > have it in one of the earlier lines?
>
> Nothing is required of any earlier lines, the only requirement is of
the
> line one is considering at any one moment, whether it has a last
> non-zero digit or not. And it always does.
>
Then consider this list 2 please:

..
0 1
0 1 0 1
0 1 0 1 0 1 0 1
....

and read from top to bottom (like the Chinese do or like we do in order
to read the diagonal number of a list).

Then you find ALL the bits of 1/3 = 0.010101... in the positions
1,2,3,6,11,22,... The recursive definition of that sequence a_n (with n
the line number and a the position in the line) is

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.

In this way ALL reals of the interval (0,1) are given by that list 2.

We have, as easy can be proved:
1) The set of all bit positions of list 2 is countable.
2) The number of all bit positions is larger than the number of all
reals in the interval (0,1).

This shows the countability of the reals. If there is another proof
showing the unountability of the reals, the notion of countability is
self-contradictive.

Regards, WM

.

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