Re: abundance of irrationals!)
- From: "Randy Poe" <poespam-trap@xxxxxxxxx>
- Date: 8 May 2005 14:14:38 -0700
mueck...@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 are talking about your list, not "any real number in
(0,1)".
> You can find it in
> the list, for the first time in line k in number l at position m = i.
What about the real number pi/4? What position m do I find
that?
> > Every number on your list has an m-th digit, b_m, for every
> > natural number m. But for every number on your list, all
> > of the b_m are equal to zero beyond some point.
>
> Beyond which point?
Certainly beyond digit m.
> Consider the list as a whole.
The list as a whole consists of lines, each of which has
the property that beyond digit m, the m-th number consists
entirely of zeros.
> Is there any natural number m such that any b_m cannot be found?
No, for every natural number m, there exists b_m.
For every number on your list, the value of b_m is zero
for infinitely many values of m.
> Why should you require to
> have it in one of the earlier lines?
One of WHAT earlier lines? I stated something about the
numbers in your list. You claim every number in (0,1) is in
your list. Pick an arbitrary number on your list, at position
m. Do you think there is a number on your list such that
its position is not a natural number?
Of course not. There exists m, the position number, for every
number on your list. And we know that all digits beyond
the m-th are zero for that m-th number on your list. For
any m we can choose.
> Don't forget: There all m e N.
Is there any member of your list not associated with
a particular value of m?
> Your induction proof is valid only for finite subsets of N.
What induction proof?
> Otherwise
> you must accept that by Card({2,4,6,...,2n}) < 2n the finiteness of N
> has been proven.
Induction on n for {2,4,...,2n} will only consider finite
subsets of N.
> Further your critique would apply to Cantor's
> enumeration of he rationals too.
Which critique?
> There is always a stage at which denominators > m+1
> are missing yet.
My "critique" is that for any number enumerated, it is associated
with a particular value of m. This is certainly true of the
rational numbers. Give me a and b, for a/b, and I can tell you
the particular value of m in which it occurs in Cantor's
enumeration.
That doesn't cause a problem, since you can give me any
a and b, any rational number, and there is associated a
unique value of m. That is a GOOD thing to show that
you've counted the rationals.
I can't do that with your list. I can't tell you what
m is associated with pi/4, mainly because there isn't one.
> Why do you consider the next one without problems in that proof?
I don't know what you mean by "the next one".
I can consider an arbitrary number in Cantor's list.
It has the desired property: it's rational. And I can
do the reverse consider an arbitrary rational: I can prove
it's in the list. I do this by considering only the
properties of that individual number. I'm not considering
the "next one". I'm looking at an arbitrary choice of
element. Every arbitrary choice from Cantor's list
has an associated m. The rational 1/3 occurs at position
6. (1/1, 2/1, 1/2, 3/1, 2/2, 1/3, ...}
Now I look at an arbitrary member of your list, at position
m. I make a true statement: that number, every number,
has the property that it terminates in zeros.
- Randy
.
- References:
- Re: abundance of irrationals!)
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