Re: Easy question: WHICH LIST CONTAINS MORE DIGITS OF Pi?

From: Ed Murphy (emurphy42_at_socal.rr.com)
Date: 02/10/05 

Date: Thu, 10 Feb 2005 04:38:22 GMT

On Wed, 09 Feb 2005 23:12:29 +1000, |-|erc wrote:

> there is NO WAY..'how many digits of an infinite sequence' has any other
> answer than a direct quantity. QED.

I believe this was the original post in this thread, just over a
month ago:

----- begin quote -----

List1 =
<314159265....>

List2 = {
<3>
<31>
<314>
<3141>
<31415>
<314159>
...

}

------ end quote ------

If I answer this question as directly as I can figure how, will you
shut the hell up about it already? Boy, do I ever hope so.

Q: Which list contains more digits of pi?

A: Neither. Both lists contain countably-infinitely-many (CIM)
    digits of pi, in the following senses of "contain":

    List 1 consists of one element with CIM digits. There is an
    obvious bijection from the digits of pi to the positive integers,
    and an equally obvious bijection from the positive integers to
    the digits of List 1's element, which leads directly to a transitive
    bijection from the digits of pi to the digits of List 1's element.

    There are CIM positive integers, thus there are CIM digits of pi
    and CIM digits of List 1's element, all of which are mapped back
    and forth to one another.

    List 2 consists of CIM elements, each with a finite number of
    digits. There is an obvious bijectino from the digits of pi to
    the positive integers, and an equally obvious bijection from the
    positive integers to List 2's elements, which leads directly to
    a transitive bijection from the digits of pi to List 2's elements.

    There are CIM positive integers, thus there are CIM digits of pi
    and CIM elements of List 2, all of which are mapped back and forth
    to one another. Furthermore, for any positive integer N, the Nth
    element of List 2 is the same as the first N digits of pi.

And now the $64,000 follow-up question:

Q: Is List 1's element a member of List 2?

A: No! This can be proved via /reductio ad absurdum/:

    Suppose that List 1's element is a member of List 2. Then it is
    mapped to some positive integer N. Then it has N digits. But it
    has CIM digits. Contradiction. Then the supposition is false;
    List 1's element is _not_ a member of List 2!

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