From: John K Clark (johnkc@well.com)
Date: Sat Nov 08 1997 - 10:38:22 MST
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Hal Finney <hal@rain.org> Wrote:
>Consider the following number, whose name I forget:
>..123456789101112131415161718192021222324...979899100101102103104105...
>This is created by writing the series 1,2,3,4,5,6,7,8,9,10,11,12,13,...
>and concatenating all the digits. It is known to be an irrational number.
It's called Champernowne's number and you're right, it is irrational. When I
said no decimal could be irrational I was suffering from brain lock and
vastly oversold my case, which was very foolish of me because my argument is
not dependent on it.
By the way, Champernowne's number is the first irrational number to be proven
"normal" in base 10. Normal means that in the limit each digit will occur
exactly 10% of the time, each block of 2 digits will occur exactly 1% of the
time, each block of 3 digits will occur exactly .1% of the time etc. It's not
known if Champernowne's number is normal in any other base but 10, it's not
known if e or PI are normal in any base.
>The number of bits in an infinite binary fraction is a countable
>infinity.
You're right.
>The formula I * I (where I = aleph-null) does not come into play here.
Agreed. I shouldn't have mentioned it.
>The number of values that can be represented by an n bit fraction is 2^n.
True.
>So, you need to disagree with one (or both?) of these two points:
>The real numbers correspond to infinite binary (or decimal) fractions
>The number of infinite binary fractions is 2^aleph-null.
Your second statement is certainly true and it's not that I disagree with the
first, it's just that nobody can prove it.
>For the first to be wrong you'd have to have a real number which did
>not correspond to any infinite decimal.
That's right, and needless to say I can't give you such a number. Nobody has
been able to produce a number on the line that can't be represented by a
decimal and so prove the Continuum Hypothesis wrong, BUT nobody has been able
to show that the existence of such a number leads to a logical contradiction
and so prove the Continuum Hypothesis correct.
>In effect, it would have no value.
It would have no decimal value, but maybe there is a number on the line that
is a little bit bigger or a little bit smaller than any decimal number, maybe
decimal notation is just too coarse a net to catch all the points in a line,
maybe there are aleph-two or aleph- three points on the line. Or maybe not.
>In fact aren't the real numbers sometimes *defined* as all the
>values taken by infinite decimals?)
If so then the question becomes, does every point on the line have its own
unique real number?
>The Continuum Hypothesis does say that C is equal to aleph-one, but
>the issue in question is what is aleph-one. There are no sets known
>to have the cardinality of aleph-one, as I understand it.
Not so, there is a set called the Power Set Of Omega (Pw) that has been
proven to contain aleph-one members. Pw is a set of sets, it's made up of all
the sets of positive integers such a {1} {1,2,3,...} {314159...} and every
other sequence of integers you can think of. To prove the Continuum
Hypothesis you'd need to find a one to one correspondence between the points
on the line and the elements in set Pw. This would be possible if every point
on the line can be represented by a decimal, but can it? As you say, a decimal
has aleph -null digits, so you could represent 2^ aleph -null (aleph -one)
points on the line with decimal notation. The trouble is, what if there are
more that aleph-one points on the line?
>Aleph-one is defined as the smallest cardinal greater than
>aleph-zero.
That's right, so if C is the number of points on the line then the Continuum
Hypothesis can be phrased as "There is no cardinal number between aleph-zero
and C". All we need now is a proof.
John K Clark johnkc@well.com
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