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| Science Forum Index » Logic Forum » THERE ARE oo DIGITS IN < 0 . 1 2 1 2 1.. > ( THAT ARE < 10 |
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| |-|erc |
 Posted: Tue Jan 25, 2005 7:39 pm |
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Guest
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--
In 100 years cardinality and incompleteness theory will only be offered under modern history as cult beliefs
Veni, vidi, vamos. |
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| valdez |
| Posted: Tue Jan 25, 2005 9:33 pm |
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Guest
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"|-|erc" <H@r.c> wrote in message news:35o77uF4pki6oU1@individual.net...
[quote:7382f1bf14]
--
In 100 years cardinality and incompleteness theory will only be offered
under modern history as cult beliefs
Veni, vidi, vamos.
**********************************************************[/quote:7382f1bf14]
WRONG. 12/99 = 0.12121....
I count only 4 digits, only 3 unique ones.
Try again....................... |
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| The Ghost In The Machine |
| Posted: Wed Jan 26, 2005 12:01 am |
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Guest
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In sci.logic, valdez
<nospam@nospam.com>
wrote
on Tue, 25 Jan 2005 20:33:33 -0600
<35odr6F4na9hjU1@individual.net>:
[quote:75e238b4ad]
"|-|erc" <H@r.c> wrote in message news:35o77uF4pki6oU1@individual.net...
--
In 100 years cardinality and incompleteness theory will only be offered
under modern history as cult beliefs
Veni, vidi, vamos.
**********************************************************
WRONG. 12/99 = 0.12121....
I count only 4 digits, only 3 unique ones.
[/quote:75e238b4ad]
Uh...I think he means digit *slots* or positions, though
I like your thinking. :-)
Not that it matters; is 1/3 in S_3 = {.3, .33, .333, ... }?
|-|erc seems to think so, but a simple proof shows otherwise,
as S_3 = { (10^k - 1) / (3 * 10^k): k >= 0, k in J} is easily
seen to generate this set (in order), and for 1/3 to be
in S_3, 3 * 10^k - 3 = 3 * 10^k for some integer k, or
0 = -3, which is absurd.
[quote:75e238b4ad]
Try again.......................
[/quote:75e238b4ad]
--
#191, ewill3@earthlink.net
It's still legal to go .sigless. |
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