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Science Forum Index » Logic Forum » Rucker on Infinitesimals (was '0.999... = 1')
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| G. A. Edgar |
 Posted: Mon Jan 05, 2004 1:29 pm |
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In article <881c8779.0401041419.1097372c@posting.google.com>, Paul
Holbach <paulholbachSPAMBAN@freenet.de> wrote:
There is another system, besides Robinson's hyperreals, namely Conway's
surreals, in which such infinitesimal calculations are common.
--
G. A. Edgar http://www.math.ohio-state.edu/~edgar/ |
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| David Canzi |
| Posted: Mon Jan 05, 2004 5:32 pm |
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In article <881c8779.0401041419.1097372c@posting.google.com>,
Paul Holbach <paulholbachSPAMBAN@freenet.de> wrote:
Quote: "In our ordinary real number system, we say that the number K with
decimal expansion .99999... is the samas 1. An informal argument for
this is sketched below:
10K = 9.999...
- K = .9999...
____________
9K = 9
K = 1
But maybe this argument is misleading. What if there is some number,
call it 1 - 1/omega, that is greater than any finite string .9...9 of
nines, yet less than 1? If K were actually equal to 1 - 1/omega, the
informal argument used in the last paragraph would not work, ...
Let t be the difference between 1 and 0.9999... ("t" stands for 'tesimal.)
K = 1 - t = 0.9999...
10K = 10 - 10t = 9.9999...
K + 9 = 10 - t = 9.9999...
Uh oh...
10 - 10t = 10 - t
10t = t
9t = 0
t = 0
--
David Canzi |
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| David Canzi |
| Posted: Mon Jan 05, 2004 5:32 pm |
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Guest
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In article <881c8779.0401041419.1097372c@posting.google.com>,
Paul Holbach <paulholbachSPAMBAN@freenet.de> wrote:
Quote: "In our ordinary real number system, we say that the number K with
decimal expansion .99999... is the samas 1. An informal argument for
this is sketched below:
10K = 9.999...
- K = .9999...
____________
9K = 9
K = 1
But maybe this argument is misleading. What if there is some number,
call it 1 - 1/omega, that is greater than any finite string .9...9 of
nines, yet less than 1? If K were actually equal to 1 - 1/omega, the
informal argument used in the last paragraph would not work, ...
Let t be the difference between 1 and 0.9999... ("t" stands for 'tesimal.)
K = 1 - t = 0.9999...
10K = 10 - 10t = 9.9999...
K + 9 = 10 - t = 9.9999...
Uh oh...
10 - 10t = 10 - t
10t = t
9t = 0
t = 0
--
David Canzi |
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| Alekzander |
| Posted: Mon Jan 05, 2004 8:24 pm |
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paulholbachSPAMBAN@freenet.de (Paul Holbach) wrote in message news:<881c8779.0401041419.1097372c@posting.google.com>...
Quote: "In our ordinary real number system, we say that the number K with
decimal expansion .99999... is the samas 1. An informal argument for
this is sketched below:
10K = 9.999...
- K = .9999...
____________
9K = 9
K = 1
But maybe this argument is misleading. What if there is some number,
call it 1 - 1/omega, that is greater than any finite string .9...9 of
nines, yet less than 1? If K were actually equal to 1 - 1/omega, the
informal argument used in the last paragraph would not work, for this
argument overlooks the fact that the difference between 10K and 10 is
ten times as great as the difference between K and 1. There is a
residual infinitesimal quantity below that does not get canceled out:
10K = 10 - 10/omega
- K = 1 - 1/omega
_________________
9K = 9 - 9/omega
K = 1 - 1/omega
but how would this string 1-1/aleph0 be represented? in every real
base b+1, .bbbbb... repeating is never bigger than .999... in base 10
..aaaaa... in base b = K, follow the same process.. x = a/(b-1). the
maximum value of a is b-1, and thus x cannot be greater than 1. this
even has the logical extension: limit b to infinity, where each
integer becomes infinitessimal, there's still no breaking of it.
multiplication by (b-1) in base b moves the decimal place one to the
right, which has _no_ effect on an infinite list of nines, and
therefore 10K still equals 10 - 1/aleph0 |
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| Alekzander |
| Posted: Mon Jan 05, 2004 8:24 pm |
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Guest
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paulholbachSPAMBAN@freenet.de (Paul Holbach) wrote in message news:<881c8779.0401041419.1097372c@posting.google.com>...
Quote: "In our ordinary real number system, we say that the number K with
decimal expansion .99999... is the samas 1. An informal argument for
this is sketched below:
10K = 9.999...
- K = .9999...
____________
9K = 9
K = 1
But maybe this argument is misleading. What if there is some number,
call it 1 - 1/omega, that is greater than any finite string .9...9 of
nines, yet less than 1? If K were actually equal to 1 - 1/omega, the
informal argument used in the last paragraph would not work, for this
argument overlooks the fact that the difference between 10K and 10 is
ten times as great as the difference between K and 1. There is a
residual infinitesimal quantity below that does not get canceled out:
10K = 10 - 10/omega
- K = 1 - 1/omega
_________________
9K = 9 - 9/omega
K = 1 - 1/omega
but how would this string 1-1/aleph0 be represented? in every real
base b+1, .bbbbb... repeating is never bigger than .999... in base 10
..aaaaa... in base b = K, follow the same process.. x = a/(b-1). the
maximum value of a is b-1, and thus x cannot be greater than 1. this
even has the logical extension: limit b to infinity, where each
integer becomes infinitessimal, there's still no breaking of it.
multiplication by (b-1) in base b moves the decimal place one to the
right, which has _no_ effect on an infinite list of nines, and
therefore 10K still equals 10 - 1/aleph0 |
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