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Topic: John Gabriel's Thread on Mathematics.
Replies: 231   Last Post: Mar 22, 2014 9:23 PM

 Messages: [ Previous | Next ]
 Karl-Olav Nyberg Posts: 1,558 Registered: 12/6/04
Re: John Gabriel's Thread on Mathematics.
Posted: Feb 26, 2014 6:41 PM

On Wednesday, February 26, 2014 11:02:07 PM UTC+1, konyberg wrote:
> On Wednesday, February 26, 2014 8:06:58 PM UTC+1, John Gabriel wrote:
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> > On Wednesday, 26 February 2014 20:20:45 UTC+2, Soap Research wrote:
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> > > On February 26, 2014, Wizard-Of-Oz wrote:
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> > > > John Gabriel <thenewcalculus@gmail.com> wrote in
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> > > > > On Wednesday, 26 February 2014 15:12:03 UTC+2, Wizard-Of-Oz wrote:
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> > > > >> But YOU claimed it always divided *exactly*
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> > > > > I stated very clearly that m+n always divides f(x+n)-f(x-m) EXACTLY.
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> > > > That's what I said you claimed.
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> > > > That means the result is an integer quotient with a zero remainder
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> > > > You agreed that that is what divides exactly means
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> > > > > Still waiting for you to show me a counterexample where this is not
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> > > > > true.
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> > > > I have
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> > > > > In order for you to do this, you must show me an example where you
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> > > > > simplify the quotient and you have a term with m+n in the denominator.
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> > > > So now you are saying that that is what you mean by 'divides exactly'. I
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> > > > did ask you what you meant, and you did't say.
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> > > > > You can't. Moron!
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> > > > YOU need to proof that noone can.
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> > > > You claim what you say is rigorous, so you need to PROVE that for any
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> > > > function f(x), you can simplify (f(x+n)-f(x-m))/(m+n) and that
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> > > > simplification will not have any term involving (m+n) in the denominator
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> > > > Try f(x) = sin(x)
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> > > I'll give it a try. Let f(x) = sin(x)
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> > > f(x+n)-f(x-m) = sin(x+n) - sin(x-m)
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> > > = sin(x) cos(n) + cox(x) sin(n)
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> > > - (sin(x) cos(m) - cos(x) sin(m))
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> > > = sin(x) (cos(n) - cos(m)) + cos(x) (sin(n) + sin(m))
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> > > = sin(x) * 2 cos((n+m)/2) cos((n-m)/2)
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> > > + cos(x) * 2 sin((n+m)/2) cos((n-m)/2)
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> > > Now, Mr. Gabriel, would you please help me and indicate how I can simplify (m+n) in the previous example? How can I reach f'(x) = cos(x) + Q(x, m, n) from here? BTW, you can't use anything from "flawed old calculus", only your new calculus.
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> > > Thanks.
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> > It's wrong from the first line. Tsk, tsk.
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> > Here you go:
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> > http://www.spacetimeandtheuniverse.com/math/4507-0-999-equal-one-442.html#post25653
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> Hi.
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> Is n and m both integers in your definition of divisible? So if some number T exactly divisible of (m+n) is an integer timed (m+n)? Or do you mean that the division leads to a rational number, or what?
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> KON

What you haven't told is this: Is 2/sqrt(2) =sqrt(2) or 1/0.5 = 2 an exact division.What do you mean? In my mathematics an division into numbers is between the natural nunbers.
KON

Date Subject Author
2/22/14 thenewcalculus@gmail.com
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2/23/14 Art Hopkins
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2/26/14 thenewcalculus@gmail.com
2/26/14 Wizard-Of-Oz
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2/27/14 David C. Ullrich
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2/28/14 Peter Percival
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2/28/14 John Gabriel
3/1/14 David C. Ullrich
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3/1/14 thenewcalculus@gmail.com
3/1/14 thenewcalculus@gmail.com
3/1/14 Karl-Olav Nyberg
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3/2/14 Wizard-Of-Oz
3/2/14 David C. Ullrich
3/2/14 thenewcalculus@gmail.com
3/2/14 David C. Ullrich
3/2/14 John Gabriel
3/3/14 John Gabriel
3/3/14 David C. Ullrich
3/3/14 thenewcalculus@gmail.com
3/4/14 David C. Ullrich
3/4/14 thenewcalculus@gmail.com
3/4/14 Karl-Olav Nyberg
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3/3/14 Roland Franzius
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3/3/14 David C. Ullrich
3/3/14 thenewcalculus@gmail.com
3/3/14 Roland Franzius
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3/8/14 Brian Q. Hutchings
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