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Topic: Konyberg - for you!
Replies: 2   Last Post: Nov 9, 2017 8:34 PM

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 Karl-Olav Nyberg Posts: 1,575 Registered: 12/6/04
Re: Konyberg - for you!
Posted: Nov 9, 2017 8:17 PM

fredag 10. november 2017 02.01.48 UTC+1 skrev John Gabriel følgende:
> On Thursday, 9 November 2017 19:56:10 UTC-5, konyberg wrote:
> > fredag 10. november 2017 01.32.07 UTC+1 skrev John Gabriel følgende:
> > > On Thursday, 9 November 2017 18:54:37 UTC-5, konyberg wrote:
> > > > torsdag 9. november 2017 00.43.28 UTC+1 skrev John Gabriel følgende:
> > > > > "It confuses me that you can set m+n equal to zero after you cancel out m+n in the ratio. Surely this is wrong?"
> > > > >
> > > > > Well, consider first that I proved the sum of all the terms in m and n must be zero using 8th grade algebra (see http://thenewcalculus.weebly.com) but also note that the slope of the tangent line does NOT depend on m or n so given the derivative as an expression in x and Q(x,m,n), it follows Q(x,m,n) must be zero, that is, m=n=0.

> > > >
> > > >
> > > > As I see it you have a transformation from p(x) = q(x) *(m+n)/(m+n).

> > >
> > > It's not a transformation that I make but a result of determining the slope of a parallel secant line or the tangent line.
> > >

> > > > Then you you cancel by setting (m+n)/(m+n) = 1.
> > > > This works only if m+n <> 0.

> > >
> > > Correct! But before you cancel, m<>0 and n<>0 because NO parallel secant line has the (m,n) pair (0,0).
> > >
> > > Your next paragraph is not relevant.
> > >

> > > > Consider this:
> > > > 0 = a * 0 This is correct! But what is a? It is correct for any a!
> > > > If you allow this in your NM -> 0/0 = a, and then decide (since cancelling out) that 0/0 suddenly equals 1 is an enigma. 0/0 could be any number (if you allow division by 0).
> > > > (m+n)/(m+n) is not cancelled out! It could be any number you decide.
> > > >
> > > > KON
> > > > KON

> >
> > So in a proof, you can get rid of division by 0 in the beginning, but use it later?

>
> Of course not. There is NEVER division by zero in the New Calculus.
>
> Before you cancel (m+n), neither m nor n are zero, so division of zero never occurs.
>
> NOW, once you cancel, you are left with f'(x) = k + Q(x,m,n) where k=f'(x).
>
> At this time, we know Q(x,m,n)=0 from the proof, so we can set m=n=0 because the tangent line slope does not depend on m or n! However, you can still use ANY one of the non-zero (m,n) pairs before canceling in Q(x,m,n) since they will result in Q(x,m,n)=0.
>
>

> > KON

I understand what you are telling me, but do you?
You are dividing by (m + n).
And you are telling me that m = n = 0. That is (m + n) = 0.
Give me one example where this is allowed and how you do it.
Give me 3/0 = ?