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Topic: Konyberg - for you!
Replies: 1   Last Post: Nov 9, 2017 7:55 PM

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Karl-Olav Nyberg

Posts: 1,575
Registered: 12/6/04
Re: Konyberg - for you!
Posted: Nov 9, 2017 7:55 PM
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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 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?

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