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




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 UTC5, konyberg wrote: > > fredag 10. november 2017 01.32.07 UTC+1 skrev John Gabriel følgende: > > > On Thursday, 9 November 2017 18:54:37 UTC5, 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 nonzero (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 = ? What is the answer?
KON



