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Topic:
Sci.math morons still struggling to prove that m+n is ALWAYS a factor of New Calculus derivative!
Replies:
1
Last Post:
Nov 8, 2017 6:28 PM



Python
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Re: Sci.math morons still struggling to prove that m+n is ALWAYS a factor of New Calculus derivative!
Posted:
Nov 8, 2017 6:28 PM


John Gabriel wrote: > On Wednesday, 8 November 2017 18:07:27 UTC5, John Gabriel wrote: >> On Wednesday, 8 November 2017 17:57:03 UTC5, John Gabriel wrote: >>> On Wednesday, 8 November 2017 17:22:37 UTC5, Python wrote: >>>> John Gabriel wrote : >>>>> On Wednesday, 8 November 2017 15:24:01 UTC5, Python wrote: >>>>> >>>>> <a lot of bullshit I don't care to address> >>>>> >>>>>> There is actually a counter example you've always evading, >>>>>> x > e^(1/x^2) at x = 0 >>>>> >>>>> That's not a counterexample you stupid! m+n is a factor of every term in the numerator. >>>> >>>> Mmmh, this is so obvious that you've shown it in your answer. >>>> >>>> Oh, dear, you didn't. >>>> >>>> How weird :) >>> >>> Oh really? Let's see how wrong you are: >>> >>> https://drive.google.com/open?id=1CgIaVtJjtzsp7h4SjeNfF6KdtHqzDzZA >>> >>> Just because YOU are not able to do it, does not mean it cannot be done. >>> >>> FAIL (again!) >> >> And just to shut your mouth: >> >> https://drive.google.com/open?id=0BmOEooW03iLWldTU1ZkTDVQR0E
> ... that article was from 2002 !!!!!!
1st fact: Last modification of document: 00:05 9 nov. 2017
2nd fact: irrelevant anyway
3rd fact: the point is about to find the derivative at x=0 of this function, with your lousy "newcalculus", this is the question you've evaded.
4th fact: by a is dividing b you claim nothing more than: it exists c so thatn b = c*a, which is trivial on the field of rationals (or reals), take c = b/a. Q and R are not factorial rings.



