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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

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Python

Posts: 352
Registered: 8/16/16
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
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John Gabriel wrote:
> On Wednesday, 8 November 2017 18:07:27 UTC-5, John Gabriel wrote:
>> On Wednesday, 8 November 2017 17:57:03 UTC-5, John Gabriel wrote:
>>> On Wednesday, 8 November 2017 17:22:37 UTC-5, Python wrote:
>>>> John Gabriel wrote :
>>>>> On Wednesday, 8 November 2017 15:24:01 UTC-5, 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=0B-mOEooW03iLWldTU1ZkTDVQR0E


> ... 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 "new-calculus", 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.



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