Re: Is continuum completely filled up?


toshiaki wrote:
"Saurav" <saurav1b@xxxxxxxxx> wrote in message
news:1168839154.302850.158470@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx


But when time would end, this problem would certainly have been solved.
dx = 0, dy= 0 , but dy / dx = 3, when y = 3*x).
I didn't know that dx and dy can be set zero. I only knew that d_{a}f
is a linear functional : h |---> h.f'(a). Where did you see such
things: dx = 0, dy = 0, yet dy/dx is is a real number?
I interpreted formulas above as abreviation of following explanation.
As for this, traditional formula of differencial is sufficient, with no
need
of unspecifiable number.
dy/dx = lim(h- > 0){3*(x+h)-3*x}/{(x+h)-x}
dy = 3*(x+h)-3*x =>0, dx = (x+h)-x =>0
This is not compatible with dx = 0, dy = 0.
I think you may have tried to write,
dx ----> 0 as h ----> 0.
But it is ->not<- as terrific a blow as dx = 0.
Karl Marx said that scholars interplit histry, but I will create histry.
And
Ludwig Witgenstein said that the work of philosophy is to correct
misunderstanding.
Isn't it the same string I harped on again and again?
funny.

Regards
Ozaki Tosiaki


.

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