Re: TI-M: Re: I think you're a weenie!!!
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Re: TI-M: Re: I think you're a weenie!!!
At 10:21 PM 5/17/2000 -0500, you wrote:
>umm.. dont you start out with different values to begin with? (1/3)x^3 !=
>(2/3)x^3... and those different values invalidates the rest of it....
More importantly, if (1/3)x^3 + C1 = (2/3)x^3 + C2, then the constants have
to be such a value that C2 = (-1/3)x^3 + C1. (Each C can be a different value.)
Since that can never be true for all x, 1=2 for only one value of x.
And I believe, if you want those two constant values to be equal, X has to
equal 0.
Therefore C1 = C2.
.......
Weren't we proving 1 = 2 here? :)
(C1 is a variable, C2 is a variable as well. So don't tell me you can
cancel out a C.)
--
Nick Disabato <nickd@ticalc.org> <http://nickd.org>
News Editor, Featured Programs, Room Service
the ticalc.org project - http://www.ticalc.org/
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