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
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