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| ``Re: Equivalence of equations''
by lvoyster on 2009-11-03 16:27:35 |
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| We have two functions here.
f(x) = (1/x)*x where x is in R-{0}
g(x) = 1 where x is in R
The short answer is that f(x)=g(x) almost everywhere.
We did not "remove" 0 from the domain of f. It was never there because, in the field of real numbers, 0 does not have a multiplicative inverse.
We can note that g(x) is continuous at x=0 and that, if we define F(x) by
F(x) = (1/x)*x when x is in R-{0}
F(x) = 1 when x=0
then we have 'extended' f to the continuous function F and that F(x) = g(x) for all x.
This is not a trivial problem. Indeed the study of the ratio 0/0 led to the formalization of the differential Calculus and the resulting theory is a template for epistemology. |
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