quasi wrote: >Hetware wrote: >> >>So Quasi was correct. > >Of course! > >(just kidding) > >>I can define a function g(x) to be continuous, > >I don't think I said that. > >>but it is not the same function as the function f(x) used >>to define g(x) everywhere f(x) is defined. > >As I've said before, the wording is key. > >Care needs to taken so as to be both sufficiently rigorous >and sufficiently clear. > >Let's take some examples. > >Define f by: > > f(x) = (x^- 9)/(x - 3), x! = 0.
f(x) = (x^- 9)/(x - 3)
>Then: > >* f(3) is undefined. > >* f is defined for all real numbers _except_ 3. > >* f is continuous on its domain. > >* f is not continuous at x = 3. > >* f has a removable discontinuity at x = 3. > >Now define g by: > > g(x) = f(x) for x != 3 > > g(3) = 6 > >Then: > >* g is continuous on R > >* g(x) = x + 3 for all x in R > >Remark: > >Note that I didn't _define_ g to be continuous. > >Rather, I defined g, then concluded that g is continuous. > >That assertion requires _proof_, which, although I didn't >give it, is easily supplied: > >(1) Since f is continuous for on its domain, g is continuous >when x != 3. > >(2) At x = 3, we have > > g(3) = 6 = lim (x -> 3) f(x) = lim (x -> 3) g(x) > >so g is continuous at x = 3. > >It follows that g is continuous on R. > >But once again, I didn't define g as continuous on R. Rather, >I defined g and then asserted that g is continuous on R, >initially without proof. Omitting the proof is acceptable >provided, for the intended audience, the truth of the claim >is obvious and the proof is easily supplied if requested. > >Let's try a similar example with a change of wording: > >Let h be a continuous function such that > > h(x) = f(x) for x != 3 > >Does the above qualify as a "definition" of h? > >For the above to be regarded as a definition, it must be the >case that there is one and only one function h which satisfies >the above conditions. But you would first have to _prove_ that >claim. Only then can you claim to have "defined" h. > >In other words, imposing conditions on h is not the same as >defining h. > >Now consider a sample test problem ... > >Suppose k(x) is continuous on R and such that > > k(x) = (x^2 - 9)/(x - 3) > >for x != 3. Must k(3) = 6? > >The answer is yes. > >But the question didn't _define_ k(x). Rather, it specified >some conditions on k(x) and then asked a question regarding >such a function k.