On 10/19/2013 6:21 PM, Arturo Magidin wrote: > On Saturday, October 19, 2013 12:55:37 PM UTC-5, Hetware wrote: >> On 10/19/2013 7:34 AM, Peter Percival wrote: >> >>> Hetware wrote: >> >>> >> >>>> If we define f(x) as >> >>>> >> >>>> f(x) = x/x and c[f(x)], >> >>> >> >>> That's cheating. You've been told at least eight million times >> >> >> >> In psychology that's called cognitive distortion, and is an example >> of >> >> how emotions cause our thoughts to be irrational. > > Hey, Mr. Pot. Have you met Ms Kettle? > >>> that you >> >>> may define f with a formula, but you cannot put "and c[f(x)]" in >>> the >> >>> definition. If c[f(x)] is true it has to be proved _from_ the >>> definition. >> >> >> >> Sayin' it don't make it so. > > > Apparently no, you have not met Ms Kettle. > > You made a mistake because you purposely chose to disregard the > convention that the author set forth.
What convention is that?
> Rather than admit this, you are going out of your way to try to prove > that you were right all along.
The proposition P(f) that "f(x) is continuous" appears to have meaning. So either P or !P.
P implies that f(x) is defined everywhere on the proposed domain. It also implies that over the proposed domain limit[f(s), s->x] exists, is finite, and f(x) = limit[f(s), s->x].
It is common to define a function using multiple rules. For example s(x) = sqrt(x^2) for x >= 0 and s(x) = -sqrt(x^2) otherwise. That is, we state a rule and the condition under which the rule applies. If our rules lead collectively to a contradiction, then our definition must be abandoned. It is possible to have two rules with overlapping conditions. For example, s(x) = sqrt(x^2) for x >= 0 and s(x) = -sqrt(x^2) for x <= 0; The condition for both rules is satisfied when x = 0. This is not a problem since both rules produce the same value.
It is possible to define a piecewise function such that p(x) = v(x) where v(x) does not produce an indeterminate form and p(x) = L(x) everywhere.
Given a mapping f(x) for which the limit[f(s), s->x] exists and is finite over the proposed domain, the proposition P(f) implies that f(x) = limit[f(s), s->x] over the proposed domain. So P(f) is true.
!P(f) implies that for some x either f(x) is undefined, lacks a finite limit, or f(x) != limit[f(s), s->x]. These rules do not permit us to use f(x) = limit[f(s), s->x] to determine the value of f(x) because we nothing implies the rule to be true. If f(x) results in an indeterminate form, we have not recourse to the implications P, and we therefore conclude !P(f) is true.
So P(f) = !P(f).
> Well on your way to becoming a crank, despite whatever intelligence > or competence you may have. Congratulations. You are fighting hard to > stay ignorant. >