Fw: can someone answer my question?
Walter Eric Johnson
wej3715 at scully.tamu.edu
Mon Dec 7 23:17:35 EST 1998
kkollins at pop3.concentric.net wrote:
: > So if g(x)=h(x), you can define a function f(x)=g(x)-h(x). Then those
: > values of x for which g(x)=h(x) also result in f(x)=0. Thus, the solutions
: > of the equation g(x)=h(x) are the values of x for which f(x)=g(x)-h(x)=0.
:
: There's only one Solution to any Equation... that which Maps its entire numerical-domain.
Nonsense. The equation f(x)=x^2-1=0 has two solutions, not one.
They are x=1 and x=-1.
: It doesn't
: matter what the books 'say"...
Of course. Not only do you make up your own buzzwords to try to
bamboozle people, but you also make up your own meanings for the
words everyone else uses.
: all one has to do to Prove the Veracity of the One-Generalized-Solution
: view is subject a piece-by-piece "problem-solver" to more individual instances of the Generalized
: Solution than the piecemeal Calculator can Calculate in its piecemeal way...
Are you absolutely positive your an expert mathematician? I'd expect
any expert mathematician to understand what it means to prove something.
: > In other words, when you are solving the equation g(x)=h(x), you are
: > finding the zeros of the function f(x)=g(x)-h(x).
:
: One zero does not a Solution make.
5 is a solution to x^2=25. 5 is a zero of f(x)=x^2-25. Are you
saying that 5 is not a solution to x^2=25? Are you really, truly
absolutely positive that you are an expert mathematician?
(By the way, x^2 means x squared.)
Eric Johnson
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