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Topic: ran(EF) contains ~EF(n)
Replies: 9   Last Post: May 1, 2013 10:16 AM

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 ross.finlayson@gmail.com Posts: 2,720 Registered: 2/15/09
ran(EF) contains ~EF(n)
Posted: Apr 30, 2013 11:11 AM

Basically this EF is the limit of a function. Not-a-real-function,
it's modeled by real functions and in fact very simply n/d. Then, the
properties of its range have that: it is symmetric about some
midpoint in its range 1/2, for f^-1(1/2), defined for even d. Just as
ran(f) starts from .000... and increases in constant monotone, in
reverse it starts with .111... (or for example in decimal, .999...)
and decreases. Then, as there is a less than finite difference
between elements of the range for consecutive elements of the domain
by their natural ordering, there is in the range an element that
starts with .1, .11, .111, ..., and each finite initial segment of
integers as expansion. This is from seeing that the elements of
ran(f) are the same elements of ran(REF) for REF the "reverse"
equivalency function.

EF: n/d
REF: (d-n)/d

As there exists d-n for each n and d, there exists here ~EF(n) =
REF(n) constructively, and each element in ran(EF) is in ran(REF).

Then, here, constructively: ~EF(n) e ran(EF), quod erat
demonstrandum.

There are replete properties of this function and its range that its
range is R_[0,1].

Regards,

Ross Finlayson