Re: Negate, Augment, Denominate

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NAD.  Note 2

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Let me just follow my nose for a while
until I can recall what this was about.

Playing around -- let's call it "experimenting" --
with a selection of not especially random test
cases from the real line R, one may observe
that the NAD operator has a period of 3
on at least some values of x in R.

For example:

(1/3)NAD   =  1/(1-(1/3))   =  1/(2/3)   =  3/2
(3/2)NAD   =  1/(1-(3/2))   =  1/(-1/2)  =  -2
(- 2)NAD   =  1/(1-(- 2))   =  1/(+ 3)   =  1/3

   (3)NAD  =  1/(1 - 3)     =  1/(- 2)   =  -1/2
(-1/2)NAD  =  1/(1-(-1/2))  =  1/(3/2)   =   2/3
 (2/3)NAD  =  1/(1-(2/3))   =  1/(1/3)   =   3

So we can write:

1/3 >~NAD~>  3/2 >~NAD~> -2  >~NAD~> 1/3

 3  >~NAD~> -1/2 >~NAD~> 2/3 >~NAD~>  3

I suppose the next thing to think about is whether
this is just coincidence, or whether there is some
algebraic reason for it.

Jon Awbrey

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inquiry e-lab: http://stderr.org/pipermail/inquiry/
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