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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