Re: Lens Algebra


vreddyp@xxxxxxxxx wrote:

Here are the sets that are defined by the particular solution I gave
(1/x, 1-x, x/(x-1)).

A: +inf to 2
B: 2 to 1
C: 1 to 1/2
A': 1/2 to 0
B': 0 to -1
C': -1 to -inf

That doesn't sound precise enough. Let me make one more attempt.

A: {+ inf >= x >= 2 }
B: { 2 >= x >= 1}
C: {1 >= x >= 0}
A': {1/2 >= x >= 0}
B': {0 >= x >= -1}
C': {-1 >= x >= -inf}

Hmm. The sets are not exactly disjoint but overapping at their common
boundaries. What if I replace +inf and -inf with a single point +/-inf
to make it a common boundary between A and C' ? The notions of 'Less
than" and "Greater than" would then become circular and directional,
meaning for example, 5 could be less than or greater than 7 depending
on which path you take. We can possibly calculate two answers for the
difference between 7 and 5 as 2 and something else in terms of these
circular operations. If "inf" is the one that spoils all this fun, then
we can replace it by sufficliently large number suitable for the
context.

Anyway, I don't want to venture into these complexities, at this point.

- venkat

.

Relevant Pages