2009/8/13 <csrabak@bol.com.br>:
Em 12/08/2009 19:30, Igor Stasenko < siguctua@gmail.com > escreveu:
2009/8/12 Ken.Dickey :
"Schwab,Wilhelm K"
Floating point is not always what it seems.
Hence my comment that IEEE floats get "the wrong answer fast". Â I have used interval math, continued fractions, and linear fractional transforms (a.k.a. exact reals). Â I agree that each representation has its challenges.
Let's talk for a second about integers.
 0 = (0+0i)  --> true  1 = (1+0i)  --> true  0 < 1      --> true  (0+0i) < (1+0i)  --> ?? which answer here gives me the least surprise ??
To put it another way
(A = a) Â --> true (B = b) Â --> true (A < B) --> true (a < b) Â --> ?? what do you expect to see here ??
Let me extend your test a little
i do expect that, if :
a < b
and
0 < x
then
a*x < b*x
Your "counter example" is mathematically flawed even in Real (non imaginary):
let a = 1; b = 2 and x = -1:
do you intentionally missed the condition 0 < x?
a < b and a*x > b*x
So there isn't this kind of transitivity for inequality operators at all.
[snipped]
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-- Best regards, Igor Stasenko AKA sig.