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| From: Thomas Bushnell BSG <tb@xxxxxxxxxx> | Date: Sat, 22 Oct 2005 17:51:06 -0700 | | Aubrey Jaffer <agj@xxxxxxxxxxxx> writes: | | > | From: "Marcin 'Qrczak' Kowalczyk" <qrczak@xxxxxxxxxx> | > | Date: Sat, 22 Oct 2005 20:52:50 +0200 | > | | > | Aubrey Jaffer <agj@xxxxxxxxxxxx> writes: | > | | > | > The total order of the reals is a crucial property for many | > | > applications. Any subset of the reals has a total order. | > | > The reals with +inf.0 and -inf.0 have a total order. | > | > | > | > But the reals with +inf.0, -inf.0, and +nan.0 do not have a total | > | > order [because (<= 5 +nan.0) and (>= 5 +nan.0) would both be false]. | > | | > | It is well known that the default order on the floating point | > | approximation of reals is not total. The floating-point approximations to the reals do not include NaN. (which real number would it be approximating?) | > From Wikipedia, the free encyclopedia. | > <http://en.wikipedia.org/wiki/Total_order> | > | > In mathematics, a total order, linear order or simple order on a set | > X is any binary relation on X that is antisymmetric, transitive, and | > total. This means that, if we denote the relation by <=, the | > following statements hold for all a, b and c in X: | > | > if a <= b and b <= a then a = b (antisymmetry) | > if a <= b and b <= c then a <= c (transitivity) | > a <= b or b <= a (totalness) | > | > Which condition does it violate? | | Totalness (as Marcin said). NaN comparisons (other than | not-equals) always evaluate false. My point exactly! NaN violates totalness. Thus it is not a real number. Having (real? +nan.0) ==> #f and +nan.0 be an illegal argument to >, <, <=, and >= is compatible with IEEE-754. Just because IEEE-754 defines a behavior for comparisons with NaN doesn't mean Scheme must redefine <, ... so that it accepts non-real arguments. Scheme already has lots of non-real numbers which are illegal arguments to >: (> 5 5.+3.i)