To me it looks that it is:irreflexivity: NaN < NaN == falsetransitivity: if a < b and b < c then a < c (the antecedent is never true for NaNs)asymmetry: if a < b then not b < a (again, the antecedent is never true for NaNs)However, it cannot be extended to non-strict partial ordering <= because there's no (reflexive) equality for doubles (NaN != NaN).
Is < binary relation a strict partial order on IEEE doubles?
floating point;partial order
Yes, it is. (Though you need to consider signed zero as well as NaNs.) For extra confirmation:So, the floating-point operator< does not form weak order and therefore does not form a total order. It is, however, a partial order. However, it cannot be extended to non-strict partial ordering <= because there's no (reflexive) equality for doubles (NaN != NaN).It can be extended, but <= as defined by IEEE isn't that extension.