Is < binary relation a strict partial order on IEEE doubles? To me it looks that it is:


irreflexivity: NaN &lt; NaN == false
transitivity: if a &lt; b and b &lt; c then a &lt; c (the antecedent is never true for NaNs)
asymmetry: if a &lt; b then not b &lt; a (again, the antecedent is never true for NaNs)


However, it cannot be extended to non-strict partial ordering &lt;= because there's no (reflexive) equality for doubles (NaN != NaN).
Yes, it is. (Though you need to consider signed zero as well as NaNs.) For extra confirmation:


So, the floating-point operator&lt; 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 &lt;= because there's no (reflexive) equality for doubles (NaN != NaN).

It can be extended, but &lt;= as defined by IEEE isn't that extension.