Is < binary relation a strict partial order on IEEE doubles?
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.