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Is < binary relation a strict partial order on IEEE doubles?
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orderdoublesieeebinarypartialstrictrelation
Problem
To me it looks that it is:
However, it cannot be extended to non-strict partial ordering <= because there's no (reflexive) equality for doubles (NaN != NaN).
- irreflexivity: NaN
- transitivity: if a
- asymmetry: if a
However, it cannot be extended to non-strict partial ordering <= because there's no (reflexive) equality for doubles (NaN != NaN).
Solution
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
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.Context
StackExchange Computer Science Q#50991, answer score: 4
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