Re: Is infinity equal to infinity?

Erik Runeson <erikr@iar.se> wrote:
>Inf == Inf ?
>The IEEE 754 standard for floating-point arithmetics does not say
>anything about this case...


Actually it does, but careful reading is needed. A good resource for
interpretation of IEEE 754 is David Goldberg's paper "What Every Computer
Scientist Should Know About Floating-Point Arithmetic", in the March 1991
issue of ACM Computing Surveys. It contains a discussion which clarifies
the somewhat cryptic initial statement of section 6.1: "Infinity
arithmetic shall be construed as the limiting case of real arithmetic with
operands of arbitrarily large magnitude, when such a limit exists." The
paper discusses 0/0 vs. 1/0, and the argument is trivially modified to
deal with comparisons.


First, a note: IEEE comparison yields one of four results -- "less than",
"equal", "greater than", and "unordered" -- and then defines at some
length how these map into true/false for various predictates. The mapping
isn't always obvious, and there are rather more useful predicates than the
conventional six.


So, what about Inf:Inf (that is, infinity compared to infinity)? The
paper explains this, as per the standard's terse wording, as the limit of
f(x):g(x), as x approaches some limit that makes both f(x) and g(x) go to
infinity. Just what is the limit of f(x):g(x)? Depending on what f(x)
and g(x) are, it could be "less than", "equal", or "greater than". So
there is no single limit... and that means the result of Inf:Inf must be
"unordered".


And the table on page 13 of the standard specifies that the .EQ. predicate
(one of the first six, which according to 5.7 are the ones that map into
the traditional six predicates of programming languages) yields false,
and does not cause an exception, for a comparison result of "unordered".
So (Inf==Inf) is false.


Sigh, I doubt that many compiler writers or silicon builders have gotten
this right... It's awfully subtle.
--
  | Henry Spencer henry@spsystems.net
  | (aka henry@zoo.toronto.edu)
--

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