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Re: Evaluating Min[2^(1/2),3^(1/2)]


Using David Jacobson's name and generalizing Todd Gayley's solution
by replacing the outer First@ with Sequence@@, we get

ExactMin[items__] :=
  Module[ {nlist = N /@ {items}},
          {items}[[ Sequence@@First at Position[nlist,Min[nlist]] ]]
        ]

ExactMin can be applied to a list or a sequence of numbers, lists.
E.g.

ExactMin[Sin[1], {Cos[1], 2 - 2^(1/2)} evaluates to Cos[1].

Thanks for the help.

Roger Kirchner





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