Suppose I want to have a function (call it NLess) that given two real numeric quantities, a and b, it returns True if a is smaller than b, ignoring precision. Here are some test cases:

NLess[3.14`1, Pi]
NLess[E, Pi]
NLess[3.14159265358979323846263`22, Pi]
NLess[4.49`1, 9/2]
NLess[E, 2.782`1]

All of these should return True. The usual Less function only returns true for Less[E, Pi]:

Less[3.14`1, Pi]
Less[E, Pi]
Less[3.14159265358979323846263`22, Pi]
Less[4.49`1, 9/2]
Less[E, 2.782`1]







I have some code to do this, but I find it unsatisfactory.

One further detail. It is acceptable to me to have both:

 N[3.14`1, 3.14`100]
 N[3.14`100, 3.14`1]

be false, or one or the other to be True, but it would be nice if this behavior could be controlled

  • $\begingroup$ I take it that 3.14`1 should be treated as though it were 3.14 exactly and that that the machine precision (MP) number 4.49999999999999` should be treated as 5066549580791797/2^50 if MP = binary64 (which might depend on machine precision)? Or should MP numbers be rounded to their decimal equivalents displayed by InputForm? $\endgroup$ – Michael E2 Jan 26 '17 at 2:17
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    $\begingroup$ Is this "satisfactory? NLess[a_, b_] := Less[Rationalize[ SetPrecision[a, Max[MachinePrecision, Precision[a]]], 0], Rationalize[SetPrecision[b, Max[MachinePrecision, Precision[b]]], 0]]. It actually only works because, to my surprise, Rationalize[Pi,0] still gives Pi (which is not a rational number). $\endgroup$ – Felix Jan 26 '17 at 2:17
  • $\begingroup$ @Felix Should NLess[0.3333333333333333`, 1/3] return True? $\endgroup$ – Michael E2 Jan 26 '17 at 2:22
  • $\begingroup$ Indeed, the working precision must be increased beyond the provided precision. How about this? NLess[a_, b_] := Less[Rationalize[ SetPrecision[a, Max[MachinePrecision, Precision[a]] + 1], 0], Rationalize[ SetPrecision[b, Max[MachinePrecision, Precision[b]] + 1], 0]] $\endgroup$ – Felix Jan 26 '17 at 2:26
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    $\begingroup$ But is 4.49999999999999` identical to double-precision floating-point fraction 5066549580791797/2^50 or to decimal input fraction 449999999999999/10^14? It can't be both. And what to do about comparing it to 4.49999999999999`15 would seem to depend on the answer. I would have assumed it was the internal binary representation, but somehow you've made me doubt that. $\endgroup$ – Michael E2 Jan 26 '17 at 2:54

How about this?

NLess[a_, b_, p_: 1000] := Order @@ N[{a, b}, p] == 1

The third (optional) parameter $p$ controls the precision used for exact numbers and numbers with precision greater than $p$. Order appears to ignore Internal`$SameQTolerance etc.

| improve this answer | |
  • $\begingroup$ Mr.Wizard what does the : 1000 near the pattern do? i have never seen it before $\endgroup$ – Alucard Sep 2 '17 at 12:44
  • $\begingroup$ @Alucard FullForm[p_: 1000] gives Optional[Pattern[p, Blank[]], 1000] so you should look at Optional $\endgroup$ – Mr.Wizard Sep 2 '17 at 12:45
  • $\begingroup$ ah got it , thank you $\endgroup$ – Alucard Sep 2 '17 at 12:47

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