I run these lines:

a = 0.833
SetPrecision[a, 20]

and this is the output:



I expected to receive


Do you have any suggestion? Thank you!

  • 2
    $\begingroup$ 0.833'20 gives what you expect, but SetPrecision[a,20] is equivalent to N[FromDigits[RealDigits[a,2],2],20] = 0.83299999999999996270. MMA works with binary numbers. $\endgroup$ – Alx Aug 25 '19 at 16:40
  • $\begingroup$ It works with 0.833'20, but not with a'20 $\endgroup$ – Andrea2810 Aug 25 '19 at 17:59
  • $\begingroup$ The back tick is used in specifying numeric input. Using it on variables thus a`20 is not syntactically valid. The problem is that a = 0.833 evaluates 0.833 in machine precision binary and obtains the binary floating point number nearest to 0.833, but that number is only approximately equal to 0.833. $\endgroup$ – Michael E2 Aug 25 '19 at 20:34
  • $\begingroup$ Related: mathematica.stackexchange.com/questions/55292/… $\endgroup$ – Michael E2 Aug 25 '19 at 20:38
  • 1
    $\begingroup$ You don't need a'20. Just use a=0.833'20 and a will have 20 digits of precision from then on or until it is reassigned. That is supposed to be a back tick, but I can't write it that way because of the way comments treat back ticks. $\endgroup$ – Bill Watts Aug 25 '19 at 21:58

I suggest SetPrecision[833/1000 , 20] or SetPrecision[.833 // Rationalize, 20].


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