Number definition and approximation

Question

I run these lines:

a = 0.833
SetPrecision[a, 20]

and this is the output:

0.833

0.83299999999999996270

I expected to receive

0.83300000000000000000

Do you have any suggestion? Thank you!

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. — Alx
– Alx, Commented Aug 25, 2019 at 16:40
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. — Michael E2
– Michael E2, Commented Aug 25, 2019 at 20:34
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. — Bill Watts
– Bill Watts, Commented Aug 25, 2019 at 21:58

Rohit Namjoshi · Accepted Answer · 2020-01-22 23:14:25Z

1

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

Rohit Namjoshi

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answered Aug 25, 2019 at 18:19

Somos

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