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I want to truncate a simple number to n decimal digits. For example, 2/3. I used

f[x_, n_] := N[IntegerPart[x 10^n]/10^n]

but I get

f[2/3, 20] = 0.666667

Is there a way to get

f[2/3, 20] = 0.66666666666666666666 ?

I have searched a lot but I haven't really found an answer. I am looking for something like N or Round for truncation.

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  • $\begingroup$ just try N[2/3, 20] $\endgroup$
    – ZaMoC
    Commented Nov 27, 2018 at 12:10
  • $\begingroup$ N rounds the last decimal digit. I get N[2/3, 20] = 0.66666666666666666667 $\endgroup$
    – Alex
    Commented Nov 27, 2018 at 12:12
  • 1
    $\begingroup$ You probably want to use RealDigits[2/3, 10, 20] $\endgroup$
    – Carl Lange
    Commented Nov 27, 2018 at 12:18
  • $\begingroup$ NumberForm[N[2/3], 30] stops after 15 to 16 decimals but does not round up. $\endgroup$
    – Titus
    Commented Nov 27, 2018 at 12:37

2 Answers 2

Reset to default
1
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ClearAll[f]
f[x_, n_] := 
Module[{s = RealDigits[x, 10, n + IntegerLength@IntegerPart[x]]}, 
N[FromDigits[s[[1]]]/10^(n), n + s[[2]]]];

f[2/3, 20]

0.66666666666666666666

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2
$\begingroup$

Trunc[x_, n_] := N[FromDigits[RealDigits[x, 10, n]], n] also works.

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  • $\begingroup$ Nice, I made a change in my code so that you always get n decimal digits like f[5000/3, 20]=1666.66666666666666666666. thanx for accepting $\endgroup$
    – ZaMoC
    Commented Nov 27, 2018 at 12:43

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