Exact result of 2^20 + 2^-20

[mohamed-180]

Based on How can I get the exact result of 10^20 + 10^-20 in Python? , I have tried this

using DecFP
using Printf
a = Dec128(10)^20
b = Dec128(10)^-20

@sprintf("%.20f", a + b)
# > "100000000000000000000.000000000000000000000"

where it must be "100000000000000000000.00000000000000000001"

Note that

@sprintf("%.20f", b)

#> "0.00000000000000000001"

Why a + b miss the fraction.

Regards;

[stevengj]

Dec128 corresponds to 34 significant digits in decimal:

julia> precision(Dec128, base=10)
34

Adding 10^-20 to 10^20 would require 41 significant digits. As a result, the 20th decimal place in the result gets rounded off to 0.

This is fundamental to how floating-point arithmetic, of any type, works — only ever have a finite number of significant digits.

To get 10^20 + 10^-20 exactly, you'd need to use a type with even more precision (more digits) than Dec128. You could use an arbitrary precision type like Julia's built-in BigFloat type:

julia> setprecision(100, base=10); # use roughly 100 decimal digits

julia> big"1e20" + big"1e-20"
1.00000000000000000000000000000000000000010000000000000000000000000000000000000000000000000000000000003e+20

Even this is not quite exact — see the 3 in the 80th digit — because BigFloat is a "binary" floating-point type, which represents integers times powers of 2 exactly (up to some number of binary digits), but not integers times powers of 10, so 1e-20 needs to be rounded to represent it as the nearest BigFloat. To do this operation exactly you'd need a high-precision decimal floating-point type, e.g. Decimals.jl.

See also PSA: Floating-point arithmetic on the Julia discussion board.