## Float is float [Software]

Hi all,

This is very trivial story... let's take

a = 0.1

next float is:

julia> nextfloat(0.1) 0.10000000000000002

so

julia> nextfloat(0.1) - 0.1 1.3877787807814457e-17

float numbers is discrete... you can't use 0.10000000000000001 ...

you can use rational numbers:

julia> 5//10 - 4//10 - 1//10 == -4//10 - 1//10 + 5//10 true

or some packages for working with big decimals

julia> using Decimals julia> decimal(0.5)-decimal(0.4)-decimal(0.1) == decimal(-0.4)-decimal(0.1)+decimal(0.5) true

I think in R it can be done... and never use Excel for cacl...

edit: or you can do float(big(1//10)) if you just need more precision...

edit2:

❝ which is just “noise”

you should be very careful with eps() because this is not an indicator of accuracy

you can use something like:

a = 1e-70 b = 2e-70 b - a

end it will work fine, problems begans whe you do somesing like:  2.0 + 1e-70 , but you can still use  2.0 * 1e-70  without problems ...

make sure not to sum numbers with difference in eps() times ...

for comparation

all.equal(target, current, tolerance = sqrt(.Machine\$double.eps))

can be used, but for small numbers you should adjust tolerance  Ing. Helmut Schütz 