# Matlab accuracy (when 1-1~=0)

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Damian Maxwell el 20 de Sept. de 2022
Comentada: Damian Maxwell el 21 de Sept. de 2022
Below please find a screenshot from a sample script that really puzzles me.
When defining simple variables and then substracting them there appear to be tiny error (3e-18) that comes in with surprising results.
It would be great if somebody could explain why this happens and how to avoid it.
w=0.026
b=0.024
i=0.001
w-0.026
b-0.024
i-0.001
w-(b+2*i)
w==(b+2*i)
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### Respuestas (1)

Eric Delgado el 20 de Sept. de 2022
It's float operation universe. :)
w=0.026;
b=0.024;
i=0.001;
w-0.026;
b-0.024;
i-0.001;
w == (b+2*i)
ans = logical
0
Use:
abs(w - (b+2*i)) <= 1e-5 % You could use 1e-17 and still will receive a true logical value
ans = logical
1
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Walter Roberson el 20 de Sept. de 2022
1/10 is not exactly representable in finite binary floating point -- for the same mathematical reason that 1/3 is not exactly representable in finite decimal.
Suppose we were working in decimal and said 1/3 = 0.3333333333 then if we add those together 3 times we get 0.9999999999 -- which in decimal is distinct from 1.0 . In decimal you need a literally infinite number of digits of precision for the 0.3-repeated added together 3 times to produce a result that is mathematically exactly equal to 1 .
So the problem is not exactly with the fact that MATLAB uses binary: the problem is that if you choose any finite-length fixed-point base, there will always be rational fractions that cannot be exactly represented in finite length. It is an unfortunate mathematical limitation of the Universe.
Damian Maxwell el 21 de Sept. de 2022
Thank you so much Walter for the clear explanation.

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