Matlab accuracy (when 1-1~=0)

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Damian Maxwell
Damian Maxwell 2022-9-20
评论： Damian Maxwell 2022-9-21
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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回答（1 个）

Eric Delgado
Eric Delgado 2022-9-20
编辑：Eric Delgado 2022-9-20
Ran in:
It's float operation universe. :)
w=0.026;
b=0.024;
i=0.001;
w-0.026;
b-0.024;
i-0.001;
Instead of:
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
Walter Roberson 2022-9-20
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
Damian Maxwell 2022-9-21
Thank you so much Walter for the clear explanation.

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