Matlab accuracy (when 1-1~=0)

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)

Answer 1

[Eric Delgado]

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

[John D'Errico]

w=0.026;
b=0.024;
i=0.001;

Then you wrote these lines, which do effectively nothing in MATLAB, except do a computation, and then dump it directly into the bit bucket, not even displayed in the command window.

w-0.026;
b-0.024;
i-0.001;

I could have removed the semi-colons, as else, the result is not even displayed. But even then w - 0.026 will still be EXACTLY zero.

(w-0.026) == 0

ans = logical
   1

(b-0.024) == 0

ans = logical
   1

(i-0.001) == 0

ans = logical
   1

As you can see, in each case, the result is exactly zero. That is to be expected. It is only the computation that produces a non-zero result due to floating point arithmetic.

w - (b + 2*i)

ans = -3.4694e-18

[Damian Maxwell]

Thanks John. 'Floating point arritmetric' is a great pointer. I am aware that the above script does absolutely nothing. I created it purely to highlight an issue I encountered in a more complex data processing script where I relied on the logic w>=(b+2i) to return a '1' which it didn't. I have found a work around.

[Damian Maxwell]

Thanks Eric, that works!

[Walter Roberson]

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]

Thank you so much Walter for the clear explanation.