MATLAB: FAQ: Why is 0.3 – 0.2 – 0.1 not equal to zero
faq_fpfaqlistfloating pointlimited precision
Why does
0.3 - 0.2 - 0.1 == 0
or
v = 0:0.1:1;
any(v == 0.3)
(or similar numbers) replyfalse?
Best Answer
0.3 - 0-2 - 0.1 returns-2.7756e-17.
As is mentioned frequently in the newsgroup, some floating point numbers can not be represented exactly in binary form. So that's why you see the very small but not zero result. See EPS.
The difference is that 0:0.1:0.4 increments by a number very close to but not exactly 0.1 for the reasons mentioned below. So after a few steps it will be off whereas [0 0.1 0.2 0.3 0.4] is forcing the the numbers to their proper value, as accurately as they can be represented anyway.
a = [0 0.1 0.2 0.3 0.4];
b = 0:.1:.4;
as = sprintf('%20.18f\n',a)
>> as =
0.000000000000000000 % ==
0.100000000000000010 % ==
0.200000000000000010 % ==
0.299999999999999990 % ~= bs !
0.400000000000000020 % ==
bs = sprintf('%20.18f\n',b)
>> bs =
0.000000000000000000 % ==
0.100000000000000010 % ==
0.200000000000000010 % ==
0.300000000000000040 % ~= as !
0.400000000000000020 % ==
and:
format hex;
hd = [a.',b.']
>> hd =
0000000000000000 0000000000000000 % ==
3fb999999999999a 3fb999999999999a % ==
3fc999999999999a 3fc999999999999a % ==
3fd3333333333333 3fd3333333333334 % ~= !
3fd999999999999a 3fd999999999999a % ==
If you're trying to compare two floating-point numbers, be very careful about using == to do so. An alternate comparison method is to check if the two numbers you're comparing are "close enough" (as expressed by a tolerance) to one another:
% instead of a == b
% use:
areEssentiallyEqual = abs(a-b) < tol
% for some small value of tol relative to a and b
% perhaps defined using eps(a) and/or eps(b)
You can see this same sort of behavior outside MATLAB. Using pencil and paper (or a chalkboard, or a whiteboard, etc.) compute x = 1/3 to as many decimal places as you want. The number of decimal places must be finite, however. Now compute y = 3*x. In exact arithmetic, y would be exactly 1; however, since x is not exactly one third but is a rounded approximation to one third, y will not be exactly 1.
For more rigorous and detailed information on floating point arithmetic, read the following paper: What Every Computer Scientist Should Know About Floating Point Arithmetichttp://docs.sun.com/source/806-3568/ncg_goldberg.html
To see what is going on you can examine the exact floating point values. Note that none of your original numbers except 0.5 can be represented in IEEE double format exactly:
Best Answer