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# Why does 1 - 2/3 - 1/3 not equal zero?

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Ned Gulley on 13 Jan 2011
Edited: James Tursa on 7 Dec 2020
Try this:
>> 1 - 2/3 - 1/3
MATLAB gets the wrong answer:
5.5511e-017
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James Tursa on 14 Jul 2011

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### Accepted Answer

Matt Fig on 20 Jan 2011
Ned! I am sure you know that the reason why (1 - 2/3 - 1/3) ~=0 is that MATLAB does its work in floating point arithmetic, not in exact arithmetic.
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Ned Gulley on 20 Jan 2011
Hi Matt: Yes, this is an obvious one to get the ball rolling. By the way, I changed the formatting of your answer slightly to make the link a little clearer.

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### More Answers (3)

Doug Hull on 21 Jan 2011
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 a readable introduction to floating point arithmetic, look at Cleve's Corner article from 1996: Floating Points (PDF)
For more rigorous and detailed information on floating point arithmetic, read the following paper: What Every Computer Scientist Should Know About Floating Point Arithmetic.
Another resource is Technical Note 1108 on the Support section of The MathWorks website.
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Amir on 29 Jul 2011
I wrote my code wondering if matlab must have it!
function [am] = cancelout(a,tol) %for small gain probelm
for k=1:size(a,1)
for m=1:size(a,2)
[z,gain] =zero(a(k,m));
if ( abs(gain)<tol)
a(k,m)=0;
end
gain=NaN;
end
end
am=a;

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Derek O'Connor on 29 Jul 2011
Ned,
Let me add to the confusion by asking
1. Why does en1 = (1 - 2/3) - 1/3 = 5.5511e-017 = 2^(-54) = eps/2^2?
2. Why does en2 = 1 - (2/3 + 1/3) = 0?
3. Why does Kahan's
a = 4/3; b = a-1; c = b+b+b; ek = 1-c = 2.2204e-016 = 2^(-52) = eps?
An explanation of Kahan's result is given on page 7 of:
The difficulties in defining machine precision are discussed by Nick Higham and Cleve Moler here:
Derek O'Connor
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Skaletki Liviu on 7 Dec 2020
how I write this expression |n-(N-1)/2| ?
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Walter Roberson on 7 Dec 2020
abs(n-(N-1)/2)

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