291 views (last 30 days)

Try this:

>> 1 - 2/3 - 1/3

MATLAB gets the wrong answer:

5.5511e-017

Matt Fig
on 20 Jan 2011

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.

Amir
on 28 Jul 2011

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;

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

Opportunities for recent engineering grads.

Apply TodayFind the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!
## 1 Comment

## Direct link to this comment

https://www.mathworks.com/matlabcentral/answers/69-why-does-1-2-3-1-3-not-equal-zero#comment_25586

⋮## Direct link to this comment

https://www.mathworks.com/matlabcentral/answers/69-why-does-1-2-3-1-3-not-equal-zero#comment_25586

Sign in to comment.