Bug in equality checks? 28.15e-3 == 28.15/1000 false

Why does this happen? It does not regocnize equality on 7.48e-3 == 7.48/1000, but it does on 7.47e-3 == 7.47/1000 as well. Another example below.

28.1e-3 == 28.1/1000

ans = logical
   1

28.15e-3 == 28.15/1000

ans = logical
   0

Answer 1

[Star Strider]

Wellcome to the world of floating-point approximation error!

See Why is 0.3 - 0.2 - 0.1 (or similar) not equal to zero? and Floating-Point Numbers

This is roughly equivalent to expressing in decimals. The result is 0.3 an infinite number of 3 following it, never being excactly equal to .

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[Gustavo Lunardon]

Thanks for the information. I thought it would consider the floating point error the same way on both sides of the equation so I thought it was a bug as for some cases worked and some did not.

If I am looking for values with decimals on tables, how do I bypass this error? Is there a smart way or I need to check if the value is within value+- 10*eps for example?

[Star Strider]

My pleasure!

The usual approach is to include a tolerance, something like this (using the requested criterion of 10*eps) —

TF1 = (28.15e-3 - 28.15/1000) == 0

TF1 = logical
   0

TF2 = (28.15e-3 - 28.15/1000) <= 10*eps

TF2 = logical
   1

So instead of testing for equality, do essentially the same thing with a subtraction, test it against the criterion, and return a logical value result.

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[Gustavo Lunardon]

Thank you again!

[Star Strider]

As always, my pleasure!

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[Paul]

Shouldn't that be:

TF2 = abs(28.15e-3 - 28.15/1000) <= 10*eps

[Stephen23]

" I thought it would consider the floating point error the same way on both sides of the equation"

Nope. In general, different operations will accumulate different floating point error.

Writing a number is quite a different operation to dividing two other numbers.

[Paul]

Then I'm very confused. If the idea is to test for "equality" then not including the abs check will lead to results like this:

TF2 = (-10 - 28.15/1000) <= 10*eps

TF2 = logical
   1

[Image Analyst]

I think the abs() is needed because one quantized expression could be above or below the other quantized expression, and I don't think you'd know which it will be in general.

[Paul]

Exactly and well said. I should have included a statement to that effect in my comment

Answer 2

[Image Analyst]

See the FAQ:

https://matlab.fandom.com/wiki/FAQ#Why_is_0.3_-_0.2_-_0.1_.28or_similar.29_not_equal_to_zero.3F

and use ismembertol() like it recommends.