 # Float equality

Hi,

sorry if this is a very naive question (I’m new to ruby), but I
haven’t found an explication yet. When comparing to floats in ruby I
came across this:

a = 0.1
=> 0.1

b = 1 - 0.9
=> 0.1

a == b
=> false

a > b
=> true

a < b
=> false

a <=> b
=> 1

I’m a bit lost here, shouldn’t (0.1) and (1 - 0.9) be equals regarding
the == operator? I also found that for example 0.3 == (0.2 + 0.1)
returns false, etc.

guille

PD: I’m using ruby 1.8.6 (2008-03-03 patchlevel 114)
[universal-darwin9.0]

On Mon, Oct 27, 2008 at 11:50 AM, guille lists [email protected]
wrote:

a == b
returns false, etc.
Most people will point you to this:
http://en.wikipedia.org/wiki/Floating_point_arithmetic

There are several ways around it (using BigDecimal, Rational, Integers,
etc.)

Totally off-topic, but has any one figured out exactly why 1/9
(0.111…) plus 8/9 (0.888…) is 1 instead of 0.999… Todd

guille lists wrote:

I’m a bit lost here, shouldn’t (0.1) and (1 - 0.9) be equals regarding
the == operator?

Nope. Floating-point is always inexact. This is a computer thing, not
a Ruby thing. The issue applies to all languages everywhere which use
floating point.

Here you can see the two values are slightly different:

irb(main):001:0> 1 - 0.9 == 0.1
=> false
irb(main):002:0> [1 - 0.9].pack(“D”)
=> “\230\231\231\231\231\231\271?”
irb(main):003:0> [0.1].pack(“D”)
=> “\232\231\231\231\231\231\271?”

guille lists wrote:

I’m a bit lost here, shouldn’t (0.1) and (1 - 0.9) be equals regarding
the == operator?

No. The result of 1 - 0.9 using floating point math is not actually 0.1.
In
irb it is displayed as 0.1, but that’s only because Float#inspect
rounds.
Using printf you can see that the result of 1-0.9 actually is
0.09999lots:

printf “%.30f”, 1-0.9
0.099999999999999977795539507497
0.1 itself isn’t actually 0.1 either - it’s
0.100000000000000005551115123126…
Because of this you should not check two floats for equality (usually
you want
to check for a delta or not use floats at all). This is so because of
the
inherent inaccuracy of floating point maths. See this for more
information:
http://docs.sun.com/source/806-3568/ncg_goldberg.html

HTH,
Sebastian

Thanks a lot for the answers and references, and sorry for not having

So I guess that if one wants to work for example with float numbers in
the range [0,1], the best way to do it is by normalising from an
integer interval depending on the precision you want, say [0,100] for
two decimal digits precision, and so on. Is there any other better
approach? Does the use of BigDecimal impose a severe penalty on
performance?

guille

Matthew M. wrote:

Totally off-topic, but has any one figured out exactly why 1/9
(0.111…) plus 8/9 (0.888…) is 1 instead of 0.999… Ummm… because 1/9 + 8/9 == (1 + 8)/9 == 9/9 == 1 ?

And… because 0.999… == 1?

``````x = 0.999...
``````

10x = 9.999…
(10x - x) = 9.999… - 0.999…
9x = 9
x = 1

in your proof. Look at this:

``````  x == 1 - 1 + 1 - 1 + 1 - 1 + 1 - ...
x ==     1 - 1 + 1 - 1 + 1 - 1 + ...
``````

`````` 2x == 1 + 0 + 0 + 0 + 0 + 0 + 0 + ...
x == 0.5
``````

Does x == 0.5? No, because x was never a number in the first place
because the given series does not converge. Your proof appears to
work because you’ve already assumed 0.99999… converges, but that is
what you are trying to prove.

P.S. Euler thought the answer was x == 0.5.

On Mon, Oct 27, 2008 at 11:09 AM, Todd B. [email protected]
wrote:

Totally off-topic, but has any one figured out exactly why 1/9
(0.111…) plus 8/9 (0.888…) is 1 instead of 0.999… I figured it out once, but I can’t remember precisely how it worked.

TwP

Totally off-topic, but has any one figured out exactly why 1/9
(0.111…) plus 8/9 (0.888…) is 1 instead of 0.999… Ummm… because 1/9 + 8/9 == (1 + 8)/9 == 9/9 == 1 ?

And… because 0.999… == 1?

``````x = 0.999...
``````

10x = 9.999…
(10x - x) = 9.999… - 0.999…
9x = 9
x = 1

-------- Original-Nachricht --------

Datum: Tue, 28 Oct 2008 04:31:06 +0900
Von: “guille lists” [email protected]
An: [email protected]
Betreff: Re: float equality

guille

Dear guille,

you could use some approximate equality check:

class Float
def approx_equal?(other,threshold)
if (self-other).abs<threshold # “<” not exact either return true
else
return false
end
end
end

a=0.1
b=1.0-0.9
threshold=10**(-5)

p a.approx_equal?(b,threshold)

Best regards,

Axel

Mathematically 1.(0) is the same thing as 0.(9). Computers simply
represent this to whatever precision they can.

On Oct 27, 2008, at 2:48 PM, The Higgs bozo wrote:

(10x - x) = 9.999… - 0.999…
x == 0.5

Does x == 0.5? No, because x was never a number in the first place
because the given series does not converge. Your proof appears to
work because you’ve already assumed 0.99999… converges, but that is
what you are trying to prove.

But 0.999… does converge, while 1 - 1 + 1 - 1 +… does not.

``````x ==     1 - 1 + 1 - 1 + 1 - 1 + ...
``````

But 0.999… does converge, while 1 - 1 + 1 - 1 +… does not.
On re-reading, I see that you weren’t so much questioning whether
0.999… converges, but that my proof uses circular reasoning. Yeah, I
suppose that’s correct.

Still, 0.999… does converge and is 1. Nyah! I just don’t
remember the better proof I once knew.

The Higgs bozo wrote:

guille lists wrote:

I’m a bit lost here, shouldn’t (0.1) and (1 - 0.9) be equals regarding
the == operator?

Nope. Floating-point is always inexact.

What you mean is that binary floating point inexactly represents decimal
fractions.

This is a computer thing, not
a Ruby thing. The issue applies to all languages everywhere which use
floating point.

…binary floating point.

On Oct 27, 2008, at 2:29 PM, Bilyk, Alex wrote:

Mathematically 1.(0) is the same thing as 0.(9). Computers simply
represent this to whatever precision they can.

Hmmm … my humor is a little to obtuse today. I was hoping the word
“precisely” would cause a mental link to the word “precision” which is
what this problem is all about – precision and computer
representations of floating point values.

Alas. I’ll stick with my day job of writing software.

Blessings,
TwP

This forum is not affiliated to the Ruby language, Ruby on Rails framework, nor any Ruby applications discussed here.