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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 <> wrote:
> 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.

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?"
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Posted via http://www.ruby-forum.com/.

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
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Jabber:
ICQ: 205544826

> 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 + /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

Thanks a lot for the answers and references, and sorry for not having
check deeper on google
(http://blade.nagaokaut.ac.jp/cgi-bin...ruby-core/9655).

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 Moss 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 + /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

While your answer is correct, you cannot subtract infinities as shown
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.

--
Posted via http://www.ruby-forum.com/.

On Mon, Oct 27, 2008 at 11:09 AM, Todd Benson <> 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

-------- Original-Nachricht --------
> Datum: Tue, 28 Oct 2008 04:31:06 +0900
> Von: "guille lists" <>
> An: ruby-
> Betreff: Re: float equality

> Thanks a lot for the answers and references, and sorry for not having
> check deeper on google
> (http://blade.nagaokaut.ac.jp/cgi-bin...ruby-core/9655).
>
> 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

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


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

-----Original Message-----
From: Tim Pease [private.php?do=newpm&u=]
Sent: Monday, October 27, 2008 1:11 PM
To: ruby-talk ML
Subject: Re: float equality

On Mon, Oct 27, 2008 at 11:09 AM, Todd Benson <> 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