Why doesn't .3 - .2 - .1 = 0

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Lucas
Lucas il 9 Nov 2012
I was bored yesterday at work and I was playing around with some decimal numbers when I stumbled across something I thought was pretty interesting.
.3 - .2 - .1 != 0
but
.2 - .1 - .1 = 0
then I began to run some other tests and I found out that only .3 and .8 are the only numbers that you use that'll not get you a 0 answer. If you take:
.4 - .2 - .1 - .1 = 0
and
.9 -.2 -.1 -.1 -.1 -.2 -.1 -.1 = 0
If you change any of the .2's to two - .1's you don't get 0. So what’s so special about those 2 decimals that make them the only ones that don't work with basic subtraction?

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Matt J
Matt J il 9 Nov 2012
Modificato: Matt J il 9 Nov 2012
What makes them special? Nothing. There are lots more numbers like that
>> isequal(0, .7 - .5 - .2)
ans =
0
It also depends on the operation you do. Sometimes 0.3 will work fine, e.g.,
>> isequal(0, .6 - .3 -.3)
ans =
1
  2 Commenti
Matt Fig
Matt Fig il 9 Nov 2012
It is not just in MATLAB. ALL numerical systems that follow IEEE 754 will behave this way. That is the standard.
José-Luis
José-Luis il 9 Nov 2012
Modificato: José-Luis il 9 Nov 2012
This is not Matlab specific. It has to do with the fact that computers use binary numbers and humans tend to feel more comfortable with decimals. Things gets lost in translation and these are the side effects.

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Più risposte (1)

Thomas
Thomas il 9 Nov 2012
Modificato: Thomas il 9 Nov 2012
This should answer your question. Look at the wiki under the section Why_is_0.3_-_0.2_-_0.1_.28or_similar.29_not_equal_to_zero? (might have to scroll on the page to locate...)
  2 Commenti
Dr. Seis
Dr. Seis il 9 Nov 2012
Modificato: Dr. Seis il 9 Nov 2012
I had similar issues when doing mod(0.4,0.4) and mod(0.8,0.4) returning a number slightly smaller than 0.4 instead of 0. Whenever I do these types of comparisons in the future I will need to overload the builtin functions.
Oleg Komarov
Oleg Komarov il 9 Nov 2012
@Lucas, if you already know that you wouldn't be asking the question in the first place. I recommend to read the FAQ more carefully.

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