Why does Visual Studio 2008 tell me .9 - .8999999999999995 = 0.00000000000000055511151231257827?

Question

When I type this into the Visual Studio 2008 immediate window:

? .9 - .8999999999999995

It gives me this as the answer:

0.00000000000000055511151231257827

The documentation says that a double has 15-16 digits of precision, but it's giving me a result with 32 digits of precision. Where is all that extra precision coming from?

Answer 1

There are only 15-16 digits in the answer. All those leading zeroes don't count. The number is actually more like 5.5511151231257827 × 10^-16. The mantissa portion has 15-16 digits in it. The exponent (-16) serves to shift the decimal point over by 16 places, but doesn't change the number of digits in the overall number.

Edit

After getting some comments, I'm curious now about what's really going on. I plugged the number in question into this IEEE-754 Converter. It took the liberty of rounding the last "27" into "30", but I don't think that changes the results.

The converter breaks down the number into its three binary parts:

Sign: 0 (positive)
Exponent: -51
Significand: 1.0100000000000000000000000000000000000000000000000000 (binary for 1.25₁₀)

So this number is 1.01₂ × 2^-51, or 1.25₁₀ × 2^-51. Since there are only three significant binary digits being stored, that would suggest that Lars may be onto something. They can't be "random noise" since they are the same each time the number is converted.

The data suggests that the only stored digit is "5". The leading zeros come from the exponent and the rest of the seemingly-random digits are from computing 2^-51.

Answer 2

You should read: What Every Computer Scientist Should Know About Floating-Point Arithmetic .

Basically it comes down to Floating Point numbers being stored with finite precision. You have to do your comparison with some delta.

if(.9 - .8999999999999995 <= 0.0001)
  //close enough to be equal

Answer 3

The leading zeros are not significant/part of the precision (as far as the floating point number is concerned -- mathematically speaking, they are significant). The leading zeros are due to the exponent part of the floating point number's internal representation.

The portion 55511151231257827 (which is the significand or mantissa) has 17 decimal digits, which is close enough to 15-16 digits.

@Lars D: What you consider to be correct, is only correct within the context of the question. .9 - .8999999999999995 works out to a float with significand 0.625 and exponent of -50. Taking 0.625 * 2^-50 results in 5.5511151231257827e-16. Now, out of the context of the original question, we have a number with 17 significant digits which does happen to be our best binary approximation of 0.0000000000000005. However, those leading zeros are still not significant as far as the representation of the floating point number is concerned.

Answer 4

? .9 - .8999999999999995

This subtraction process, with 15-16 significant digits, gives

0.0000000000000005

The rest of the digits are just rounding errors. However, since the computer always stores 15-16 significant digits after the first non-zero digit, the rounding errors are shown, and you get a lot of trailing random digits produced by rounding errors. So the result has 16 significant digits from the subtraction operation plus 16 digits from the storage of the result, which gives 32 digits.

Answer 5

The "floating" part of "floating point" means that you are getting something closer to 5.5511151231257827 * 10^(-16). That's not exactly how it's represented, because of course it's all done in binary under the hood, but the point is, the number is represented by the significant digits, plus a number which represents how far to move the radix (decimal point). As always, wikipedia can give you more detail:

• http://en.wikipedia.org/wiki/Floating_point
• http://en.wikipedia.org/wiki/Double_precision

(The second link is more specifically focused on your particular case.)

Answer 6

I think its because in the binary system, 5 is periodic as it is not dividable by 2. And then what Mark Rushakoff said applies.