question on addition of numbers.

I have 2 numbers. a=11.699124064044344*2^192 and b=-17.935606820123120*2^252. adding a and b in MATLAB gives me the c=-1.2980e+77 (c=a+b). c-b gives me zero instead of 'a' value.

How to get back 'a' from 'c'

Answer 1

[FirefoxMetzger]

The simple answer is that a double is not precise enough to represent a + b .

A double is represented using IEEE® Standard 754 . Which (very roughly) stores variables as (sign) * 1.abcdef...*10^(XYZ) where all the numbers are binary.

Representing variable a as X * 10^252 means that X has a lot of leading zeros (0.000...00001169912...). At some point this takes more then the 52 bit to represent the number, so the standard tells us to round the 53nd bit, which is 0. All leading digits are also 0. Thus a = 0 * 10^252 (due to lack of precision). Logically a + b = b if we account for this lack of precision.

To still calculate correctly concider digits which are variable precision numbers. However be advised that they might be a lot slower when computing

[Ashwini]

Thanks a lot!!!!

Answer 2

[Azzi Abdelmalek]

when I run

a=11.699124064044344*2^192 
b=-17.935606820123120*2^254
c=a+b 
isequal(c,b)

c and b are equal, that's why c-b is 0. To understand the reason, read this http://matlab.wikia.com/wiki/FAQ#Why_is_0.3_-_0.2_-_0.1_.28or_similar.29_not_equal_to_zero.3F

[Ashwini]

Iam finding difficult to understand the concept. The content in the link says "some floating point numbers can not be represented exactly in binary form". Does that mean i cannot recover 'a' from 'c' in the above example. Please let me know if there is any method to recover 'a'. Please share the code for it if possible.

Thanks ....

[John D'Errico]

You CANNOT recover a from c, when using double precision. NEVER. A double lacks sufficient precision to do so.