# About Numerical Precision Related with Zeros [duplicate]

I encounted the following precision issue. I define c = a + b, but the result for 'c - a - b' does not equal zero. I am not sure how to resolve this.

'''

In:=
a = N[Exp];
b = N[Exp];
c = a + b;
c - a - b

Out= 7.10543*10^-15


'''

• What happens when you do Chop[%] on the final output? does it become zero? see help on Chop or you can use infinite precision. a = N[Exp, Infinity]; b = N[Exp, Infinity]; and you will get exact zero without doing Chop Sep 27 at 7:47
• But note that N[Exp, Infinity] just keeps it as Exp and will not make it real number. Once you convert things to approximate numbers, you'll get approximate numbers as result of computation. Sep 27 at 7:56
• notice also that N[...] use machine precision (unless you give second argument to N). on same PC, I typed the same thing on Matlab, and got same output: a=exp(3); b=exp(4); c=a+b; c-a-b gives 7.1054e-15 This is the nature of using real numbers. !Mathematica graphics Sep 27 at 8:05
• You can avoid some of the problems by using c - (a + b) instead of c - a - b. Now you will get 0. as output. This is because the order makes difference in terms of real numbers computations (large number - small number, vs. difference of two numbers close to each others). see What Every Computer Scientist Should Know About Floating-Point Arithmetic Sep 27 at 8:11
• Everything is indeed correct. You can see the same phenomenon even on a scientific calculator. As @Nasser said, this is the nature of floating point computations. Sep 27 at 9:51

This phenomenon occurs for simple cases too:

0.3 + 1.1 - 1.4

2.22045*10^-16


These numbers do not have finite length in binary and so they're truncated when expressed as an inexact number. We can see this with RealDigits:

RealDigits[1.1, 2]

{{1, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0,
1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1,
0, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0}, 1}


The identity $$1.1_{10} = 1.0001100110011_2...$$ gets truncated to $$53$$ digits to the right of the decimal (on a $$64$$ bit machine). The same happens for $$0.3$$ and $$1.4$$ — and in fact only dyadic rationals can be represented exactly in floating point arithmetic.

We can see the error by expressing the truncated reals as rational numbers:

SetPrecision[0.3 + 1.1, ∞] - SetPrecision[1.4, ∞]

1/4503599627370496

N[%]

2.22045*10^-16
`