Editing Round-off error (section)

=== Subtraction ===

Absorption also applies to subtraction.

* For example, subtracting <math>2^{-60}</math> from <math>1</math> in IEEE double precision as follows, <math display="block">\begin{align}
1.00\ldots 0 \times 2^{0} - 1.00\ldots 0 \times 2^{-60} &= \underbrace{1.00\ldots 0}_\text{60 bits} \times 2^{0} - \underbrace{0.00\ldots 01}_\text{60 bits} \times 2^{0}\\
&= \underbrace{0.11\ldots 1}_\text{60 bits}\times 2^{0}.
\end{align}</math> This is saved as <math>\underbrace{1.00\ldots 0}_\text{53 bits}\times 2^{0}</math> since round-to-nearest is used in IEEE standard. Therefore, <math>1-2^{-60}</math> is equal to <math>1</math> in IEEE double precision and the roundoff error is <math>-2^{-60}</math>.

The subtracting of two nearly equal numbers is called '''subtractive cancellation'''.<ref name="Forrester_2018"/> 
When the leading digits are cancelled, the result may be too small to be represented exactly and it will just be represented as <math>0</math>. 

* For example, let <math>|\epsilon| < \epsilon_\text{mach}</math> and the second definition of machine epsilon is used here.  What is the solution to <math>(1+\epsilon) - (1-\epsilon)</math>?{{Break}} It is known that <math>1+\epsilon</math> and <math>1-\epsilon</math> are nearly equal numbers, and <math>(1+\epsilon) - (1-\epsilon)=1+\epsilon-1+\epsilon=2\epsilon</math>.  However, in the floating-point number system, <math>fl((1+\epsilon) - (1-\epsilon))=fl(1+\epsilon)-fl(1-\epsilon)=1-1=0</math>.  Although <math>2\epsilon</math> is easily big enough to be represented, both instances of <math>\epsilon</math> have been rounded away giving <math>0</math>.

Even with a somewhat larger <math>\epsilon</math>, the result is still significantly unreliable in typical cases.  There is not much faith in the accuracy of the value because the most uncertainty in any floating-point number is the digits on the far right. 

* For example, <math>1.99999 \times 10 ^{2}- 1.99998 \times 10^{2} = 0.00001\times10^{2} =1 \times 10^{-5}\times 10^{2}=1\times10^{-3}</math>. The result <math>1\times10^{-3}</math> is clearly representable, but there is not much faith in it.

This is closely related to the phenomenon of [[catastrophic cancellation]], in which the two numbers are ''known'' to be approximations.