Machine epsilon: why is $(1 + \epsilon) + \epsilon = 1$?
My book on real analysis has the following statement:
I don't understand how the first equation can possibly be true, by definition of machine epsilon.
Machine epsilon is defined as the smallest number that, when added to the number $1$, will yield the next representable machine number in our floating-point system that is $> 1$. In other words, it is the distance between the number $1$ and the next machine number to the right of $1$.
By this definition, that means:
$(1 + \epsilon) > 1$
So, how is it possible that adding $\epsilon$ yet again will produce $1$ itself? That seems paradoxical.
$\endgroup$ 51 Answer
$\begingroup$The quoted text states that $\epsilon$ is slightly smaller that $\epsilon_{\text{mach}}$. This implies $1+\epsilon=1$ and hence also $(1+\epsilon)+\epsilon=1$, but if $\epsilon$ is sufficiently large, i.e. $\epsilon\geq\tfrac12\epsilon_{\text{mach}}$, then$$1+(\epsilon+\epsilon)\geq1+\epsilon_{\text{mach}}>1.$$
$\endgroup$ 1
