# machine epsilon value for IEEE double precision standard alternative proof using relative error

From the textbook, I know that the machine epsilon number for IEEE double precision standard $$F(\beta=2, t = 53, L = -1022, U = 1023)$$ is:

$$\epsilon_{M} = 2 \mu$$

where $$\mu$$ is the unit round-off which equals to $$2^{-53}$$

I was trying to prove this using the definition of relative error, but I could not deduce that relation for $$\epsilon_{M}$$:

$$\frac{|x - fl(x)|}{|x|} \leq \mu$$

How can I attempt to justify this relation of machine epsilon?

• It was never really clear to me what you were looking for. By definition, $1+\epsilon_M$ is the first floating point number after $1$. The fact that $\epsilon_m = 2\mu$ where $\mu$ is the unit roundoff hinges on the analysis of the floating point representation given below. Oct 17, 2020 at 22:33

Let $$x \in \mathbb{R}$$ be in the representational range. Without loss of generality we can assume that $$x>0$$. If $$x$$ is a floating point number, then there is nothing to show, so we can assume that $$x$$ is not the largest floating point number. By assumption, the binary representation of $$x$$ takes the form $$x = (1.f_1f_2f_3,\dotsc)_2 \times 2^e, \quad f_i \in \{0,1\}, \quad L \leq e \leq U.$$ The number $$x_{-}$$ given by $$x_{-} = (1.f_1f_2f_3,\dotsc,f_{t-1})_2 \times 2^e$$ is the largest floating point number which is smaller than $$x$$. The next floating point number is given by $$x_{+} = \left[ (1.f_1f_2f_3,\dotsc,f_{t-1})_b + 2^{-(t-1)} \right] \times 2^e = x_- + 2^{e-(t-1)}.$$ By construction, $$x_{-} \leq x \leq x_{+}, \quad x_{+} - x_{-} = 2^{e-(t-1)}.$$ Let $$\hat{x}$$ denote the number which is closest to $$x$$, i.e., either $$\hat{x} = x_{-}$$ or $$\hat{x} = x_{+}$$. Then the error $$x - \hat{x}$$ satisfies $$|x-\hat{x}| \leq \frac{1}{2} 2^{e-(t-1)}.$$ Since $$x \ge 2^e$$ we can bound the relative error $$r = (x-\hat{x})/x$$ as follows $$|r| \leq 2^{-t} = u$$ where $$u$$ is the unit roundoff. By definition, machine epsilon $$\epsilon$$ is the difference between $$1$$ and the next floating point number, i.e., $$1+2^{-(t-1)}$$. It follows that $$\epsilon = 2u$$.