# Alternative expression for Rounding Function on Floating Point Arithmetic

Let $$\mathbb{F}(\beta,t,e_{\min},e_{\max})$$ be a Floating Point Arithmetic. Let $$\text{domain}(\mathbb{F}) = [x_{\min},x_{\max}] \subseteq \mathbb{R}$$ for minimal and maximal elements $$x_{\min},x_{\max} \in \mathbb{F}$$. Let $$\text{rd}(x): \text{domain}(\mathbb{F}) \longrightarrow \mathbb{F}$$ be a rounding operation in accordance with the IEEE standards for rounding.

Under the above assumptions it is known, that for all $$x \in \text{domain}(\mathbb{F})$$ a $$\delta \in [-u(\mathbb{F}),u(\mathbb{F})]$$ exists such that $$rd(x) = x(1+\delta),$$ where $$u(\mathbb{F}) := \frac{1}{2}\beta^{1-t}$$ is the unit roundoff of $$\mathbb{F}$$. Now, throughout the literature it is often mentioned that the above expression can equivalently formulated as $$rd(x) = \frac{x}{1+\delta^{\ast}}$$ for some $$\delta^{\ast} \in [-u(\mathbb{F}),u(\mathbb{F})]$$. While conceptually the alternative expression does make sense to me, I am unable to formalise the equivalence of the expressions. How might one progress in order to so do?

• Would you add a few words about what is unclear at this point? – Carl Christian Apr 30 '20 at 17:09

The two representations are not equivalent, but they can be derived using the same basic principles. Let $$x = (1.f_1f_2\dots)_2 \cdot 2^m$$ denote a positive real number in the representational range. Let $$x_- = (1.f_1f_2\dots f_k)_2 \cdot 2^m$$ denote the largest machine number which is less than or equal to $$x$$ and let $$x_+ = x_1 + 2^{m-k}$$ denote the next floating point number. The floating point representation $$\text{fl}(x)$$ of $$x$$ is $$x_-$$ when $$x_-$$ is closer to $$x$$ than $$x_+$$ and vise versa. Any ties are resolved as dictated by the rounding mode. In any case, we have $$|x- \text{fl}(x)| \leq \frac{1}{2} 2^{m-k} = 2^m u,$$ where $$u = 2^{-k-1}$$ is the unit roundoff.
1. Since $$2^m \leq x$$ we have the relative error bound $$\frac{|x- \text{fl}(x)|}{|x|} \leq u.$$ It follows that $$fl(x) = x(1 + \delta)$$ where $$\delta = \frac{\text{fl}(x) - x}{x}$$ satisfies $$|\delta| \leq u$$.
2. Since $$2^m \leq \text{fl}(x)$$ we also have $$\frac{|x- \text{fl}(x)|}{|\text{fl}(x)|} \leq u.$$ It follows that $$x = \text{fl}(x)(1 + \epsilon)$$ where $$\epsilon = \frac{x - \text{fl}(x)}{\text{fl}(x)}$$ satisfies $$|\epsilon| \leq u$$. We now conclude that $$\text{fl}(x) = \frac{x}{1 + \epsilon}.$$