Directed dyadic rounding encloses with explicit underflow and overflow bounds

If $x=0$, then $0\in \mathbb{F}$, so $\mathrm{rd}(0)=\mathrm{ru}(0)=0$ and each endpoint error equals ${ϵ}_{\mathrm{a}\mathrm{b}\mathrm{s}}(0)=0$. If $x\mathrm{\ne }0$ is representable, both directed roundings equal $x$. Otherwise there are consecutive representables $a<b$ with $a<x<b$: finiteness of $\mathbb{F}$ and the definitions of greatest lower and least upper values give these neighbors. Hence $\mathrm{rd}(x)=a\le x\le b=\mathrm{ru}(x)$.

If $0\le x<2^{e_{\min }}$, both neighboring nonnegative representables are in the subnormal lattice with step $\eta$, so $x-a<\eta$ and $b-x<\eta$. If $2^e\le x<2^{e+1}$ is normal, consecutive representables within the bin are spaced by $2^{e-p+1}$. A bracket reaching the upper bin endpoint has no larger gap; in all cases each endpoint error is at most the stated, more conservative $2^{e-p+2}$. Symmetry of $\mathbb{F}$ gives the same bounds for negative $x$.

For $x>N_{\max }$, $N_{\max }<x<+infty$, and for $x<-N_{\max }$, $-\mathrm{\infty }<x<-N_{\max }$; the prescribed extended endpoints therefore bracket $x$. The status and infinite absolute-error bound explicitly record that no finite error estimate is available outside the finite range.