Dyadic square-root bisection gives outward endpoints

For $x>0$, the initial lower candidate is $0$ and its square is at most $x$. The initial upper candidate is $2^{n+1}\delta =2^{m+1}$; if $j=2m$ its square is $2^{j+2}>x$, and if $j=2m+1$ its square is $2^{j+1}>x$. Thus initially $(L\delta {)}^2\le x<(U\delta {)}^2$.

At each iteration, $Q$ lies strictly between the integer endpoints. The exact comparison sets $L=Q$ only when $(Q\delta {)}^2\le x$, preserving the left inequality; otherwise it sets $U=Q$, preserving the strict right inequality. The square function is strictly increasing on nonnegative reals by [L3], so the invariant implies $L\delta \le \sqrt{x}<U\delta$. If the current integer gap is $g>1$, the next gap is at most $\lceil g\mathrm{/}2\rceil$. Starting from $g=2^{n+1}$, after at most $n+1$ iterations the positive integer gap is at most one, so it equals one and the loop stops. The returned width is then $\delta$.

For $x=0$ the returned point interval is exact. For $x>0$, the algorithm returns the exact dyadic pair from step 2.1. If a caller stores these endpoints in a finite format, outward rounding preserves both inequalities by [L2], so the stored result still contains $\sqrt{x}$.

On a fixed exponent bin, step 2.1 gives endpoints $\delta \lfloor \sqrt{x}\mathrm{/}\delta \rfloor$ and $\delta (\lfloor \sqrt{x}\mathrm{/}\delta \rfloor +1)$, both nondecreasing in $x$. At the transition from exponent $2m$ to $2m+1$, $\delta$ is unchanged. At the transition from exponent $2m+1$ to $2m+2$, the square-root threshold is $2^{m+1}$ and the grid spacing doubles from $\delta$ to $2\delta$. Just below the threshold the lower and upper endpoints are at most $2^{m+1}$; at or above it they are at least $2^{m+1}$. Thus both exact endpoint functions are nondecreasing across each transition. At $x=0$ both are zero. The directed rounding maps are nondecreasing by their greatest-lower/least-upper definitions, so storage preserves this order.