Ceiling of Root of Ceiling equals Ceiling of Root

Theorem
Let $x \in \R$ be a real number.

Let $\ceiling x$ denote the ceiling of $x$.

Then:
 * $\ds \ceiling {\sqrt {\ceiling x} } = \ceiling {\sqrt x}$

Proof 2
The square root is defined on an interval $[0,\infty]$. Square Root is Strictly Increasing and from Continuity_of_Root_Function, the square root is continuous.

So we can apply McEliece's Theorem (Integer Functions) to get the statement is equivalent to proving $\sqrt{x} \in \mathbb{Z} \implies x \in \mathbb{Z}$.

Let $y = \sqrt{x}$ such that $y \in \mathbb{Z}$. From Definition:Square Root, $x = y^2$.

From Integer Multiplication is Closed, we obtain $x \in \mathbb{Z}$ as required.

Also see

 * Floor of Root of Floor equals Floor of Root