Ceiling of Root of Ceiling equals Ceiling of Root

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Theorem

Let $x \in \R_{\ge 0}$ be a positive real number.

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

Then:

$\ds \ceiling {\sqrt {\ceiling x} } = \ceiling {\sqrt x}$


Proof 1

\(\ds n\) \(=\) \(\, \ds \ceiling {\sqrt x} \, \) \(\ds \)
\(\ds \leadstoandfrom \ \ \) \(\ds n - 1\) \(<\) \(\, \ds \sqrt x \, \) \(\, \ds \le \, \) \(\ds n\) Integer equals Ceiling iff Number between Integer and One Less
\(\ds \leadstoandfrom \ \ \) \(\ds \paren {n - 1}^2\) \(<\) \(\, \ds x \, \) \(\, \ds \le \, \) \(\ds n^2\) Order is Preserved on Positive Reals by Squaring
\(\ds \leadstoandfrom \ \ \) \(\ds \paren {n - 1}^2\) \(<\) \(\, \ds \ceiling x \, \) \(\, \ds \le \, \) \(\ds n^2\) Number not greater than Integer iff Ceiling not greater than Integer
\(\ds \leadstoandfrom \ \ \) \(\ds n - 1\) \(<\) \(\, \ds \sqrt {\ceiling x} \, \) \(\, \ds \le \, \) \(\ds n\) Order is Preserved on Positive Reals by Squaring
\(\ds \leadstoandfrom \ \ \) \(\ds n\) \(=\) \(\, \ds \ceiling {\sqrt {\ceiling x} } \, \) \(\ds \) Integer equals Ceiling iff Number between Integer and One Less

$\blacksquare$


Proof 2

The square root is defined on the interval $\hointr 0 \to$.

We have that Square Root is Strictly Increasing.

From Continuity of Root Function, the square root is continuous.

Hence the conditions are fulfilled for for McEliece's Theorem (Integer Functions) to be applied:

$\forall x \in A: \paren {\map f x \in \Z \implies x \in \Z} \iff \ceiling {\map f x} = \ceiling {\map f {\ceiling x} }$


It remains to be proved that $\sqrt x \in \Z \implies x \in \Z$.

Let $y = \sqrt x$ such that $y \in \Z$.

By definition of square root:

$x = y^2$

From Integer Multiplication is Closed:

$x \in \Z$

as required.

$\blacksquare$


Also see


Sources

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