Ceiling of Root of Ceiling equals Ceiling of Root/Proof 1
Jump to navigation
Jump to search
Theorem
- $\ds \ceiling {\sqrt {\ceiling x} } = \ceiling {\sqrt x}$
Proof
| \(\ds n\) | \(=\) | \(\, \ds \ceiling {\sqrt x} \, \) | \(\ds \) | |||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds n - 1\) | \(<\) | \(\, \ds \sqrt x \, \) | \(\, \ds \le \, \) | \(\ds n\) | 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 |
$\blacksquare$