Floor of Root of Floor equals Floor of Root/Proof 1
Jump to navigation
Jump to search
Theorem
- $\ds \floor {\sqrt {\floor x} } = \floor {\sqrt x}$
Proof 1
| \(\ds n\) | \(=\) | \(\, \ds \floor {\sqrt x} \, \) | \(\ds \) | |||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds n\) | \(\le\) | \(\, \ds \sqrt x \, \) | \(\, \ds < \, \) | \(\ds n + 1\) | Integer equals Floor iff Number between Integer and One More | ||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds n^2\) | \(\le\) | \(\, \ds x \, \) | \(\, \ds < \, \) | \(\ds \paren {n + 1}^2\) | Order is Preserved on Positive Reals by Squaring | ||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds n^2\) | \(\le\) | \(\, \ds \floor x \, \) | \(\, \ds < \, \) | \(\ds \paren {n + 1}^2\) | Number not less than Integer iff Floor not less than Integer | ||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds n\) | \(\le\) | \(\, \ds \sqrt {\floor x} \, \) | \(\, \ds < \, \) | \(\ds n + 1\) | Order is Preserved on Positive Reals by Squaring | ||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds n\) | \(=\) | \(\, \ds \floor {\sqrt {\floor x} } \, \) | \(\ds \) | Integer equals Floor iff Number between Integer and One More |
$\blacksquare$