Floor Function/Examples/Floor of Root 10
Jump to navigation
Jump to search
Theorem
- $\floor {\sqrt {10} } = 3$
where $\floor x$ denotes the floor of $x$.
Proof
| \(\ds \sqrt 9\) | \(\le\) | \(\ds \sqrt 10\) | ||||||||||||
| \(\ds \) | \(<\) | \(\ds \sqrt 16\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds 3\) | \(\le\) | \(\ds \sqrt 10\) | |||||||||||
| \(\ds \) | \(<\) | \(\ds 4\) |
Hence $3$ is the floor of $\sqrt {10}$ by definition.
$\blacksquare$
Sources
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 10$: The well-ordering principle