Floor Function/Examples/Floor of 5 over 2
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Theorem
- $\floor {\dfrac 5 2} = 2$
where $\floor x$ denotes the floor of $x$.
Proof
We have that:
| \(\ds \dfrac 5 2\) | \(=\) | \(\ds 2 + \dfrac 1 2\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds 2\) | \(\le\) | \(\ds \dfrac 5 2\) | |||||||||||
| \(\ds \) | \(<\) | \(\ds 3\) |
Hence $2$ is the floor of $\dfrac 5 2$ by definition.
$\blacksquare$
Sources
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 10$: The well-ordering principle