Floor plus One
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Theorem
Let $x \in \R$.
Then:
- $\floor {x + 1} = \floor x + 1$
where $\floor x$ is the floor function of $x$.
Proof
\(\ds \floor {x + 1}\) | \(=\) | \(\ds n\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds n\) | \(\le\) | \(\, \ds x + 1 \, \) | \(\, \ds < \, \) | \(\ds n + 1\) | Definition of Floor Function | ||||||||
\(\ds \leadsto \ \ \) | \(\ds n - 1\) | \(\le\) | \(\, \ds x \, \) | \(\, \ds < \, \) | \(\ds n\) | |||||||||
\(\ds \leadsto \ \ \) | \(\ds \floor x\) | \(=\) | \(\ds n - 1\) | Definition of Floor Function | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \floor {x + 1}\) | \(=\) | \(\ds \floor x + 1\) |
$\blacksquare$