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
\(\ds \leadsto \ \ \) \(\ds \floor {x + 1}\) \(=\) \(\ds \floor x + 1\)

$\blacksquare$


Also see