Floor Plus One

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Theorem

Let $x \in \R$.

Then:

$\left \lfloor {x + 1} \right \rfloor = \left \lfloor {x} \right \rfloor + 1$

where $\left \lfloor {x} \right \rfloor$ is the floor function of $x$.

$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle \left \lfloor {x + 1} \right \rfloor$	$=$	$\displaystyle n$	$\displaystyle $	$\displaystyle $	$\displaystyle $
$\displaystyle $	$\displaystyle \implies$	$\displaystyle $	$\displaystyle n$	$\le$	$\displaystyle x + 1 < n + 1$	$\displaystyle $	$\displaystyle $	$\displaystyle $	by definition of floor function
$\displaystyle $	$\displaystyle \implies$	$\displaystyle $	$\displaystyle n - 1$	$\le$	$\displaystyle x < n$	$\displaystyle $	$\displaystyle $	$\displaystyle $
$\displaystyle $	$\displaystyle \implies$	$\displaystyle $	$\displaystyle \left \lfloor {x} \right \rfloor$	$=$	$\displaystyle n - 1$	$\displaystyle $	$\displaystyle $	$\displaystyle $	by definition of floor function
$\displaystyle $	$\displaystyle \implies$	$\displaystyle $	$\displaystyle \left \lfloor {x + 1} \right \rfloor$	$=$	$\displaystyle \left \lfloor {x} \right \rfloor + 1$	$\displaystyle $	$\displaystyle $	$\displaystyle $

In general:

$\forall n \in \Z: \left \lfloor {x} \right \rfloor + n = \left \lfloor {x + n} \right \rfloor$

\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \left \lfloor {x + 1} \right \rfloor\)	\(=\)	\(\displaystyle n\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle n\)	\(\le\)	\(\displaystyle x + 1 < n + 1\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	by definition of floor function
\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle n - 1\)	\(\le\)	\(\displaystyle x < n\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle \left \lfloor {x} \right \rfloor\)	\(=\)	\(\displaystyle n - 1\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	by definition of floor function
\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle \left \lfloor {x + 1} \right \rfloor\)	\(=\)	\(\displaystyle \left \lfloor {x} \right \rfloor + 1\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)