Real Number is between Floor Functions
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Theorem
- $\forall x \in \R: \floor x \le x < \floor {x + 1}$
where $\floor x$ is the floor of $x$.
Proof
$\floor x$ is defined as:
- $\floor x = \sup \set {m \in \Z: m \le x}$
So $\floor x \le x$ by definition.
From Floor plus One:
- $\floor {x + 1} > \floor x$
Hence by the definition of the supremum:
- $\floor {x + 1} > x$
The result follows.
$\blacksquare$