> [!NOTE] **Definition/ Theorem** (Floor is well-defined) > For any [[Real Numbers|real number]] $x\in \mathbb{R}$, $\exists!n\in \mathbb{N}$ such that $n\leq x<n+1$ (i.e. $n$ is the largest integer with $n\leq x$) denoted $\lfloor x \rfloor :=n$ known as the *floor (or integer part)* of $x$. >*Proof*. Let $S=\{ m \in \mathbb{Z}: m \leq x \}$. The set is bounded above by its definition. Using [[Archimedean Property of Real Numbers]], $\exists m \in \mathbb{Z} \text{ such that } m > -x \implies -m < x \implies -m \in S$ so $S$ is non-empty. >By [[Least Upper Bound Property]], $\exists r:=\sup S$. >We have $r \leq x$ since $x$ is an upper bound for $S$. >Since $r$ is the least upper bound, $\exists n \in S$ such that $r-1<n$. >This means that $r<n+1$, so $n+1 \not \in S \implies n+1>x$. >So we have $n \leq x<n+1$ which with $\lfloor x \rfloor:=n$ rearranges to give $x-1<\lfloor x \rfloor\leq x.$ > >Note this proof is similar to that of [[Ceiling Function]]. # Properties - [[Floor Function is Discontinuous at Every Lattice Point]].