text: # Integer equals Floor iff Number between Integer and One More
  
  ## Theorem
  
  Let $x \in \R$ be a real number.
  
  Let $\floor x$ denote the floor of $x$.
  
  Let $n \in \Z$ be an integer.
  
  Then:
  
  $\floor x = n \iff n \le x < n + 1$
  
  ## Proof
  
  ### Necessary Condition
  
  Let $n \le x < n + 1$.
  
  $n \le x \implies n \le \floor x$
  
  By definition of the floor of $x$:
  
  $\floor x \le x$
  
  and so by hypothesis:
  
  $\floor x < n + 1$
  
  We have that:
  
  $\forall m, n \in \Z: m \le n \iff m < n + 1$
  
  and so:
  
  $\floor x \le n$
  
  So we have:
  
  $n \le \floor x$
  
  and:
  
  $\floor x \le n$
  
  Thus:
  
  $n \le x < n + 1 \implies \floor x = n$
  
  $\Box$
  
  ### Sufficient Condition
  
  Let $\floor x = n$.
  
  By definition of the floor of $x$:
  
  $\floor x \le x$
  
  and so by hypothesis:
  
  $n \le x$
  
  Also by definition of the floor of $x$:
  
  $x < \floor x + 1$
  
  and so by hypothesis:
  
  $x < n + 1$
  
  Thus:
  
  $\floor x = n \implies n \le x < n + 1$.
  
  $\Box$
  
  Hence the result:
  
  $\floor x = n \iff n \le x < n + 1$
  
  $\blacksquare$
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