Proof
(related to Proposition: Floor Function and Division with Quotient and Remainder)
 Let $a,b\in\mathbb Z$ with $a > 0,$ be integers with $a > 0$ and let $q,r$ be uniquely determined by the division with quotient and remainder $b=qa+r,$ with $0\le r < a.$
 It follows that $qa \le b=qa+r < (q+1)a.$
 This is equivalent to $q\le \frac ba < q+1.$
 By definition of the floor function, $q=\lfloor \frac ba\rfloor.$
∎
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References
Bibliography
 Landau, Edmund: "Vorlesungen über Zahlentheorie, Aus der Elementaren Zahlentheorie", S. Hirzel, Leipzig, 1927