◀ ▲ ▶Branches / Number-theory / Proof

Proof

(related to Proposition: Properties of Floors and Ceilings)

In the following, et $x\in\mathbb R$ be a real number.

Ad $1,2,3$

• These propositions follow immediately from the definitions of the floor and ceiling functions.

Ad $4$

• For every integer $n\in\mathbb Z$, $n\le n < n + 1.$
• By definition, $\lfloor x\rfloor\le x < \lfloor x\rfloor+1.$
• Adding both inequations results in $\lfloor x\rfloor+n\le x+n < \lfloor x \rfloor+n + 1.$
• On the other hand, we have $\lfloor x+n\rfloor\le x+n< \lfloor x+n\rfloor+1.$
• This is equivalent to $\lfloor x+n \rfloor = \lfloor x \rfloor+n.$

Ad $5$

• For example, take $n=2$ and $x=\frac 12.$
• Then $\lfloor 2\cdot\frac 12\rfloor=1$ but $2\cdot\lfloor \frac 12\rfloor=2\cdot 0=0.$

Ad $6$

• Ad $(6a)$ If $x < n$ then since $\lfloor x\rfloor\le x$ we have $\lfloor x \rfloor < n.$ Conversely, if $\lfloor x \rfloor < n$ then since $x < \lfloor x \rfloor + 1\le n$ we have $x < n.$
• Ad $(6b)$ If $n < x$ then since $x \le \lceil x\rceil$ we have $n < \lceil x\rceil.$ Conversely, if $n < \lceil x\rceil$ then because of $(3)$ we have $n < x.$
• Ad $(6c)$ If $x \le n$ then because of $(2)$ and $(3)$ and $(6a)$ we have $\lceil x\rceil\le n,$ and vice versa.
• Ad $(6d)$ If $n\le x$ then because of $(2)$ and $(3)$ and $(6b)$ we have $n\le \lfloor x \rfloor,$ and vice versa.

Ad $7$

• If $g\le \frac xn < g+1$ then $ng\le x < n(g+1).$
• It follows $ng\le \lfloor x\rfloor < n(g+1),$
• and finally $g\le \frac {\lfloor x\rfloor}n < g+1.$
  ∎

Thank you to the contributors under CC BY-SA 4.0!

Github:

References

Bibliography

1. Graham L. Ronald, Knuth E. Donald, Patashnik Oren: "Concrete Mathematics", Addison-Wesley, 1994, 2nd Edition
2. Landau, Edmund: "Vorlesungen über Zahlentheorie, Aus der Elementaren Zahlentheorie", S. Hirzel, Leipzig, 1927