I have some trouble proving that if $x,y\in\mathbb{R}$ then $\lfloor x\rfloor \geq y$ if, and only if, $x\geq\lceil y\rceil$.

I have tried some different approaches, the most recent being a proof by contradiction: Assume (for contradiction) that $x<\lceil y\rceil$, then $\lfloor x\rfloor < y +1$. However, I can not (of course) get rid of the $+1$ term because in general $y \leq\lceil y\rceil$.

One can perhaps prove that both the floor- and ceiling function preserve inequalities? If so, the result become trivial.

Any ideas are highly appreciated.


4 Answers 4


Note that $ \lfloor x \rfloor \ge y$ implies $ \lfloor x \rfloor \ge \lceil y\rceil $, so $x\ge \lceil y\rceil$.

The other direction is similar.

If the above is not clear, recall that $\lceil y \rceil $ is defined as the least integer not smaller than $y$, i.e. if $C(y)$ is the set $ C(y):= \{n\in\mathbb Z:n\ge y\}$, then $$ \lceil y\rceil = \inf C(y).$$ The assumption $\lfloor x\rfloor \ge y $ is exactly that $\lfloor x\rfloor \in C(y)$. The only thing left is to use the defining property of an infimum, $$c \in C(y) \implies \inf C(y) \le c.$$

  • $\begingroup$ Thanks, but I can not see why the implication is true in general. $\endgroup$ Sep 28, 2018 at 12:02
  • $\begingroup$ @Wuberdall This is because $\lfloor x \rfloor \in \mathbb Z$. For example, $5\ge y$ implies $5 \ge \lceil y\rceil $. This holds for every integer in place of 5. In words - the ceiling of $y$ is the least integer larger than $y$, so any integer larger than $y$ is larger than the ceiling. $\endgroup$ Sep 28, 2018 at 12:03
  • $\begingroup$ (You should replace "larger than" with "not smaller than" but that's harder to say) $\endgroup$ Sep 28, 2018 at 12:10
  • $\begingroup$ Yes, I see it know. This is of course true due to the result "number not greater than integer iff ceiling is not greater than integer" exactly as you explained. – Thank you for your help. $\endgroup$ Sep 28, 2018 at 12:12

Hint: $$ \lfloor x\rfloor\geq y\iff \lfloor x\rfloor\geq\lceil y \rceil\iff x\geq \lceil y\rceil. $$

  • $\begingroup$ Thanks, but why is the first bi-implication true? $\endgroup$ Sep 28, 2018 at 12:01
  • $\begingroup$ @Wuberdall Because $\lfloor x\rfloor$ is an integer that is greater than or equal to $y$, so $\lfloor x\rfloor\geq\lceil y\rceil$. For the reverse direction, just note $\lceil y\rceil \geq y$. $\endgroup$
    – yurnero
    Sep 28, 2018 at 12:04

We need to prove both direction, that is

  • $\lfloor x\rfloor \geq y \implies x\geq\lceil y\rceil$


  • $x\geq\lceil y\rceil \implies \lfloor x\rfloor \geq y $

For the first one we have that

$$\lfloor x\rfloor \geq y \implies \lfloor x\rfloor \geq \lceil y\rceil $$

and therefore

$$x\ge \lfloor x\rfloor \geq \lceil y\rceil \implies x\geq\lceil y\rceil$$

For the second one we have that

$$x\geq\lceil y\rceil \implies \lfloor x\rfloor \geq \lceil y\rceil $$

and therefore

$$\lfloor x\rfloor \geq \lceil y\rceil \ge y \implies \lfloor x\rfloor \ge y$$

  • $\begingroup$ Thank you, it is completely clear why it is true now. $\endgroup$ Sep 28, 2018 at 12:20
  • $\begingroup$ @Wuberdall You are welcome! Bye $\endgroup$
    – user
    Sep 28, 2018 at 12:26

We need the following elementary properties of the floor and ceiling functions: $a\ge\lfloor a\rfloor$, $\lceil\lfloor a\rfloor\rceil=\lfloor a\rfloor$, $a\ge b\implies\lceil a\rceil\ge\lceil b\rceil$, and $\lceil a\rceil=-\lfloor-a\rfloor$.

From the first three elementary propeties, we have, for any real $u$ and $v$,

$$\lfloor u\rfloor\ge v \implies u\ge\lfloor u\rfloor=\lceil\lfloor u\rfloor\rceil\ge\lceil v\rceil$$

so letting $u=x$ and $v=y$ gives

$$\lfloor x\rfloor\ge y\implies x\ge\lceil y\rceil\qquad(*)$$

while letting $u=-y$ and $v=-x$ gives


Invoking now the fourth elementary property turns that last implication into

$$-\lceil y\rceil\ge-x\implies-y\ge-\lfloor x\rfloor$$

which can be re-expressed as

$$x\ge\lceil y\rceil\implies\lfloor x\rfloor\ge y\qquad(**)$$

Putting $(*)$ and $(**)$ together, we have

$$\lfloor x\rfloor\ge y\iff x\ge\lceil y\rceil$$


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