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.


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. $\endgroup$ – Wuberdall Sep 28 '18 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. $\endgroup$ – Calvin Khor Sep 28 '18 at 12:03
  • $\begingroup$ (You should replace "larger than" with "not smaller than" but that's harder to say) $\endgroup$ – Calvin Khor Sep 28 '18 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. $\endgroup$ – Wuberdall Sep 28 '18 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? $\endgroup$ – Wuberdall Sep 28 '18 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 '18 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. $\endgroup$ – Wuberdall Sep 28 '18 at 12:20
  • $\begingroup$ @Wuberdall You are welcome! Bye $\endgroup$ – gimusi Sep 28 '18 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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.