# Prove that $\lfloor x\rfloor \geq y$ if, and only if, $x\geq\lceil y\rceil$

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, $$c \in C(y) \implies \inf C(y) \le c.$$

• Thanks, but I can not see why the implication is true in general. Sep 28, 2018 at 12:02
• @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. Sep 28, 2018 at 12:03
• (You should replace "larger than" with "not smaller than" but that's harder to say) Sep 28, 2018 at 12:10
• 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. Sep 28, 2018 at 12:12

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

• Thanks, but why is the first bi-implication true? Sep 28, 2018 at 12:01
• @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 $\lceil y\rceil \geq y$. Sep 28, 2018 at 12:04

We need to prove both direction, that is

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

and

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

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

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

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

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

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

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