Proofs with floor and ceiling: $\lceil -x \rceil = -\lfloor x \rfloor$

Question

I am trying to prove that $\lceil -x \rceil = -\lfloor x \rfloor$ using the following definitions:

$y = \lceil x \rceil$ means $y \in \mathbb{Z} \land y \ge x \land (\forall z \in \mathbb{Z}\; z\ge x \implies z \ge y)$

$y = \lfloor x \rfloor$ means $y \in \mathbb{Z} \land y \le x \land (\forall z \in \mathbb{Z}\;z\le x \implies z \le y)$

Since $\lceil -x \rceil = -\lfloor x \rfloor$ then $-\lceil -x \rceil = \lfloor x \rfloor$

I can substitute the $-x$ into the above definition for $\lceil x \rceil$

This gets me

$y = \lceil -x \rceil$ means $y \in \mathbb{Z} \land y \ge -x \land (\forall z \in \mathbb{Z}\; z\ge -x \implies z \ge y)$

How would I incorporate the $-$ on the outside of $-\lceil -x \rceil$ into the above statement, to show that it is equal to $\lfloor x \rfloor$?

I have looked at other questions discussing this proof, but they do not have anything to do with the two predicate definitions I wrote above.

Answer 1

To go directly from your definitions:

$y=-\lfloor x\rfloor$ means $-y=\lfloor x\rfloor$ and therefore we can again directly insert to get: $$-y \in \mathbb{Z} \land -y \le x \land (\forall z \in \mathbb{Z}\;z\le x \implies z \le -y)$$

Now obviously $-y\in\mathbb Z$ iff $y\in\mathbb Z$. Also $-y \le x$iff $y\ge -x$. Furthermore, $z\le x$ iff $-z\ge -x$, and $z\le -y$ iff $-z\ge y$. So we get: $$y \in \mathbb{Z} \land y \ge -x \land (\forall z \in \mathbb{Z}\;-z\ge -x \implies -z \ge y)$$ Finally, we observe that for any proposition $P(z)$, $\forall z\in\mathbb Z\;P(-z)$ is equivalent to $\forall z\in\mathbb Z\;P(z)$. Using this with $P(z)=(-z\ge -x \implies -z \ge y)$, we finally arrive at $$y \in \mathbb{Z} \land y \ge -x \land (\forall z \in \mathbb{Z}\;z\ge -x \implies z \ge y)$$ But that is exactly the expression you correctly derived as meaning $y=\lceil -x\rceil$.

So we have $y=-\lfloor x\rfloor \iff y=\lceil -x\rceil$ and thus $-\lfloor x\rfloor=\lceil -x\rceil$.

Answer 2

Concerning your final answer I would proceed as follows.

Some preliminary definitions first.

• $y = \left\lfloor y \right\rfloor + \left\{ y \right\}$
  which is a definition of the fractional part $0 \leqslant \left\{ y \right\} < 1$,
• $\left\lceil y \right\rceil = \left\lceil {\left\lfloor y \right\rfloor + \left\{ y \right\}} \right\rceil = \left\lfloor y \right\rfloor + \left\lceil {\left\{ y \right\}} \right\rceil = \left\lfloor y \right\rfloor + 1 - \left[ {0 = \left\{ y \right\}} \right]$
  which is the relation between ceil and floor functions,
• and where the square brackets indicates the Iverson bracket $$ \left[ \text{P} \right] = \left\{ {\begin{array}{*{20}c} 0 & {\text{P} = \text{FALSE}} \\ \text{1} & {\text{P} = \text{TRUE}} \\ \end{array} } \right. $$

Then $$ \begin{gathered} \left\lfloor { - y} \right\rfloor = \left\lfloor { - \left\lfloor y \right\rfloor - \left\{ y \right\}} \right\rfloor = - \left\lfloor y \right\rfloor + \left\lfloor { - \left\{ y \right\}} \right\rfloor = \hfill \\ = - \left\lfloor y \right\rfloor - 1 + \left[ {0 = \left\{ y \right\}} \right] = \hfill \\ = - \left( {\left\lfloor y \right\rfloor + 1 - \left[ {0 = \left\{ y \right\}} \right]} \right) = - \left\lceil y \right\rceil \hfill \\ \end{gathered} $$ and replacing $y$ with $-x$, you get the complementary $$ \left\lfloor x \right\rfloor = - \left\lceil { - x} \right\rceil $$ Of course you could proceed like that right from the beginning.

Answer 3

Let $y = \lceil - x \rceil$. So $y \ge -x$ so $-y \le x$. $\forall z\in \mathbb Z, z \ge -x \implies z \ge y$ so $\forall -z \in \mathbb Z -z \le x \implies -z \le -y$. So by definition $-y = \lfloor x \rfloor$.

Although, to tell the truth, I'm not sure we can accept your conditions as definitions unless we can also prove i) such a $y$ exists and ii) such a $y$ is unique.