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.

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    $\begingroup$ "since ⌈−x⌉=−⌊x⌋..." Stop!!!!!!! You haven't proven that yet. That is what you are trying to prove. You can never assume what you are supposed to prove. NOTHING you say from here on out is going to have any validity. $\endgroup$ – fleablood Oct 11 '16 at 21:16
  • $\begingroup$ @fleablood: Wrong: Nothing that uses that claim has any validity. Anything following that does not use that claim is perfectly valid. In particular, the substitution of $-x$ in the first definition is not invalidated by following that claim, as it doesn't use that claim. $\endgroup$ – celtschk Oct 11 '16 at 21:31
  • $\begingroup$ Anything that follows that doesn't use the claim may or may not be valid but I'm not going to bother to continue reading as that claim was invalid so I have no faith anything that follows is of merit. $\endgroup$ – fleablood Oct 11 '16 at 21:35
  • $\begingroup$ @fleablood: Well, you can of course choose not to read on, but that doesn't make your claim that everything said afterwards is invalid more correct. $\endgroup$ – celtschk Oct 11 '16 at 21:43
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    $\begingroup$ "all he was stating was that ceiling(-x) = -floor(x) is an equivalent statement to -ceiling(-x) = floor(x)" Ah. I hadn't read it that way. That would make sense. I'm sorry. People assuming what the need to proof as part of the proof are unfortunately common and a particular irritant of mine (because it just doesn't make sense!!!!!!) $\endgroup$ – fleablood Oct 12 '16 at 18:12

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 $-\lfloor x\rfloor=\lceil -x\rceil$.

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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.

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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.

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  • $\begingroup$ Well, since he says he's using those definitions (rather than proposing them), I assume that he didn't invent them himself but got them from some source (which he however didn't specify), and I'd assume that the source also proves that those definitions are meaningful (or refers to some other source that proves it). $\endgroup$ – celtschk Oct 11 '16 at 21:39
  • $\begingroup$ Yeah, I agree. And I gave an answer using just them. I'd be remiss if I didn't state my reservations. I've only shown if $y$ is ceiling of $-x$ the therefore $-y$ satisfies the conditions for floor x. Usually showing something is a candidate is not enough to show it is a specific value. If the definitions are valid (and they can be proven to be) then I've proved equality. If they are not I haven't proven uniqueness. $\endgroup$ – fleablood Oct 11 '16 at 21:49
  • $\begingroup$ It's also good practice to not take definitions at face value. Many, many false proofs start with "Let x be the ...." when no such thing exists or may not be unique. $\endgroup$ – fleablood Oct 11 '16 at 21:51
  • $\begingroup$ But you could at least have formulated your reservations in a less aggressive way. $\endgroup$ – celtschk Oct 11 '16 at 22:04
  • $\begingroup$ "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" is hardly aggressive. It's a clarification as to what a good definition is. And as you point out "your" definitions are presumably, not the definitions, the op came up with but merely the ones she is using. $\endgroup$ – fleablood Oct 11 '16 at 23:26

Let $F = \lfloor x \rfloor$. Then $F$ is an integer, $F \le x$, and $\forall m \in \mathbb{Z},\; m\le x \implies m \le F$.

Then $F+1 \le x \implies F+1 \le F$. It follows that

$$ F \le x \lt F+1$$ $$ -F - 1 \lt -x \le -F $$

Let $-C = \lceil -x \rceil$. Then $-C$ is an integer, $-x \le -C$, and $\forall m \in \mathbb{Z}, \; -x \lt m \implies -C \le m)$.

By reasoning similar to that used above, it follows that

$$-C -1 < -x \le -C$$

Then $-C-1 < -F$ and $-F-1 < -C$. Hence $-1 < F-C < 1$. Since $F$ and $C$ are integers,it follows that $F=C$; that is

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

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