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My homework assignment has asked me to prove that $\lfloor-x\rfloor = -\lceil x\rceil$.

Conceptually this makes perfect sense to me, I just am at a loss for how to start actually proving it. I figure that it has something to do with these definitions: $$\lfloor x \rfloor: a \le x \lt a+1$$ $$ \lceil x \rceil : a \lt x \le a+1 $$

But I'm unsure of how to apply these to devise a proof for this. Any help that anyone could give would be incredibly appreciated, I've spent way too much time staring at the question trying to figure it out.

nonuser
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Anerdson
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8 Answers8

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\begin{eqnarray}-x-1< &\lfloor -x \rfloor &\leq -x \\ x\leq &\lceil x \rceil &< x+1 \end{eqnarray}

Adding these two we get $$(-x-1)+x< \lfloor -x \rfloor + \lceil x \rceil <-x+(x+1)$$

$$-1< \lfloor -x \rfloor + \lceil x \rceil <1$$ so $$\lfloor -x \rfloor + \lceil x \rceil =0$$

nonuser
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Hint

$$k\le -x<k+1\iff -k-1<x\le -k\\k\in\Bbb Z$$

Mostafa Ayaz
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If $x$ is an integer than an obvious equality holds. Otherwise we have $k-1\lt x\lt k$ for some $k\in\mathbb{Z}$. So we have $$\lfloor -x\rfloor=-k$$ $$-\lceil x \rceil = -k$$ which are equal. The first equality is true because $-k\lt-x\lt1-k$ so $\lfloor -x\rfloor$ takes the lower value. The second is true as $\lceil x \rceil$ takes the higher value and it becomes negated.

Peter Foreman
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The left-hand side says ($p_1=\lfloor{-x}\rfloor) \quad$ $-x=p_1+\varepsilon$ for some $0 \le \varepsilon \le 1$.

The right-hand side says ($p_2=-\lceil{x}\rceil)\quad $ $p_2=-(x+\delta)$ for some $0 \le \delta \le 1$.

Since both $\varepsilon$ and $\delta$ are the distances from $-x$ to nearest integer less than $x$ they are equal and $p_1 = p_2$.

mjw
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With most elementary floor and ceiling proofs, the way to start is: Let $x = k +r$ where $k$ is an integer and $0\leq r <1.$ That usually fixes everything.

You have, if $r\neq 0$:

$$\lfloor -x \rfloor = \lfloor -k-r \rfloor = -k +\lfloor -r \rfloor = -k -1.$$

and:

$$-\lceil x \rceil = -\lceil k+r \rceil = -(k+\lceil r \rceil) = -k -1.$$

And if $r=0$, a similar thing happens.

B. Goddard
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Just do it:

We know that $\lfloor -x \rfloor \le -x < \lfloor -x \rfloor + 1$ and that $\lfloor -x \rfloor$ is an integer.

So $-\lfloor -x \rfloor \ge x > -\lfloor -x \rfloor-1$ and so $-\lfloor -x \rfloor- 1< x \le -\lfloor -x \rfloor$ and $-\lfloor -x \rfloor$ is an intger.

So by definition that means $\lceil x \rceil = -\lfloor -x \rfloor$.

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

fleablood
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Let $x \in \mathbb{R}$ be given. By definition, $\lfloor -x \rfloor$ is the unique integer $a$ such that $$ a \leq -x < a+1. $$ Now, this implies that $$ -(a+1) < x \leq -a $$ where $-a \in \mathbb{Z}$ also. By definition of $\lceil x \rceil$, we must then have $\lceil x \rceil = -a$. In short, we have proven that \begin{align*} \lfloor -x \rfloor = a = -\lceil x \rceil. \end{align*}

rolandcyp
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The map $\varphi\colon x\mapsto -x$ is an order isomorphism of $(\mathbb{R},\le)$ onto $(\mathbb{R},\ge)$.

Since $\lfloor x\rfloor=\sup_{\le}\{z\in\mathbb{Z}:z\le x\}$ and $\varphi$ maps $\mathbb{Z}$ onto $\mathbb{Z}$, you're done, taking into account that $\lceil x\rceil=\sup_{\ge}\{z\in\mathbb{Z}:z\ge x\}$ and, obviously, $\sup_{\ge}$ is the same as $\inf_{\le}$.

egreg
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  • Is $\lceil x \rceil$ a supremum or an infimum? – mjw Mar 27 '19 at 15:43
  • @mjw It's a supremum with respect to $\ge$ and an infimum with respect to $\le$. – egreg Mar 27 '19 at 15:46
  • I would have expected $\lceil x \rceil$ and $\lfloor x \rfloor$ to have dual descriptions. – mjw Mar 27 '19 at 15:48
  • @mjw They have: one is a supremum with respect to $\ge$, the other one is a supremum with respect to $\le$. Do you see the duality? – egreg Mar 27 '19 at 15:49
  • Yes, I would have thought, though, that $\lceil x\rceil=\inf_{\ge}{z\in\mathbb{Z}:z\ge x}$. – mjw Mar 27 '19 at 15:51
  • @mjw $\inf_{\ge}=\sup_{\le}$. – egreg Mar 27 '19 at 16:09
  • Yes, that is fine. Please look again at your expressions for $\lfloor x \rfloor$ and $\lceil x \rceil$. I believe you have a proof. I just think there possibly may be a typo in the second expression. – mjw Mar 27 '19 at 16:16
  • @mjw No, there isn't. Don't be fooled by the unusual use of symbols. – egreg Mar 27 '19 at 16:17
  • What exactly does $\sup_{\ge}$ mean? – mjw Mar 27 '19 at 16:19
  • @mjw The supremum with respect to the order relation $\ge$ (which is not the same as $\le$, but its opposite) – egreg Mar 27 '19 at 16:55
  • This is a supremum over integers greater than $x$, or an infimum over integers greater than $x$? Asking about $\lceil x \rceil$. – mjw Mar 27 '19 at 17:04