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I'm stuck on this for quite of few hours now. Can someone please explain me how would I prove this? TIA

$\lceil x + n \rceil = \lceil x \rceil + n $ (x is a real number and n is an integer)

Steven Alexis Gregory
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John Casey
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2 Answers2

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Use this alternative definition of ceiling:

For $x \in \Bbb R$, $\lceil x \rceil$ is an integer $n$ such that:

  1. $x \le n$, and
  2. for every other integer $m$ such that $x \le m$, we have $n \le m$.

To show that $\lceil x \rceil + n = \lceil x + n \rceil$, I will show that $\lceil x \rceil + n$ satisfies the property.

  1. By (1) we have $x \le \lceil x \rceil$, so $x + n \le \lceil x \rceil + n$.
  2. Let $x + n \le m$ for some $m$. So $x \le m - n$, so $\lceil x \rceil \le m - n$ by (2), i.e. $\lceil x \rceil + n \le m$.
Kenny Lau
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  • In words, #1 and #2 say that $n$ is the smallest integer greater than or equal to $x$. (#1 says that $n\ge x$ and #2 says that it's the smallest such number.) – Akiva Weinberger Mar 22 '18 at 02:43
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$\lceil x\rceil$ satisfies $x\le\lceil x\rceil< x+1$. Then $$x+n\le\lceil x\rceil+n< x+n+1$$ where $\lceil x\rceil+n$ is integer.

But $\lceil x+n\rceil$ is the only integer that satisfies $x+n\le\lceil x+n\rceil< x+n+1$, so $$\lceil x+n\rceil=\lceil x\rceil+n$$

Tiago Emilio Siller
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