# Possible circular reasoning in textbook proof that $\lceil x+m\rceil=\lceil x\rceil +m$

The goal is to prove that $\lceil x+m\rceil=\lceil x\rceil +m$, where $x$ is a real number and $m$ is an integer. The book outlines the following proof:

Write $x=n-\epsilon$, where $n$ is an integer and $0\leq\epsilon<1$; thus, $\lceil x\rceil=n$. Then $\lceil x+m\rceil=\lceil n-\epsilon+m\rceil=n+m=\lceil x\rceil+m$.

If we "read between the lines," it really seems like the following is being communicated:

\begin{align} \lceil x+m\rceil&=\lceil n-\epsilon+m\rceil\\ &= \lceil-\epsilon\rceil+(n+m)\tag{circular reasoning?}\\ &= 0+(n+m)\tag{since $\epsilon\in[0,1)$}\\ &= \lceil x\rceil+m \end{align}

Is the book's proof fine and I'm just not seeing something clearly or is there a subtle error somewhere (if so, what could be done to fix the proof?)?

• how do you define $\lceil x \rceil$ if it's not with $\lceil n - \epsilon \rceil = n$ for any integer $n$ and $\epsilon \in [0;1[$ ??? – reuns Jan 11 '16 at 4:29
• It's not quite circular since we know that $n+m$ is an integer. Since $\epsilon < 1$, we can evaluate $\lceil n + m - \epsilon \rceil$. – amcerbu Jan 11 '16 at 4:33
• I don't think it's circlular. It is presumed that if $0 \le e < 1$ then ceil(n - e) = n for integer n. (either this is the definition or a very basic proposition). Thus there is no circularity at all ceil(x + m) = ceil(n - e + m) = ceil((n+m) - e) = n + m = ceil(n) + m. (because (n+m) is an integer so ceil((n+m) -e) = n + m). – fleablood Jan 11 '16 at 4:34
• @user1952009 Yeah, I think that makes sense. The restriction on $\epsilon$ makes the argument non-circular; I guess I was just seeing the explication of the argument in a somewhat circular fashion when that's really not the case – interrogative Jan 11 '16 at 4:37

As is pointed out in the comments, there is no circular reasoning here. To clear things up substitute $u=n+m$ (which will be an integer) and then use the theorem you already used and accepted, that $\lceil n-\epsilon \rceil = n$ where $n$ is any integer (and replace $n$ by $u$)