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I have to prove that

$\forall x \in \mathbb{R},\exists\text{ exactly ONE }n \in \mathbb{Z} \text{ s.t. }n \leq x < n+1\;.$

I'm done with proving that there are at least one integers for the solution.

I couldn't prove the "uniqueness" of the solution, and so I looked up the internet, and here's what I found:

Let $\hspace{2mm}n,m \in \mathbb{Z} \text{ s.t. }n \leq x < n+1$ and $m \leq x < m+1$.

Since $n \leq x \text{ and } -(m+1) < -x$, by adding both, $n + (-m-1) < (x-x) = 0$. And (some steps here) likewise, $(x-x) < n+m+1$.

Now, can I add up inequalities like that, even when the book is about real analysis (and in which assumptions are supposed to be really minimal)?

Or should I also prove those addition of inequalities?

Thank you :D

2 Answers 2

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The usual proof in the context of real analysis goes like this:

Let $A= \{ n \in \mathbb Z : n \le x \}$. Then $A$ is not empty. Indeed, there is $n\in \mathbb N$ such that $n>-x$, because $\mathbb N$ is unbounded. But then $-n\in A$.

Let $\alpha=\sup A$. Then there is $n\in A$ such that $\alpha-1. But then $\alpha and so $n\le x$ and $n+1\notin A$, that is, $n\le x < n+1$.

If $m\in A$ then $m\le n$ because $m>n$ implies $m\ge n+1>x$. If $m then $m+1\le n\le x$ and $m$ cannot be a solution. Hence the solution is unique.

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    Thanks for the outline of the proof :D2012-03-08
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As to your "adding inequalities," I do not see any issue. The basic properties of $<$ are used frequently in proofs, ordinarily without explicit mention, except at the very beginning, when ordered field has just been defined.

Multiplication by $-1$, followed by addition, is cute, but has a somewhat magical character. I think it is more natural to argue as follows. Suppose that $m$ and $n$ are integers such that $m\le x Then $m \le x So we have $m and $n. From $m we conclude that $m-n<1$. From $n we conclude that $m-n>-1$. Thus $m-n=0$.

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    @user269334: I noticed that a TeX error may have cut o$f$f the end, the usual missing dollar sign.2012-03-08