Show given any $x\in\mathbb{R}$ show there exists a unique $n\in\mathbb{Z}$ such that $n-1\leq x <n$. I already know that there exists a $n\in\mathbb{N}$ such that $x<n$ by the archimedean property. But to prove that $n-1\leq x$ part I'm not sure. I think I have to create some set such that $n-1$ is a lower bound. But I'm not sure.
[Math] Show given any $x\in\mathbb{R}$ show there exists a unique $n\in\mathbb{Z}$ such that $n-1\leq x
real-analysis
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Best Answer
Hint: Let $x\ge 0$. By what you wrote, the set $A_x$ of positive integers such that $n\gt x$ is non-empty. Thus $A_x$ has a smallest element. Call that $n$, and show that it does the job.
For negative $x$, use the previous argument to show there is an integer $m$ such that $m-1\le |x|\lt m$. Then $-m\lt x\le -m+1$. If $x\ne -m+1$, we are finished. If $x=-m+1$, make a minor adjustment.
We leave dealing with uniqueness to you.
Remark: We don't need to give special treatment to negative $x$ if we first prove the following small extension of the Least Number Principle: Any non-empty set of integers which is bounded below has a smallest element.