I'm trying to show that
given any $x \in \mathbb{R}$, prove that there exists a unique $n \in \mathbb{Z}$ s.t. $n-1 \leq x < n$.
my solution:
Consider the set $A = \{ n \in N : x < n \}$. Since this set is nonempty( it is bounded above by ceiling(x+1)) , by the well ordering principal of the natural numbers, A must have a least element. Call it $m$. Then $m$ is the smallest natural number greater than $x$. Now, note that $ x \geq m-1$ or else m wouldn't be the least element of $A$. Thus, $m-1 \leq x < m$.
It seems to me that the uniqueness of m should be obvious. Is there a rigorous way to show it here?
Does my solution work?
Thank you.
Best Answer
As for the uniqueness, suppose that $m$ and $n$ both satisfy the requirement.
If $n<m$, then $n-1<m-1\leq x$ so $n\leq m-1\leq x$ and $n\leq x$. But $x<n$ by hypothesis, a contradiction.
A similar argument shows that $m<n$ is impossible.
Hence, $m=n$ by trichotomy.
Note. The existence of such an $n$ (noted $\lfloor x+1\rfloor$) is equivalent, in an ordered field, to the Archimedean property.