1

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.

PiStrich
  • 67
1233211
  • 797
  • Related: http://math.stackexchange.com/questions/496733/show-given-any-x-in-mathbbr-show-there-exists-a-unique-n-in-mathbbz-such?rq=1 arguably a duplicate if not for the proof-verification tag. – JMoravitz Dec 20 '16 at 08:14
  • I think that you can simply assign $n=\lfloor{x+1}\rfloor$. – barak manos Dec 20 '16 at 08:16
  • This was an exercise from a book i'm reading and i was trying to do something similar to what they did on a similar problem to try to understand the process better. – 1233211 Dec 20 '16 at 08:20
  • What if x<0?.................... – DanielWainfleet Feb 20 '19 at 06:27

1 Answers1

2

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.

Guest
  • 4,158