# Thread: show there is exactly one n such that n<x<n+1

1. ## show there is exactly one n such that n<x<n+1

If x$\displaystyle \notin$ Z (the integers) show that there is exactly one n$\displaystyle \in$Z such that n<x<n+1.
Would I show this by contadiction stating x$\displaystyle \in$Z?

2. We can suppose that $\displaystyle x>0$ first.
We know that $\displaystyle \left( {\exists n \in \mathbb{Z}} \right)\left[ {x < n} \right]$. By well ordering let $\displaystyle K$ be the least to have that property.
What can be said about $\displaystyle K-1?$

If $\displaystyle x<0$ then do it for $\displaystyle -x$.

3. Originally Posted by Plato
We can suppose that $\displaystyle x>0$ first.
We know that $\displaystyle \left( {\exists n \in \mathbb{Z}} \right)\left[ {x < n} \right]$. By well ordering let $\displaystyle K$ be the least to have that property.
What can be said about $\displaystyle K-1?$

If $\displaystyle x<0$ then do it for $\displaystyle -x$.
Would it be that its not in the section? Or am I straying away? I'm looking at my notes and it goes from well ordering to countable without much detail.

4. If $\displaystyle K$ is the least positive integer having the property that $\displaystyle x<K$ then $\displaystyle K-1<K$ so $\displaystyle K-1<x<K$.

You know that there is no integer between $\displaystyle K-1~\&~K$.