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

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

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

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

If $x<0$ then do it for $-x$.
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 $K$ is the least positive integer having the property that $x then $K-1 so $K-1.

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