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

**tn11631** · September 19th 2010, 11:01 AM

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?

**Plato** · September 19th 2010, 11:14 AM

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 $-x$ .

**tn11631** · September 19th 2010, 11:36 AM

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.

**Plato** · September 19th 2010, 01:05 PM

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

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

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