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

  1. #1
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    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?
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  2. #2
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    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 $\displaystyle -x$.
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  3. #3
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    Quote Originally Posted by Plato View Post
    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.
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  4. #4
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    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 $\displaystyle K-1~\&~K$.
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