I would appreciate some help with this proof!

Let n be an element of the integers. There exists no integer x such that n < x < n+1.

Thanks!

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- Feb 16th 2011, 03:16 PMjstarks44444Integer Corollary Proof
I would appreciate some help with this proof!

Let n be an element of the integers. There exists no integer x such that n < x < n+1.

Thanks! - Feb 16th 2011, 03:32 PMdwsmith
Here is a proof that there is no integer between 0 and 1 as an example for you.

$\displaystyle S=\{n\in\mathbb{Z} | 0<n<1\}$

Now, suppose $\displaystyle a\in (0,1)$ and a is an integer. Now, since 0 < a < 1 is in S, then S is non-empty. By the well-ordering principle, S has a least element, l, so 0 < l < 1.

Multiple the inequality by l.

$\displaystyle 0*l < l*l < 1*l\Rightarrow 0<l^2<l$. We have reached a contradiction since $\displaystyle l^2<l$ - Feb 17th 2011, 09:24 AMjstarks44444
How would I go about doing this for n and n+1 though? using this method i arrive at

ln < l^2 < ln + l

but can any contradiction be made there? - Feb 17th 2011, 09:42 AMPlato
You have us at a clear disadvantage. We have no way to know what text material you are using. In the title of this post is the word ‘corollary’. That means it is in addition to some theorem. What theorem? You just have let in on the what basis you are proving things.

- Feb 17th 2011, 09:56 AMjstarks44444
Well the theories that come before this corollary are,

*For all k in the Naturals, k >= 1

*There exists no integer x such that 0<x<1 - Feb 17th 2011, 10:18 AMPlato
I suspect that you also have the following definition:

$\displaystyle m>n$ if and only if $\displaystyle m-n\in \mathbb{N}$.

That means that if $\displaystyle n\in \mathbb{N} $ then $\displaystyle n-0=n\in \mathbb{N} $ so that $\displaystyle n>0$

If the theorem to which the current question is a corollary is $\displaystyle n\ge1$ for all $\displaystyle n\in \mathbb{N} $ the how could there be an integer between $\displaystyle 0~\&~1~?$ - Feb 17th 2011, 02:40 PMjstarks44444
But n is an element of the integers, not of the naturals....

where is the inductive step in this, or do you not need one? contradiction? - Feb 17th 2011, 02:50 PMPlato
- Feb 17th 2011, 02:55 PMjstarks44444
Sorry..axioms for the natural numbers:

If m,n are in the naturals then m+n is in the naturals

If m,n are in the naturals then mn is in the naturals

0 is not in the naturals

For every m in the integers, we have m in the naturals or m=0 or -m in the naturals - Feb 17th 2011, 03:04 PMPlato