1. ## Integer 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!

2. Originally Posted by jstarks44444
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!
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 $\displaystyle l^2<l$

3. How would I go about doing this for n and n+1 though? using this method i arrive at

ln < l^2 < ln + l

4. Originally Posted by jstarks44444
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.
Originally Posted by jstarks44444
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?
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.

5. 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

6. Originally Posted by jstarks44444
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
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 $\displaystyle 0~\&~1~?$

7. 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?

8. Originally Posted by jstarks44444
But n is an element of the integers, not of the naturals....
If you continue to refuse to post a complete list of axioms, definitions, and theorems you will not receive help.

9. 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

10. Originally Posted by jstarks44444
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
Thank you for that.
BUT what is the definition for $\displaystyle m>n$ where $\displaystyle \{m,n\}\subset\mathbb{Z}$.

In case you do no know this, your text material is non-standard.
We need to know the axiom set for the integers.