Thread: 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!

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.



Now, suppose 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.

. We have reached a contradiction since

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?

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.

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

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:
if and only if .
That means that if then so that
If the theorem to which the current question is a corollary is for all the how could there be an integer between

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?

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.

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

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 where .

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