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 $\displaystyle l^2<l$
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.
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 $\displaystyle 0~\&~1~?$
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