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.
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
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