I'm defining real numbers from the field axioms plus 2 axioms stating that:
-
The set of positive real numbers is closed for addition and multiplication.
-
All real numbers except zero either belong to the set of positive real numbers or are symmetrics of the real numbers and no positive real is equal to its symmetric.
(Also the supremum axiom but I'm required not to use it yet.)
The definition I'm given for the set of natural numbers is the intersection of all inductive subsets of real numbers which contain $0$. (Inductive meaning that if $x$ belongs to that set then so does $x+1$).
I havent yet proved the well ordering theorem. From the definition, it was easy to prove that induction can be used to prove statements for all natural, and also very simple to prove that the naturals are closed for multiplication and addition. However whatever I do I cant prove the following:
$$\forall m,n \in N : m>n \implies m-n \in N$$
My attempt was the following:
Induction basis: $\forall m \in N: m>0 \implies m-0 \in N$
Which is obvious.
Induction hypothesis: $\forall m \in N: m>k \implies m-k \in N$
So we have 3 possibilities: either $k+1>m$ or $k+1=n$ or $k+1<m$. The first two evidently mean the condition holds for k+1. But if $k+1<m$ then I have to prove that $m-k-1 \in N$. I have no idea how to proceed.
Best Answer
Well, it's nice that you attacked this problem using induction, but it's better to do it for $m$. You should first fix $n$, for $m = n+1$ we have $(m-n) \in \Bbb{N}$, clearly. Now, let $m\in \Bbb{N_{\ge n+1}}: m-n \in \Bbb{N}$.
We have $m-n \in \Bbb{N}$ Therefore $S(m-n)=m+1-n \in \Bbb{N}$ With S the successor function We have then $$\forall m\in \Bbb{N_{\ge n+1}},m-n \in \Bbb{N}$$ So $$\forall n\in \Bbb{N},\forall m\in \Bbb{N}, m>n\implies m-n \in \Bbb{N}$$