I am learning proofs with $\mathbb N$. Here is my proposition:
For each $n \in\mathbb N$ there exists $m \in\mathbb N$ such that m > n. Here are my axioms:
a)If $m,n \in\mathbb N$ then $m + n \in\mathbb N$
b)If $m,n \in\mathbb N$ then $mn \in\mathbb N$
c) $0 \ne\ \mathbb N$
d) For every $m \in\mathbb Z$, we have $m \in\mathbb N$ or $m = 0$ or $-m \in\mathbb N$
Definition: $m > n = m - n \in\mathbb N$ Btw, I would greatly appreciate if someone could please explain to me why this is. My strategy is to use a contradiction.
Here is another proposition that I have proven: For $m \in\mathbb Z$, one and only one of the following is true: $m \in\mathbb N$, $-m \in\mathbb N$, $m = 0$.
Btw, we haven't seen any number (in class) that belongs to $\mathbb N$, only the properties of $\mathbb N$. Hence, I can't use 1,2,..
Proof: Let $m, n \in\mathbb N$, assume that $m - n \in\mathbb N$ is false, namely, $m - n \notin\mathbb N$. Given that addition is a binary operation, it will give a number $p \in\mathbb Z$. $m - n \notin\mathbb N$ holds if $-p \in\mathbb N$ or $p = 0$ (i.e. $0 \notin\mathbb N$) (axiom d). However, according to the proposition that I have proven, one and only one of the following is true: $m \in\mathbb N, -m \in\mathbb N, m = 0$. There is a contradiction. Thus, $m - n \in\mathbb N$ is true.
What do you think? Thank you!
abiessu correctly identified in his/her comment what the problem is with your approach. So let's try a contradiction proof using his/her suggestion that avoids this problem: Let m=n+1.We know this number exists since n is in N and 1 is in N so the existence of m follows from axiom (a). Assume $m - n \notin\ N$. Then $m - n =1\notin\ N$, which is impossible. So there is at least one number where m>n. Q.E.D.
I notice the existence of 1 in N is not explicitly included in the axioms.We can define 1 as the first nonzero member of N. Clearly,the fact that N is nonempty is assumed here. In more complete treatments, the number 1-or more precisely,a smallest natural number-is given by the axioms,as is the inductive property. But we'll just go by what you got and assume 1 comes with the package.
To answer your other question,the definition of "greater then" in the axioms is a consequence of axiom (d). This definition allows one to make sense of the idea that one number being "more positive" or "larger" then another means when one takes the difference between them, the result is a number larger then 0. In more abstract treatments, the set which uses N here is referred to as a positive class.Positive classes are critical to establishing the order properties of the number systems and other ordered sets.