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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!

Johnathan
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    The difficulty with your proof is that $-n$ is not a member of $\Bbb N$. Suppose you tried a direct approach and used $m=n+1$. Would that work? – abiessu Feb 10 '15 at 04:57
  • If the OP proof were correct, then every natural number is greater than every other natural number, which can't be true. – vadim123 Feb 10 '15 at 04:58
  • @abiessu Thank you! I see what you mean. I will re-think it! :) – Johnathan Feb 10 '15 at 05:12
  • I don't think the theorem can be proved from these axioms. The axioms are consistent with a model in which $\mathbb N = \mathbb Z = {}$; that is, none of the axioms actually requires any numbers to exist. They only say what properties the numbers would have if they existed. I think there must be a missing axiom. – David K Feb 10 '15 at 22:38
  • Also, if $>$ is defined in terms of subtraction, subtraction must first be defined. – David K Feb 10 '15 at 22:41
  • @DavidK There is another proposition: Let $m,n,p \in\mathbb Z$. If $m < n$ and $n < p$ then $m < p$. Subtraction is defined as $m - n = m + (-n)$. My prof said the idea of contradiction is to assume that a statement is true (e.g. $m -n \notin\mathbb N$) and then see what derives from that to (hopefully) see a contradiction. Here, my contradiction (I think) is that in order for $m -n \notin\mathbb N$ to be true, $m-n$ is either equal to zero or $-(m - n) \in\mathbb N$. However, (m - n) can't be two different options (according to the proposition). I think it's a contradiction. – Johnathan Feb 11 '15 at 00:58

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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.

Mathemagician1234
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  • Thank you for your input! :) In my course, we have seen $ \mathbb Z$ (Chapter 1). In chapter 2, we are introducing $\mathbb N$ so we can sort $\mathbb Z$. The problem is that we haven't yet seen any number (e.g. 1) to belong to $\mathbb N$. It's still abstract. Hence, I am looking for another approach. :) – Johnathan Feb 10 '15 at 16:29