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I am trying to prove the following proposition:

For each $n \in\mathbb N$ there exists $m \in\mathbb N$ such that m > n.

Here are my axioms:

  1. If $m,n \in\mathbb N$ then $m + n \in\mathbb N$

  2. If $m,n \in\mathbb N$ then $mn \in\mathbb N$

  3. $0 \notin\ \mathbb N$

  4. For every $m \in\mathbb Z$, we have $m \in\mathbb N$ or $m = 0$ or $-m \in\mathbb N$

Definition: $m > n = m - n \in\mathbb N$.

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: $m \in\mathbb N$, $-m \in\mathbb N$, $m = 0$.

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?

user642796
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Johnathan
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  • The final paragraph is not right. You should know it is wrong intuitively because at the start you merely assume $m$ and $n$ are in $\mathbb N$ and then proceed to "show" that $m > n$. Counterexample: choose $n = 2$, $m = 1$. The problem is there was really never any contradiction in your argument; you started with the assumption $m\in\mathbb N$, and that satisfies your "one and only one" theorem just fine. – David K Feb 10 '15 at 22:10
  • @DavidK Thank you for your input. In my class, we have not yet seen any number that is natural. We simply have assumed that there exists a subset $N$ within $Z$. Hence, I am not allowed to use numbers. I assume the opposite and then try to show a contradiction from the axioms and the propositions. :) Could you suggest a different approach? Thank you! – Johnathan Feb 10 '15 at 22:17
  • I wasn't actually suggesting you write $1$ or $2$ in your proof. The counterexample was just a way for you to check your proof; it should be a tipoff that something is wrong if an argument shows that $1\in\mathbb N$ and $2\in\mathbb N$ implies $1 > 2$, even if you haven't actually constructed either of those numbers yet. – David K Feb 10 '15 at 22:26
  • BTW, there should be a link to the previous question http://math.stackexchange.com/q/1141649/139123 (and now there is). It's laudable that you're making multiple attempts at this rather than giving up. – David K Feb 10 '15 at 22:34
  • @DavidK Thank you! :) I am being obsessed lol I really want to understand. – Johnathan Feb 11 '15 at 00:48
  • Question: can you prove that there exists something in $\Bbb N$? – Greg Martin Feb 13 '15 at 04:38
  • can you please explain what you mean by $0\neq \mathbb N$ – Elaqqad Feb 14 '15 at 14:36
  • @Elaqqad Hi! It is an axiom: it means that zero is not a natural number (I wrote it wrong actually.) – Johnathan Feb 16 '15 at 02:03

1 Answers1

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Since you refer to $\mathbb{Z}$ in your axiom d), I assume that $\mathbb{Z}$ is somehow "known" while $\mathbb{N}$ is not.

The first order of business is to prove that $1\in \mathbb{N}$. You can do this using your trichotomy axiom d), and arithmetic facts you know from $\mathbb{Z}$. Let me know if you need more hints.

Once you know $1\in \mathbb{N}$, for any given $n\in \mathbb{N}$, can you think of how to construct an $m$ with $m>n$ (using your definition of $>$)?

user7530
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  • Hi! Thank you for your reply. Here is my strategy for proving that 1 $\in\mathbb N$: Given m $\in\mathbb N$, $-m \notin\mathbb N$. $-m = (-1)m$. $-1$ is obviously not zero but isn't a natural number either because if it were, (-1)m would yield a natural number, which is not the case. A number is exclusively 0, natural, or non-natural. Hence, -1 is not natural. If you take the inverse of -1(i.e -(-m)), you get one which must be natural. What do you think? I am still thinking about the other proof. I'll get back to you! :) – Johnathan Feb 16 '15 at 02:54
  • Yes, this will work. You can also argue about 1 directly: since $1\neq 0$ (we know this from $\mathbb{Z}$) we must have that $1\in \mathbb{N}$ or $-1\in\mathbb{N}$. If the former, we are done. If the latter, then $-1\cdot -1=1$ must be a natural number after all. – user7530 Feb 16 '15 at 05:43
  • I think I have the rest. Ok: so assume that $n \in\mathbb N$, I can construct another number $m \in\mathbb N$ by adding 1 to $n$. If $m > n$, then $m - n \in\mathbb N$. This is the case because $m - n = 1$. Hence, it is always possible to generate a natural number greater than another natural number. What do you think? Thank you! – Johnathan Feb 17 '15 at 01:53