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Intuitively, one can say that $S(n) > n$. But how do we prove it using the Peano Axioms. It seems like I need a formal statement as to what $>$ means.

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$n>m$ if $n=m+a$ for some positive integer $a$ –  Holdsworth88 Oct 16 '12 at 16:55
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One could define $y>x$ to mean $\exists a (y = x + a)\land (a\neq 0)$, this would of imply that $S(n) > n$ because it is easy to see in PA that $S(x) = x + S(0)$ –  Deven Ware Oct 16 '12 at 16:57
    
> is a subset of $\Bbb N\times \Bbb N$. It contains all ordered pairs (n,m) where n = m + k for some $k\in\Bbb N$. –  Shahab Oct 16 '12 at 16:58
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@Deven: Or even simpler: $\exists a:y=x+S(a)$. –  Henning Makholm Oct 16 '12 at 16:59
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@Shahab: That phrasing doesn't work if we're working in Peano Artithmetic, since PA is a separate first-order theory that doesn't let you speak about "sets", "ordered pairs" and so forth. –  Henning Makholm Oct 16 '12 at 17:01

2 Answers 2

up vote 6 down vote accepted

Usually, it's $\leq$ which gets defined first, not $>$. In the case of PA, you can define $\leq$ as $$ a \leq b \leftrightarrow \exists c\: (b = a+ c) $$

But of cource, once you've defined one of the relations $\leq$, $<$, $>$, $\geq$, definitions for the others follow immediately. You e.g. have $$ a > b \leftrightarrow (a \neq b) \land (b \leq a) $$

Or you can define $>$ directly as $$ a > b \leftrightarrow \exists c\: (c \neq 0) \land (a = b + c) $$

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I'm not sure what your first sentence means. The axioms themselves don't define any ordering at all; that's something that the user of the axioms is free for himself to do or don't, according to his needs. –  Henning Makholm Oct 16 '12 at 17:03
    
@HenningMakholm What I meant is that if you define an order, you usually define it in terms of $\leq$. Will try to make it say that more clearly... –  fgp Oct 16 '12 at 17:24
    
Shouldn't you add that $c$ is nonnegative? Because for all $a,b$ there exists a $c$ such that $b = a + c$, not only when $a \leq b$. –  timvermeulen Jul 20 '13 at 16:54
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@timvermeulen: we're working in PA. Everything's nonnegative. –  Chris Eagle Jul 20 '13 at 18:31

Here is a definition that uses the order axiom:

For an ordered field $\mathbb{F}$, there is a unique subset $P$ satisfying the following conditions.

  1. if $a,b\in{P}$ then $a+b,ab\in{P}$.

  2. for all $a$ in $\mathbb{F}$, one and only one of the following is true: $a\in{P}$, $-a\in{P}$, or $a=0$.

We say that $p \in \mathbb{F}$ is positive iff $p \in P$.

The subset $P$ is the positive numbers of the field. From here, the relation of 'strictly greater than' can be defined as follows.

$$a>b\iff{a-b}\in{P}$$

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Hey is $P$ always unique? Btw, I believe the OP was specifically asking about the Peano Axioms approach to arithmetic. But, this answer is still interesting. –  goblin Jul 20 '13 at 15:54
    
@user18921 Yes, good question. See my latest edit, I had forgotten an important part of the definition of $P$. I noticed that the user asked for Peano axioms, but (and I could be wrong) it sounded like from the way the question was stated, the OP was more concerned with finding a formal definition than with finding it under that particular constraint. –  Ataraxia Jul 20 '13 at 16:08
    
Zetta, is the second part an additional condition on $P$, or is it a consequence of the definition? If its the former, it would be better to make that clearer. Perhaps use a numbered list, with one condition per number. –  goblin Jul 20 '13 at 16:11
    
@user18921 It's an additional condition on P. It's part of the definition. I tried doing a numbered list, but I couldn't get it inside the quotation box. Any idea how I would do that? –  Ataraxia Jul 20 '13 at 16:15
    
I did some edits so tell me what you think. You're free to roll it back if you wish. –  goblin Jul 20 '13 at 16:19

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