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.
Is there a formal definition of "Greater Than"
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4$n>m$ if $n=m+a$ for some positive integer $a$ – 2012-10-16
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1One 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 $S(x) = x + S(0)$ – 2012-10-16
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0> is a subset of $\Bbb N\times \Bbb N$. It contains all ordered pairs (n,m) where n = m + k for some $k\in\Bbb N$. – 2012-10-16
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1@Deven: Or even simpler: $\exists a:y=x+S(a)$. – 2012-10-16
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1@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. – 2012-10-16
2 Answers
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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1I'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. – 2012-10-16
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0@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... – 2012-10-16
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0Shouldn't you add that $c$ is nonnegative? Because for all $a,b$ there exists a $c$ such that $b = a + c$, not only when $a \leq b$. – 2013-07-20
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1@timvermeulen: we're working in PA. Everything's nonnegative. – 2013-07-20
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.
if $a,b\in{P}$ then $a+b,ab\in{P}$.
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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0Hey 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. – 2013-07-20
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0@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. – 2013-07-20
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0Zetta, 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. – 2013-07-20
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0@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? – 2013-07-20
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0I did some edits so tell me what you think. You're free to roll it back if you wish. – 2013-07-20
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0@user18921 Perfect. Thanks! – 2013-07-20
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0You're welcome! – 2013-07-20