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The following is a well-known proposition. $$\forall n \in \mathbb{N} :\sum_{j=1}^n j = \frac{n(n+1)}{2}$$

How do we actually express a sentence like this in the language of Peano Arithmetic?

goblin GONE
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  • What do you mean with the language of Peano Arithmetic? Do you mean just using the symbols $S, 0, +$ and $\cdot$, (and the logic symbols, of course)? – Git Gud Aug 15 '13 at 14:58
  • @GitGud, yeah that's what I meant. Other relation symbols are also acceptable, so long as its made clear what formula they're being used as a shorthand for. – goblin GONE Aug 15 '13 at 15:02
  • So basically you're asking to define $\sum \limits _{j=1}^n$ and the fraction, right? – Git Gud Aug 15 '13 at 15:06
  • @GitGud, yeah. I already know how to do the fraction so I can probably fill that detail in myself. Edit. Indeed, it occurs to me that we can just take the $2$ to the LHS, so the division becomes a product. – goblin GONE Aug 15 '13 at 15:08
  • You can define the sum recursively. If $n=0$, then it's just $0$. I suppose you can do the rest. – Git Gud Aug 15 '13 at 15:10
  • @GitGud, yes, but I'm not sure how translate the idea of 'defining something recursively' into an actual sentence of PA expressing that idea. – goblin GONE Aug 15 '13 at 15:13
  • Related: http://math.stackexchange.com/questions/340111/recursion-schema-and-the-arithmetical-hierarchy – Asaf Karagila Aug 15 '13 at 15:17
  • Why can't we just write $2(1+...+n)=n(n+1)$? Why do we need to invoke induction? –  Aug 15 '13 at 15:20
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    @Colin because $\ldots$ mean nothing, that's just a sort of device people use to summarize things. – Git Gud Aug 15 '13 at 16:02
  • @Colin: How do you prove that $2(1+\ldots+n)=n(n+1)$, for every $n$, without using induction? – Asaf Karagila Aug 15 '13 at 17:25
  • Ah, I see. There is a universal quantifier at the beginning of the given statement. –  Aug 17 '13 at 11:34

2 Answers2

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The expression $n \mapsto \sum_{j=1}^{n} j$ can be regarded as defining a certain primitive recursive function $f$: \begin{align} f (0) & = 0 \\ f (n + 1) & = f (n) + n + 1 \end{align} Gödel has shown that every primitive recursive function is definable in Peano arithmetic. In our case, that means there exists a formula $\varphi (x, y)$ such that PA proves:

  • $\varphi (x, y)$ is functional.
  • $\varphi (0, 0)$.
  • $\forall x . \varphi (x, y) \to \varphi (x+1, y + x + 1)$

The formula $\varphi$, even for such a simple function, is rather complicated. The hardest part is to develop a coding for finite lists of natural numbers. Once we have this, $\varphi (x, y)$ essentially says that there exists $L$ coding a finite list such that:

  • The $0$-th entry of $L$ is $0$.
  • If the $n$-th entry of $L$ is $a$, and $L$ has an $(n+1)$-th entry, then the $(n+1)$-th entry is $a + n + 1$.
  • $L$ has an $x$-th entry.
  • The $x$-th entry of $L$ is $y$.

The theorem you wish to state can then be formalised as follows:

$$\forall x. \exists y. \varphi (x, y) \land (2 \times y = x \times (x + 1))$$

Zhen Lin
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You are interested in $f(n) = \sum_{j=0}^n j$. But evidently $f$ is primitive recursive, and there is a standard trick due to Gödel for expressing any primitive recursive function in the language of PA by deploying his so-called $\beta$-function [now outlined by Zhen Lin].

The clearest fuller explanation of that which I know is in P*t*r Sm*th's Introduction to Gödel's Theorems, Ch. 15 of the second edition (which is a bit too long to cut'n'paste here)!

Peter Smith
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  • Thanks, Ch15 has been enlightening. +1 – goblin GONE Aug 16 '13 at 02:14
  • Peter, what do you think of this? -- It occurs to me that, if instead of axiomatizing $\mathbb{N},$ if we instead axiomatized the hereditarily finite sets, then we wouldn't need a $\beta$ function. (Because a sequence of hereditarily finite sets can be described as a function in the usual way). Thoughts? – goblin GONE Aug 16 '13 at 02:21