TITLE: Why is Peano arithmetic undecidable?
QUESTION [21 upvotes]: I read that Presburger arithmetic is decidable while Peano arithmetic is undecidable. Peano arithmetic extends Presburger arithmetic just with the addition of the multiplication operator. Can someone please give me the 'intuitive' idea behind this?
Or probably a formula in Peano arithmetic that cannot be proved. Does it have something to do with the self reference paradox in Gödel's Incompleteness Theorem?

REPLY [2 votes]: @hardmath:
"It's a good point, that undecidability of a formal first-order theory implies incompleteness but not the converse" - that is simply not true.
The theorem (that is actually an easy consequence of Post's theorem) we have here is:
For any FO-theory T, if T is recursively axiomatizable and complete, then it is decidable.
This does not mean that an undecidable theory is necessarily incomplete - it can be complete, but lack  recursive axiomatization. That is e.g. the case of ${Th(N, +, \times, 0, 1)}$. Moreover, there are theories which are both recursive and decidable, but incomplete, e.g. $T=\{0=0\}$.