Why is it impossible to define multiplication in Presburger arithmetic? Peano arithmetic defines multiplication recursivly as:
$$\begin{gather}a\cdot0=a\\a\cdot S(b)=a+(a\cdot b)\end{gather}$$
Why is this not possible in Presburger arithmetic?

Peano arithmetic doesn't actually define multiplication. The existence of multiplication is an axiom in the Peano system. The recursive definition of multiplication you provide in the question cannot prove the existence of multiplication function for all numbers (even though it can for any finite value of $n$ and $m$). There are two ways around it:


*

*Use the recursive definition that you provide, and assume that a function with these properties exists (i.e. existence of multiplication is an axiom).

*Add the Recursion Theorem as an axiom and use that prove the existence of multiplication.


Neither of these axioms exist in Presburger arithmetic and so it is not possible to define multiplication in Presburger arithmetic without extending it.
Note: The reason the Recursion Theorem is called a theorem and not an axiom is because it can be proved in ZF set theory. But if your only axioms are the Peano axioms, it can't be proved and needs to be it's own axiom.
A good explanation of the intricacies of defining multiplication in Peano arithmetic can be found in this blog: How multiplication is really defined in Peano arithmetic
