I'll try to answer your questions one by one hoping to be clear enough without making reference to very deep aspects (although your questions are already very deep):
- "So, since it is complete, decidable and consistent why don't we use Presburger arithmetic instead? I know that it may has something to do with it's weakness, but it's not very clear what is the meaning of this weakness, presumably one can prove less things with it perhaps?"
You are right in saying that Presburger Arithmetic proves less things. Mathematicians working in Number Theory usually like to talk about multiplication of natural numbers in general: an easy example is that they would like to have the property that "multiplication commutes for all natural numbers"; another example is that they like to talk about all the prime numbers at once. However with Presburgeer Arithmetic you cannot do that, if you check the axioms (http://en.wikipedia.org/wiki/Presburger_arithmetic) they don't talk about multiplication. So, you can only go number by number to define it. Short answer: Presburguer Arithmetic has many cute logical properties, but not enough mathematical richness, and that's why it is not "used" practically.
- "My other doubt about Presburger arithmetic is why is it a problem of it having no multiplication? I always see multiplication being defined after the operation of sum: The multiplication is just a given number of sums, can't we just make the multiplication and use it anyway or the multiplication causes problems?"
Yes we can, but as said earlier, we can only do it one number at a time. We would have to prove something like "the successor of the successor of zero (two) multiplied (imagine we just defined it as you suggested) with the successor of zero (one) is equal to the successor of the successor of zero." We would have to prove a theorem like this for every natural number, and it is not physically possible. We would also have to prove commutativity and associativity for multiplication by 2 for each natural number. In summary, this is not a very practical thing to do.
- "Also, from the first theorem of incompleteness: "Any effectively generated theory capable of expressing elementary arithmetic cannot be both consistent and complete." Doesn't Presburger arithmetic have an elementary arithmetic?"
Allow me to assume you read that in a Wikipedia article. Usually this kind of articles are written in a way that the general public understands it. However in mathematical terms a "theory capable of expressing elementary arithmetic" can be read as "a theory that at least can prove all of the axioms of PA". Hence, the answer to your question is "no", Presburger arithmetic doesn't have an elementary arithmetic. For more specific information of what you could interpret as "elementary arithmetic" you can check the MO question: https://mathoverflow.net/q/118183/32592
- "Why can't we define $a\times b$ for all natural numbers?"
This is rather hard to answer as I have not worked with the axioms of Presburger Arithmetic before (so take into account that the rest of the paragraph might not be very accurate). But let's try to attack another problem to get an idea of why this can't be done: How do you write in first order logic that there are two objects? $\exists x\exists y(x\neq y)$. How do you write that there are three objects? $\exists x\exists y\exists z(x\neq y\wedge x\neq z\wedge y\neq z)$. And so on. How do you write that there are infinite things? You would have to use an infinitely long formula which is physically impossible. If you want to define multiplication in Presburger Arithmetic, you would have to do something similar and you would need an infinitely long formula too.
One could also think that if Presburger arithmetic lacks multiplication, then we can do arithmetic without addition and just having multiplication. It turns out that such a system is also complete and consistent but it still has the practicality issue pointed above: https://mathoverflow.net/a/19874/32592
There are various ways to say $\mathsf{ZFC}$ is stronger than $\mathsf{PA}$.
One way to compare them is to measure their arithmetical consequences. Both $\mathsf{ZFC}$ and $\mathsf{PA}$ can express statements on arithmetic, and we can see that $\mathsf{ZFC}$ proves more arithmetic statements than $\mathsf{PA}$. ($\mathsf{Con(PA)}$ is an example.) Some subsets of arithmetical consequences (for example, $\Pi^0_2$-consequences of a theory) are adopted to measure the proof-theoretic strength of a given theory.
However, the above method is only applicable when the given theories are able to express arithmetic. There is a more direct (perhaps more akin to looking one theory implies another) way to see it: interpretation. Let me introduce its formal definition, as finding its definition online seems not easy.
Definition. Let $T_0$ and $T_1$ be theories over a language without function symbols (but not necessarily over the same language.) Then an interpretation $\mathfrak{t}:T_0\to T_1$ is a map which sends a formula to a formula as follows:
- $\mathfrak{t}$ preserves $\land$, $\lor$, $\to$ and $\lnot$, e.g., $(\phi\land\psi)^\mathfrak{t}$ is $\phi^\mathfrak{t}\land\psi^\mathfrak{t}$,
- There is a formula $\delta(x)$ over $T_1$ (which means domain of an interpretation) such that $(\forall x\phi(x))^\mathfrak{t}$ is $\forall x \delta(x)\to\phi^\mathfrak{t}(x)$, and $(\exists x\phi(x))^\mathfrak{t}$ is $\exists x \delta(x)\land\phi^\mathfrak{t}(x)$,
- For each relation symbol $R$ over $T_0$, there is a formula $\phi_R$ (with the same arity of $R$) such that $\mathfrak{t}$ assigns $R$ to $\phi_R$.
- Furthermore, if $T_0\vdash\phi$ then $T_1\vdash \phi^\mathfrak{t}$.
For example, there is an interpretation from the theory of $(\mathbb{Z},+)$ to the theory of $\mathbb{N}$: we can code integers and the addition operation by the standard method. Another example is an interpretation from $\mathsf{ZFC}$ to $\mathsf{ZF}$: taking a constructible universe yields this interpretation.
We may call $T_1$ is stronger then $T_0$ if there is an interpretation from $T_0$ and $T_1$, since $T_1$ can simulate $T_0$ inside itself.
We can see that $\mathsf{ZFC}$ can interpret $\mathsf{PA}$: we know that $\mathsf{ZFC}$ can define the set of natural numbers $\mathbb{N}$ and operations over $\mathbb{N}$. This gives a natural interpretation of arithmetic into $\mathsf{ZFC}$.
Best Answer
Peano arithmetic is relatively strong, and decidable theories of arithmetic like Presburger Arithmetic are fairly unusual. Here's a "tower" of three theories between the two, of increasing strength:
Note how I've distinguished them by which arithmetic operators thay have, and which they cannot prove total. This is a very natural measure of the strength of theories of arithmetic: an essential part of proving the incompleteness theorem is providing a diagonalisation operation, which we can do only if we have multiplication. These theories have this, and so all of these theories admit the incompleteness theorems.