How do I characterize the addition operator, without defining its properties using set theory or the peano axioms, so that someone with special needs could understand and apply it in the context of arithmetic and at most algebra? What is the definition of the addition operator in the simple context of algebra and arithmetic, and how is it understood?
The essence of the Addition operator within arithmetic
algebra-precalculusarithmeticeducationsoft-question
Related Solutions
Delving into the proof of completeness theorem, as Noah suggests, is one good way to gain intuition about nonstandard models of PA. Here's another approach.
First, we should see why there are nonstandard models of PA. It's not hard to find proofs that "Peano's Axioms" have a unique model, up to isomorphism. (Peano's predecessor Dedekind may have given the first such proof, in his paper "Was sind und was sollen die Zahlen" ("The nature and meaning of numbers")). This doesn't contradict the existence of nonstandard models of PA, because "PA" nowadays doesn't mean Peano's Axioms, but the first-order theory of arithmetic. Peano's induction axiom says that an arbitrary subset of $\mathbb{N}$ is $\mathbb{N}$ if it contains 0 and is closed under succession. This is a second-order axiom, meaning it talks about arbitrary sets of natural numbers.
The induction axioms of PA talk only about subsets of the system that can be defined by a first-order formula: a formula built up with +, $\times$, etc., and the basic constructs of logic ($\forall$, $\neg$, $\wedge$, etc.) A nonstandard model, call it $N$, will have subset that is isomorphic to $\mathbb{N}$; we might as well call it $\mathbb{N}$. This subset violates the second-order version of induction in $N$, since it's a subset closed under succession and containing 0 but not all of $N$. But it's "invisible" to first-order logic: it can't be defined by a first-order formula.
There is a weaker set of first-order axioms known as PA$^-$. (See here for a list of its axioms.) Basically this includes all the "algebraic" and "ordering" properties of PA, but omits all the induction axioms. So for example we have $x+y=y+x$ and $x<y\rightarrow x+z<y+z$ as axioms. Here it's much easier to construct a nonstandard model. Start with $\mathbb{N}$, and add a constant $c$; you may think of $c$ as a kind of "infinite number". Because you have $c$, you also must have $c\pm n$ for all $n\in\mathbb{N}$, $c^2$, $2c^2$, etc. In fact, it's not hard to see that the model must contain elements corresponding to all polynomials of the form $$a_nc^n+\cdots+a_0,\qquad a_n,\cdots,a_0\in\mathbb{Z},\;a_n>0$$ And that's enough! The set of such polynomials, together with 0, with the natural definitions for $+,\times,<$, is a model of PA$^-$ (call it $M$).
The induction axioms do not hold in $M$. For example, you can prove by induction that $$\forall x\exists y(y+y=x\vee y+y+1=x)$$ in other words, every number is even or odd. But there is no such $y$ in $M$ when $x=c$.
What to do, if we want a nonstandard model of PA? One idea: extend $M$ to a model $N$ of PA. Say we add a new constant $d$ and stipulate that $d+d+1=c$. (Or $d+d=c$, but not both!) Just as adjoining $c$ forced us to add lots of other elements, adjoining $d$ will force us to add a lot more. And that's just for one instance of one induction axiom! Still, the basic idea is to keep adjoining new elements whenever PA proves that something should exist, and it doesn't yet.
It is kind of remarkable that this approach can be made to work. With the model $M$, each polynomial defined a unique element, and different polynomials defined different elements. Also it was not hard to decide on the meaning of $+,\times,<$ for the elements of $M$. This becomes much more problematic with the wholesale introduction of constants designed to satisfy the induction axioms.
That it does work is the beauty of Henkin's construction, that Noah sketched. But we have to sacrifice something. The model $M$ of PA$^-$ is recursive, i.e., you could write a computer program to implement it. Tennenbaum's theorem says that no recursive nonstandard model of PA exists. In fact, you can't even make a model where the + operation is computable, not to say $\times$. (And vice versa.)
Plug: I've been blogging about nonstandard models of PA with a couple of other people (John Baez and Bruce Smith).
Best Answer
I will try to write what I understood from Prof. Herbert Gross's teachings and innovation.
The addition operator operates on numbers, so a good knowledge of numbers can give us some insights. If we follow this PowerPoint by Prof. Gross, we have something very interesting to look at
And from this presentation we have
If the students are asked to put two tiles, they will probably do this
And if they are asked to put those two tiles along with three more tiles, they will probably do this
$$ {\Huge \color{blue}{\blacksquare\blacksquare \qquad \blacksquare \blacksquare \blacksquare}} $$
$$ \begin{aligned} {\Large \mathbf{\text{How to apply it in Algebra}}} \end{aligned} $$
Let's say we have $3~\color{red}{apples}$ and $2~\color{blue}{oranges}$ and if I ask you how many fruits do I have, you will answer $5$. So, what you have basically done is that you did the following translation $$ 3~\color{red}{apples} \rightarrow 3 ~\mathbf{fruits} \\ 2 ~\color{blue}{oranges} \rightarrow 2~\mathbf{fruits} $$ And then you added "3 fruits" and "2 fruits" in the same way as we added the tiles above.
Now, let's say we have any noun $x$, such that $3x$ means we have $3$ of those $x$(in the same way as $3$ fingers, or $3$ inches) and $2x$ means we have $2$ of those. If we say that we are given $3x$ and $2x$ then how many $x$ do I have in total? We can surely apply that tiles example once again to see things clearly, because it is stated that $x$ is any noun. Let's visualize $x$ by some strange looking figure and draw $3$ and $2$ of them
As the question strictly doesn't allow Set Theory's or Peano's Axiom (works of Gottlob Frege, Bertrand Russell) definition of addition so I wouldn't touch on that and end my answer here.