A clarification of this model theory question regarding when two theories are the same I was told in an answer to a previous question that for all non-zero real numbers $r$ and $r'$, the theory of the structures $(\mathbb{R};+,r)$ and $(\mathbb{R};+,r')$ are the same. But how can this be? For example, the theory of $(\mathbb{R};+,2)$ is different from the theory of $(\mathbb{R};+,3)$, because, for example, the first contains the statement $2=2$ while the second one doesn't, because it doesn't have a $2$ symbol in the first place. So, are those theories really the same, or merely equivalent, and if so, what is the definition of two theories being equivalent?
I would like a clarification of this issue.
 A: Neither theory contains the statement $2 = 2$ because $2 = 2$ is not a statement in the vocabulary.
You have discussed structures, but structures must be a structure over some language. The language here is a language consisting of one binary operator (which we'll call $+$) and one constant symbol (which we'll call $c$).
The informal statement $2 = 2$ corresponds to the formal proposition $c = c$, which is clearly satisfied by both structures.
Because there is an isomorphism between the structures $(\mathbb{R}; +, r)$ and $(\mathbb{R}; +, r')$, they have the same theory.
A: Technically, something like $(\mathbb{R};+,2)$ is not a structure. A structure is a pair $(X; \mathfrak{I})$ where $\mathfrak{I}$ is a function assigning appropriate symbols to their interpretations. So, for example, using $\color{red}{\mathsf{red}}$ to denote symbols, if we have a binary function symbol $\color{red}{\star}$ and a constant symbol $\color{red}{c}$ then $$(\mathbb{R}; \{\color{red}{\star}\mapsto +, \color{red}{c}\mapsto 2\})$$ is a structure. However, we usually elide the symbolic part, abbreviating the above as simply "$(\mathbb{R}; +,2)$."
So the fully-precise statement is that the structures $$(\mathbb{R}; \{\color{red}{\star}\mapsto +, \color{red}{c}\mapsto 2\})\quad\mbox{and}\quad (\mathbb{R}; \{\color{red}{\star}\mapsto +, \color{red}{c}\mapsto 3\})$$ are elementarily equivalent (indeed, isomorphic), but this is abbreviated by simply saying $(\mathbb{R};+,2)\equiv(\mathbb{R};+,3)$. Note that their languages are the same: $\{\color{red}{\star},\color{red}{c}\}$.

As an aside, note that it can sometimes feel a bit hair-splitting to focus on the details of the language. Say that a prestructure is a pair $$(X; (A_i)_{i\in\mathbb{N}})$$ where $X$ is a set and $A_i\subseteq X^i$ satisfies a couple simple properties:

*

*Each $A_i$ is a sub-Boolean algebra of $\mathcal{P}(X^i)$.


*$\{(x,x): x\in X\}\in A_2$.


*For each permutation $\sigma\in S_i$ and each $Z\in A_i$ the set $\{(x_{\sigma(0)},...,x_{\sigma(i-1)}): (x_0,...,x_{i-1})\in Z\}$ is in $A_i$.


*For each $Z\in A_{i+1}$ the "projection" $\{(x_0,...,x_{i-1}): \exists y\in X[(x_0,...,x_{i-1}, y)\in Z]\}$ is in $A_i$.
Every structure $\mathcal{X}=(X;\mathfrak{I})$ has an associated prestructure gotten by letting $A_i$ be the set of parameter-freely-definable $i$-ary relations on $X$ in $\mathcal{X}$.
(Contra an incredibly stupid claim of mine earlier - my best theory is that I'd accidentally switched to decaf - the converse is not true: for example, if $R,S$ are $\mathcal{X}$-parameter-freely-definable binary relations, then so must be their composition $R\circ S=\{(x,y):\exists z[xRz\wedge zSy]\}$. And that's just the start of things - consider something like "$\{(x,y):$ $xSy$ iff there are at least seven distinct $z$ with $xRz\}$." However, this won't affect what I say below.)
Effectively, given a structure $\mathcal{X}$ the associated prestructure $\hat{\mathcal{X}}$ has all the information that $\mathcal{X}$ does ... except it doesn't remember what the "primitive" relations/functions/etc. are, nor which symbol corresponds to which relation. There's an analogy here with the (term) clones of universal algebra.
(Sadly, the above terminology is nonstandard.)
