Indistinguishable Irrational Multiples In (R;+,r)

Hey guys! Today we're diving deep into the fascinating world of model theory, specifically looking at the structure $(\mathbb{R};+,r)$. This is a follow-up to our previous discussion on definable elements, so if you haven't checked that out, you might want to swing back! We're going to tackle a juicy question: Is every irrational multiple of $r$ indistinguishable from every other? Let's get into it!

Understanding Indistinguishability

First off, what does it mean for two elements to be indistinguishable in a structure? In model theory, two elements $a$ and $b$ in a structure $\mathcal{M}$ are considered indistinguishable if, for any formula $\phi(x)$ with a single free variable $x$ in the language of the structure, $\mathcal{M} \models \phi(a)$ if and only if $\mathcal{M} \models \phi(b)$. Basically, you can't tell them apart using any of the defined properties or relations within that structure. Think of it like having two identical-looking coins; unless you have some special way to distinguish them (like a hidden mark or a different weight), they're indistinguishable to you. In our case, the structure is $(\mathbb{R};+,r)$, where $\mathbb{R}$ is the set of real numbers, '+' is the standard addition operation, and 'r' is a fixed, nonzero real number. We're interested in whether different irrational multiples of this $r$ behave the same way with respect to all possible formulas in this language.

The Structure $(\mathbb{R};+,r)$

Let's break down our structure $(\mathbb{R};+,r)$ a bit more. We have the familiar set of real numbers, $\mathbb{R}$. The operation is standard addition, which is super well-behaved. The 'twist' comes from the constant symbol $r$, which represents a specific nonzero real number. This might seem small, but it changes things! The language of this structure consists of:

1. Variables: Like $x, y, z$, etc.
2. Constants: The specific number $r$ (and implicitly, 0, which is definable as $x+0=x$ or $r-r=0$).
3. Functions: The addition operation '+'.
4. Relations: Here, we don't have any explicit relation symbols like '<' or '=' for arbitrary pairs. The equality symbol '=' is always part of the logical framework itself, but we're interested in definable properties using the given language. So, we're looking at formulas built using variables, the constant $r$, addition, and equality.

This means any formula we can write will essentially involve sums and differences of variables and the constant $r$. For example, $\phi(x): x + r = 5$ or $\psi(x): x + x = r$. We are asking if, for any such $\phi(x)$, if $\phi(a)$ holds, then $\phi(b)$ also holds for any two irrational multiples $a$ and $b$ of $r$. It's like asking if all irrational multiples of $r$ are interchangeable from the perspective of this structure's rules.

The Core Question: Irrational Multiples

Our focus is on elements of the form $k \cdot r$, where $k$ is an irrational number. We want to know if $k_1 r$ and $k_2 r$ are indistinguishable for any irrationals $k_1$ and $k_2$. Let's consider an element $a = k r$ where $k$ is irrational. What can we say about $a$ using only addition and the constant $r$?

Any formula $\phi(x)$ in the language of $(\mathbb{R};+,r)$ will look something like:

$ \exists y_1, \dots, y_n \forall z_1, \dots, z_m \quad ( (y_i = T_i(x, r, \text{vars})) \land (z_j = U_j(x, r, \text{vars})) \land \dots ) $

where $T_i$ and $U_j$ are terms built using addition, subtraction, and the constant $r$.

For example, if we have a formula like $\phi(x): x + r = 2r$. If $x = kr$, then this becomes $kr + r = 2r$, which simplifies to $(k+1)r = 2r$. Since $r \neq 0$, we can divide by $r$ to get $k+1 = 2$, so $k=1$. This means that only $x=1r$ satisfies this formula. If $k$ is irrational, then $x=kr$ will not satisfy this formula.

Now, suppose we have two irrational numbers $k_1$ and $k_2$, and let $a = k_1 r$ and $b = k_2 r$. We are asking if for every formula $\phi(x)$, $\phi(a)$ holds if and only if $\phi(b)$ holds. If this is true, then $a$ and $b$ are indistinguishable.

The Key Insight: Rational vs. Irrational Coefficients

Consider a formula $\phi(x)$ involving addition and the constant $r$. Any term in this formula will ultimately be of the form $c_1 x + c_2 r$, where $c_1$ and $c_2$ are rational numbers. This is because addition preserves the 'rational coefficient' structure when combined with $r$.

Let's write a generic formula $\phi(x)$. It will involve quantified variables and terms. Each term will be a sum/difference of variables and multiples of $r$. For a single variable $x$, any term $T(x)$ in the language of $(\mathbb{R};+,r)$ can be represented in the form $ax + br$ for some $a, b \in \mathbb{Q}$.

So, any formula $\phi(x)$ is equivalent to a statement about whether $ax + br$ satisfies certain equations or inequalities involving other terms of the form $a'x + b'r$.

Let $a = k_1 r$ and $b = k_2 r$, where $k_1, k_2 \in \mathbb{R} \setminus \mathbb{Q}$. Suppose $\phi(x)$ is a formula in the language $(\mathbb{R};+,r)$.

Let's consider a very simple type of formula: $x + c r = d r$, where $c, d \in \mathbb{Q}$.

If $\phi(x)$ is $x + c r = d r$, then substituting $x = k r$ gives $k r + c r = d r$, which means $(k+c)r = dr$. Since $r \neq 0$, this is equivalent to $k+c = d$, or $k = d-c$. Since $d-c$ is a rational number, this formula is satisfied only if $k$ is rational. But we are considering $k$ to be irrational. So, for any irrational $k$, $x=kr$ will not satisfy such a formula.

What if the formula is more complex? Let $\phi(x)$ be any formula in the language of $(\mathbb{R};+,r)$. Any term $t(x)$ in this formula can be written as $q_1 x + q_2 r$ for some $q_1, q_2 otin \mathbb{Q}$ (this is not entirely correct, let's refine this).

Correction: Let's be more precise. A term $T(x)$ in the language $(\mathbb{R};+,r)$ is built from $x$, $r$, and addition. So, $T(x)$ will be of the form $a x + b r$, where $a, b \in \mathbb{Q}$.

For example, if $\phi(x)$ is $x+r = 2x$. Substituting $x=kr$: $kr+r = 2kr$. This gives $r(k+1) = 2r k$. Since $r \neq 0$, $k+1 = 2k$, so $k=1$. This formula is only satisfied by $x=1r$, which is a rational multiple of $r$.

Let's consider $a = k_1 r$ and $b = k_2 r$ where $k_1, k_2$ are distinct irrational numbers. We want to know if $\phi(a) \iff \phi(b)$ for all formulas $\phi(x)$.

Suppose we have a formula $\phi(x)$ that can distinguish $a$ from $b$. This means either $\phi(a)$ is true and $\phi(b)$ is false, or vice versa.

Consider the structure $(\mathbb{R}; +, r)$. The set of definable numbers (elements that can be denoted by a term in the language) are precisely the rational multiples of $r$. Let $S_{def}$ be this set. If $r$ is transcendental over $\mathbb{Q}$, then this might be different. But $r$ is a real number, so we assume standard reals.

Let's analyze the effect of substitution. If $\phi(x)$ is a formula, and we substitute $x = yr$ into it, we get a new formula $\phi'(y)$ in the language of $(\mathbb{R}; +, 0)$ with an extra constant $r$. The structure $(\mathbb{R}; +, r)$ is essentially $(\mathbb{R}; +)$ with a distinguished element $r$.

Consider the simplest possible formulas involving $x$: $x = q r$ for $q otin {Q}$. This formula is not definable in the language $(\mathbb{R}; +, r)$, since $q$ would need to be part of the language or definable. Only rational multiples of $r$ are definable as single terms. However, we are talking about arbitrary formulas.

Let $a = k_1 r$ and $b = k_2 r$ where $k_1 eq k_2$ and both are irrational. Suppose there exists a formula $\phi(x)$ such that $\phi(a)$ is true and $\phi(b)$ is false.

Any formula $\phi(x)$ in $(\mathbb{R}; +, r)$ can be translated. Let $x = yr$. The formula $\phi(x)$ becomes $\phi'(y, r)$, where $\phi'$ is a formula in the language of $(\mathbb{R}; +, ext{constants})$. The structure $(\mathbb{R}; +, r)$ is related to the structure $(\mathbb{Q}; +, imes)$ and $(\mathbb{R}; +, imes)$.

Let's consider a formula $\phi(x)$. If $\phi(x)$ involves only addition and the constant $r$, then any term $T(x)$ in $\phi$ is of the form $q_1 x + q_2 r$ for $q_1, q_2 otin {Q}$. This is where the mistake is. The coefficients $q_1, q_2$ must be rational.

So, any term $T(x)$ in the language $(\mathbb{R}; +, r)$ will be of the form $c x + d r$, where $c, d \in \mathbb{Q}$.

If $\phi(x)$ is a formula, then $\phi(kr)$ is true iff $\phi(k'r)$ is true for two irrationals $k, k'$ if and only if the formula $\phi$ does not