I'm facing some difficulties in understanding what its mean to prove that a set is a definable set. Below is an example of a question that was given in our class, with the solution for this question. Could someone please help me understand the solution, and how to deal with similar questions?
Let τ = <f,R> be a dictionary such that f is a 2-local function, and R is 1-local relation.
M defined as M = ⟨ℚ+, . , <1⟩ is a structure for τ.
A. Show that 1 is definable in M
B. Show that the relation { (a,b) ∈ ℚ² | a is the opposite of b } is definable in M.
The answers are as follows:
A. Create a formula α1(x) such that M ⊨_z α1(x) if and only if z(x) = 1:
α1(x) = ∀y(f(x,y) = y)
B. Create a formula α2(x,y) such that M ⊨_z α2(x,y) if and only if z(x),z(y) are opposite :
α2(x,y) = ∃z(α1(z) ∧ (f(x,y) = z )
You need to find a formula $\varphi(x)$ with a free variable $x$ such that your structure models $\varphi(x)$ iff $x$ is assigned to $1$. Since $\mathbb{Q}$ is a field, a unique and easy way to define $1$ is \begin{equation} \varphi(x)\equiv \forall y (f(x,y) = y) \end{equation} and then $\varphi(x)$ holds iff $x$ is the multiplicative identity, i.e. $x$ is assigned to $1$. As for the relation, you may first need to define $0$: you can do it with $Z(x) \equiv \forall y(f(x,y)=x)$ and then you can define the relation using the fact that $a$ is the opposite of $b$ is the same as $ab < 0$, so a possible route would be $\exists v(Z(v) \wedge R(f(a,b),v))$ where you say “$v = 0$ and $ab < v$”.