Given a $\mathcal{L}=\{0,1,+,\cdot , R\}$ and structure $C=(\mathbb{C},...)$ give a formula so that $C\models \phi(x)$ iff $x$ is a real number

Question

Given a $\mathcal{L}=\{0,1,+,\cdot , R\}$ and structure $C=(\mathbb{C},...)$ give a formula so that $C\models \phi(x)$ iff x is a real number

The interpretations of $0,1,+, \cdot$ are the usual interpretations. $R$ is a unary relation and the subset which consists of strictly imaginary numbers $0+bi$

Having a hard time coming up with this formula since I need to somehow take arbitrary complex numbers $x$ and identify that in the form $a+bi$ that $b=0$

I think the statement I want to symbolize would be, if $x$ is $Rx$ then $r+x$ is complex. Since adding any non zero number should give me something that isn't a real number.

Answer 1

$$\phi(x) := \forall y (R(y) \Rightarrow R(x \cdot y))$$

If $x = bi$ is strictly imaginary, then $R(x)$ but we don't have $R(x^2)$ since $x^2 = -b^2$. If $x = a + bi$ is fully complex, then taking $y = bi$ gives $R(y)$ but not $R(x \cdot y) = R(abi - b^2)$. But if $x = a$ is strictly real, then for every strictly imaginary $y = bi$, we have $x\cdot y = abi$, strictly imaginary, and thus $R(x\cdot y)$. So the implication holds exactly when $x$ real.

Answer 2

Nicholas Viggiano gave a universal definition of $\mathbb{R}$ in $(\mathbb{C};0,1,+,\cdot,R)$. Here's an existential definition: $$\exists z\, (z\cdot z + 1= 0 \land R(z\cdot x))$$

The point is that the witness for the existential quantifier must be either $i$ or $-i$ (the two roots of the polynomial $z^2+1$), and a number $x$ is real if and only if $xi$ is purely imaginary if and only if $-xi$ is purely imaginary.

Yet another alternative: $$\exists z\,(R(z)\land (z\cdot z=x\lor z\cdot z\cdot z\cdot z=x)).$$

This works because for any purely imaginary $z$, the square of $z$ is a negative real number or $0$, and the fourth power of $z$ is a positive real number or $0$, and every real number has a purely imaginary square root or fourth root.