I'm reading David Marker's book "Model Theory: An Introduction" and I'm trying to solve Exercise 3.4.24 which is stated as follows:
Let $x$ and $y$ be algebraically independent over $\mathbb{R}$.
a) Show that $\mathbb{R}(x,y)$ is formally real and that we can find orders $<_{1}$ and $<_{2}$ of $\mathbb{R}(x,y)$ such that $x<_{1}y$ and $y<_{2}x$.
b) Use a) to show that the ordering $<$ is not quantifier-free definable in $\mathbb{R}$ in the language of rings.
I would prove b) not using a) as follows. Let $\mathcal{R}=(\mathbb{R},+,-,\cdot,0,1)$. We know that the theory of real closed fields admits elimination of quantifiers (in the language of ordered rings). Hence, if there is a quantifier-free formula $\psi(v_0,v_1)$ in the language of rings such that for all $x,y\in\mathbb{R}$ $$\mathcal{R}\models \psi(x,y) \iff x<y$$then $Th(\mathcal{R})$ would have quantifier elimination which is not the case.
My questions are:
1) Is my proof correct?
2) Can someone give me a hint how to prove b) by using a).