I need help for this task, that I just can't solve since I have no idea where to start. It is connected with this task.
Let $K$ be the field of rational expressions with real coefficients i.e. expressions of the form $\frac{p}{q}$ where $p$ and $q\neq0$ are polynomials with real coefficients. Show that $K$ is ordered field. (Hint: Use $a$ ) with $$P:=\left\{\frac{a_0 +a_1x+...+a_{n-1}x^{n-1}+a_nx^n}{b_0+b_1x+...+b_{n-1}x^{m-1}+b_mx^m}\in K : a_nb_m > 0, m, n \in \mathbb{N}\right\}.$$
I am thankful for any advice.
My Approach
In order to show that $K$ is an ordered field, we need to either define a total order $\leq$ on it that is compatible with the field structure, or equivalently define a positive cone, which tells you what the positive elements are.
Adjusting the hint slightly to make this answer fit more nicely with your linked question, let's define $$P:=\left\lbrace\frac{a_0+a_1x+\cdots+a_{n-1}x^{n-1}+a_nx^n}{ b_0+b_1x+\cdots b_{m-1}x^{m-1}+b_mx^m}\in K:a_nb_m\geq 0\right\rbrace.$$ In words, we could say that $P$ consists of the rational expressions where the product of the leading terms $a_n$ and $b_m$ of the numerator and denominator is non-negative.
If we can show that $P$ is a positive cone, then $K$ is an ordered field with the ordering $$X\leq Y\iff Y-X\in P$$ by part (a) in your linked question. So we need to check that $P$ is closed under addition and multiplication, and that if $X\in P$ and $-X\in P$, then $X=0$ (the zero fraction).
I provide some hints, since this is ultimately just dealing with fractions from hereon in. Let $$X=\frac{a_0+a_1x+\cdots+a_{n-1}x^{n-1}+a_nx^n}{ b_0+b_1x+\cdots b_{m-1}x^{m-1}+b_mx^m}\;\;\;\text{ and }\;\;\;Y=\frac{c_0+c_1x+\cdots+c_{n-1}x^{n-1}+c_nx^n}{ d_0+d_1x+\cdots d_{m-1}x^{m-1}+d_mx^m}$$ be rational expressions in $P$. Can you express the leading terms of the numerator and denominator of $X+Y$, $XY$ and $-X$ in terms of those for $X$ and $Y$? Remember, you can treat these rational expressions in much the same way as ordinary fractions.