Denote $\mathbb Q[x]$ = set of polynomials with coefficients $c_1$, $c_2$, $...$ ,$c_n$ in $\mathbb Q$.
A number $a$ is algebraic if there exists a polynomial $f(x)$ in $\mathbb Q$[x] such that $f(a)$ $=$ $0.$ (it is a solution to the polynomial).
1) Show that $\mathbb Q[x]$ is denumerable.
My attempt:
I claim that the set Q is denumerable from a theorem which I proved previously. Thus I claim that either:
$\mathbb Q$[x]$ is denumerable or $\mathbb Q$[x]$ is uncountable.
Because a polynomial has $n$ finitely many terms, we are able to count the class of sets that contain polynomials with rational roots. (For every $n$th term, we are able to find its corresponding coefficient $c_n$ and vice versa).
Thus $\mathbb Q$$[x]$ is a denumerable union of disjoint finite sets, which is denumerable.
2) Show that the set $A$ of algebraic numbers $a$ is denumerable.
We have established that the set $\mathbb Q$[x] is denumerable. $A$ $:=$ {$a$: $a$ is a solution to $f(x)$ $=$ $0$}
A polynomial of degree $n$ may have at most $n$ roots, or $n$ many $a$ terms. That is, each set of roots for a particular polynomial is finite.
Thus $A$ is a union of these finitely many sets of roots for particular polynomials. This is a denumerable union of finite sets, which must be denumerable.
3) Show that the set of transcendental numbers is equipotent to $\mathbb R$.
Proof by Contradiction.
Assume that the set $T$ of transcendental numbers is countable.
Then $T$ U $A$ $=$ $\mathbb R$. But clearly, the set $\mathbb R$ is uncountable.
A union of 2 countable sets cannot be uncountable; since $A$ is denumerable, then $T$ cannot be denumerable and must thus be uncountable.