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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.

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    I am almost certain that I have answered all these questions in several forms on the site before. You might want to search for such questions before posting. The answers may suffice to tell you what you want to know.2012-11-11

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