Field extension $\mathbb{R/Q}$ is transcedental? Can somebody explain to me ( or give a proof ) why the field extension $\mathbb{R/Q}$, that is the field of real numbers as an extension of the field of rational numbers, is transcendental and not algebraic, which would mean that each element of $\mathbb R$ would be a root of some polynomial with rational coefficients only if it is transcendental?
 A: You are confusing the quantifiers when you negated the statement.

$K$ is an algebraic extension of $F$ if every element of $K$ is the root of a polynomial with coefficients in $F$. So $\forall k\exists p\in F[x]: p(k)=0$.
$K$ is a transcendental extension of $F$ if it is not algebraic. So $\lnot(\forall k\exists p\in F[x]:p(k)=0)$.

Let's parse this negation into the statement, we have $\exists k\forall p\in F[x]:p(k)\neq 0$. So $K$ is transcendental over $F$ if there is at least one element in $K$ which is not the root of any polynomial with coefficients in $F$.
Now a simple counting argument shows that $\Bbb Q[x]$ is a countable set, and every polynomial has finitely many roots. So there are only countably many algebraic numbers in any given extension of $\Bbb Q$. Since $\Bbb R$ is uncountable it means that "most" of its elements are indeed transcendental.
A: Any algebraic extension of $\mathbb{Q}$ is denumerable, since its elements are roots of polynomials with coefficients in $\mathbb{Q}$, but $\mathbb{R}$ is not denumerable so it contains elements that are not algebraic and this elements are said transcendental.
