Proof that $\Bbb{Q}$ ⊆ set of algebraic numbers

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A number is called algebraic if it is the root of a polynomial $p(x) = a_nx^ n+ a_{n−1}x^{n−1} + \cdots + a_1x + a_0$, where each $a_i \in \Bbb{Z}$.

Let $\Bbb{A}$ denote the set of algebraic numbers.

(a) Prove that $\Bbb{Q} \subseteq \Bbb{A}$.

(b) Prove that the set of all algebraic numbers is countably infinite. (Hint: First consider the possible roots of polynomials of degree k. Then use a union argument).)1

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$\forall q\in\mathbb{Q}$, we can write $q=\frac{a}{b}$ in reduced form where $a,b\in\mathbb{Z}$.

We consider the simple linear equation $bz-a=0$.

Note that $p(z)=bz-a$ is a polynomial, its coefficients are in the integers, and its root is clearly $q=\frac{a}{b}$.

Since $q\in\Bbb Q$ was chosen arbitrarily, and an integer coefficient polynomial was constructed such that $q$ was a zero, then $q$ is an algebraic number. Thus $\Bbb Q\subseteq\Bbb A$.

$\blacksquare$