(a) Let $x$ be an algebraic number of degree $n$ if it is a root of a polynomial of degree $n$ with rational coefficients. Then, the set of all algebraic numbers is the countable union of the sets of algebraic numbers of a finite degree, and if we show that the latter sets are all countable, we are done. Now let us map the set of all algebraic numbers of degree $n$ to the set $\mathbb{Z}_{+}\times\mathbb{Q}^{n}$ by $f(x)=(k,a_{0},\ldots,a_{n-1})$ , where $x$ is the $k$ th root of the polynomial $x^{n}+a_{n-1}x^{n-1}+\cdots+a_{1}x+a_{0}$ (we pick any such polynomial for $x$ , and we know that for each such polynomial there are only finite number of roots, so we order different roots of the polynomial, and determine $k$ for $x$ ). $f$ is clearly injective, and since $\mathbb{Z}_{+}\times\mathbb{Q}^{n}$ is countable, we have a composite of two injections being an injective correspondence from the set of all algebraic numbers of degree $n$ to $\mathbb{Z}_{+}$ . Note, that, in fact, the set of all algebraic numbers is countably infinite, as it contains $\mathbb{Q}$ , being algebraic numbers of degree $1$ .

(b) If it were countable, then the set of the real numbers would be countable as the union of the countable sets of algebraic and transcendental numbers.