Solution by Henry Woody
Problem 2.15
Problem Statement: A real number is called algebraic if it is a root of an algebraic equation f (x) = 0,
where f (x) = a0 + a1 x + a2 x2 + · · · + an xn is a polynomial with integer coefficients. Prove that the set of
all polynomials with integer coefficients is countable and deduce that the set of algebraic numbers is also
countable.
Proof. Begin by proving the set of all integer polynomials is countable. Let Ai be the set of all integer
polynomials of degree i. Define φ : Ai → Zi+1 by φ(a0 + a1 x + a2 x2 + · · · + ai xi ) = (a0 , a1 , a2 , · · · , ai ), which
is a bijection. To show that Zk is countable, for all finite k, induction on k is employed. For the base case,
let k = 1, then Z is countable by Lemma 1 (see below). Now assume that Zk is countable for all k ≤ n,
and it will be shown that Zn+1 = Zn × Z is countable. Since Zn and Z are countable, there exist bijections
f : Zn → Z+ and g : Z → Z+ . Define h : Zn × Z → Z+ by h(x1 , x2 , · · · , xn+1 ) = 2f (x1 ,x2 ,··· ,xn ) · 3g(xn+1 ) .
Since f and g are bijections, and due to the structure of this function, h is a bijection. Therefore Zk is
countable for all k.
From the above it is clear that Zi+1 is countable for all i ∈ Z+ . It follows that Ai is countable for all
∞
S
i ∈ Z+ through the composition of φ and h. Let A =
Ai , which is a countable union of countable sets
i=0
and is therefore countable by Theorem 2.27. So the set of all integer polynomials A is countable.
The information above leads to the conclusion that the set of all algebraic numbers is countable. To show
∞
S
this, let Rij be the set of roots of the jth polynomial of degree i. Then R =
Rij is the set of all roots
i,j=0
of all integer polynomials. The number of roots of any polynomial is countable since there can be at most
m roots of a polynomial of degree m, so Rij is countable for all i and j. The set of algebraic numbers R is
therefore a countable union of countable sets and is countable by Theorem 2.27.
Lemma 1. Z is countable.
Proof. Define γ : Z → Z+ by γ(x) =

2x
if x > 0
−2x + 1
if x ≤ 0.
Since γ is a bijection, Z is countable.
1