Proof:

$$a$$ is algebraic if there exists a polynomial $$f(x) = c_0 + c_1 x + \dots + c_n x^n$$, $$c_0, \dots, c_n \in \mathbb(Q)$$, such that f(a) = 0.

Multiply f(x) with M, where $$M = LCM (c_0, \dots, c_n)$$.

$$M*f(x) = M*c_0 + M*c_1 x + \dots + M*c_n x^n$$ will have all integer coefficients. Define $$I_i = M*c_i$$.

f(x) and g(x), where $$g(x) = I_0 + I_1 x + \dots + I_n x^n$$, both have same roots.

Now, consider the following number,

$P_{g} = p_1^{I_0} * p_2^{I_1} * \dots * p_n^{I_n}$

where $$p_i$$ is the $$i$$th prime number.

Note that the number $$P_g$$ associated with g(x) is unique for the given set of integers $$I_0, \dots, I_n$$.

This gives us a bijection between the set of algebraic numbers and the set of integers.

The proof is complete.