\(a\) 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.