Algebraic numbers are (complex) numbers that can be expressed as roots of polynomials with integer coefficients. The set of algebraic numbers is denumerable, i.e., it is equivalent to the set of natural numbers. Since the set of natural numbers is equivalent to the set of rational numbers it suffices to show that the set of algebraic numbers is equivalent to the set of rational numbers. Furthermore, since every rational number is also algebraic (write p/q as the solution of qx=p ) we only have to show that we can associate a rational number with every algebraic number such that no two algebraic numbers give rise to the same rational number.
To this end consider the polynomial
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A few examples may be helpful in explaining this association, Of course, since any algebraic number can be expressed as the solution of infinitely many polynomials, the number r(x) depends on the polynomial p as well as on the number x. To remove the dependence on the polynomial we pick q(x) to be the smallest rational number than can be written as r(x). Since the prime factorization of a natural number is unique, q(x) is a unique positive rational number associated with the algebraic number x (and with no other algebraic number). The set of algebraic numbers is therefore equivalent to a subset of the rational numbers, and, as we wish to show, the set of algebraic numbers is equivalent to the set of rational numbers.
As an exercise you may want to modify the above way to associate rational numbers with algebraic numbers such that it associates natural numbers instead.
Fine print, your comments, more links, Peter Alfeld, PA1UM
[16-Aug-1996]