*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.

.

We may assume that the integer coefficients have no common factors, and that the leading coefficient is positive. The polynomialdenote the sequence of prime numbers. Let

.

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]