A complex number z is called algebraic if there exists a polynomial f(x) = a0 +

Question

A complex number z is called algebraic if there exists a polynomial f(x) = a0 + a1x + a2x 2 + · · · + anx n with integer coefficients such that f(z) = 0. A complex number which is not algebraic is called transcendental. The set of algebraic numbers includes all of those which are expressible in terms of radicals and rationals, such as 2, p3 5/19, 1 2 + 7 42 3 5 q 7+6 1/2 2 3 7 . There are many algebraic numbers which are not expressible in this way; for example, the roots of x 5 x1 are not expressible in this way 1 . On the other hand, some familiar numbers such as , e, ln 2 have been shown to be transcendental. The point of this problem is to show that there are far more real (and complex) numbers which are transcendental than algebraic. (i) Show that if the sets A and B are countable, then so is A × B. Hint: you may use the result from Example 8.11(d). (ii) Use the previous result to deduce that if A is countable and n N then An is countable. (iii) Show that for each positive integer n, the set of all polynomials of degree n with integer coefficients is countable. (iv) Show that the set of all algebraic numbers is countable.

Explanation / Answer

i) For a fixed a A, let Ba = {(a, b) A × B | b B}. Since B is countable, each Ba is countable. Note that UaABa is the countable union of countable sets, and hence is countable . Since A × B = UaABa, we have that A × B is countable.

ii) Now let us show that show that An is countable, where An = A x A x A ..........x A (n times). Applying what we see that A × A is countable. Applying the same result to A and A × A , we see that A x A x A is also countable. Then we deduce (by induction on n), than An is countable for every n N.(from part i)

iii) iv) The set of integers is countable, we have this following theorem:

Let A be a countable set, and let Bn be the set of all n-tuples (a1,...,an)), where akA,k=1,...,n,akA,k=1,...,n, and the elements a1,...,an need not be distinct. Then Bn is countable.

So by this theorem, the set of all (k+1)-tuples (a0,a1,...,ak) with a00 is also countable.

Let this set be represented by Zk. For each a aZk consider the polynomial

a0zk+a1zk1+...+ak=0

From the fundamental theorem of algebra, we know that there are exactly k complex roots for this polynomial.

We now have a series of nested sets that encompass every possible root for every possible polynomial with integer coefficients. More specifically, we have a countable number of Zks, each containing a countable number of (k+1)-tuples, each of which corresponds with k roots of a k-degree polynomial. So our set of complex roots (call it R) is a countable union of countable unions of finite sets. This only tells us that R is at most countable: it is either countable or finite.

To show that R is not finite, consider the roots for 2-tuples in Z1. Each 2-tuple of the form (1,n)corresponds with the polynomial -z+n=0 whose solution is z=n. There is clearly a unique solution for each nZ, so R is an infinite set. Because R is also at most countable, this proves that R is countable.

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