Homework assignments for Math 28b, Spring 2003
Problem Set 6. Due Thursday, April 3
Read Chapter 20, through p. 350 (example 7). We will be doing Chapter 21, through p. 366, so you may want to look at that as well.
Discussion of this last problem: The point of the construction is to generalize our method for rationalizing the denominator in expressions of the form 1/(a+b √n). In that case, we multiplied numerator and denominator by a-b √n, and used the difference-of-two-squares identity to eliminate the roots in the denominator. In this case, we start with the polynomial p(z) = a + bz +cz2, and multiply numerator and denominator by p(zω)p(zω2). In the problem, you verify by direct calculation that this works. The real explanation for this procedure lies in Galois theory. A bit of this is described in chapter 32 (see page 551, especially).
- Page 357:2,3,6,17,23
- In this problem, you will work through something like problem 17, but a little more general. The goal is to see explicitly how to find inverses in Q(3√2). Let z denote the cube root of 2, so z is a root of x3-2. Also, let ω be a root of the polynomial x2 + x +1.
- Show that ω3=1. You may find it useful to have a formula for ω but it is not necessary. (Note, however, that ω is a complex number.)
- Show that the other roots of x3-2 are given by zω and zω2.
- Now we are going to find the inverse of a + bz +cz2, where at least one of a, b or c is non-zero. Step 1: show that the quantity (a + bz +cz2)(a + bzω +cz2ω2)(a + bzω2 +cz2ω) is a rational number.
- Finish the problem, by using the result from the previous step to give a formula for the inverse. Your formula should look like A + Bz +Cz2 where A,B, and C are rationals. (Ie the ω s have disappeared.)