## 305 -Homework set 2

February 9, 2009

This set is due February 20 at the beginning of lecture. Consult the syllabus for details on the homework policy. I do not think this set is particularly difficult, but it is on the longish side of things, so make sure you leave yourself enough time to work on it.

1. Gauß’ fundamental theorem of algebra states that any equation $p(x)=0,$ where $p$ is a polynomial with complex coefficients, has at least one complex root $z.$ This means that $z$ is a complex number and $p(z)=0.$ Show that $p$ has at most $n$ roots, where $n$ is its degree, and that if we count roots up to multiplicity, then it has exactly $n$ roots. Since the multiplicity of a root $z$ is by definition the largest $m$ such that $(x-z)^m$ is a factor of $p(x),$ you may want to verify that $p(z)=0$ iff $(x-z)$ is a factor of $p.$

2. Let $p(x)$ be a polynomial with real coefficients, and let $z$ be a complex root of $p.$ Show that $p(\bar z)=0$ as well. Conclude that if the degree $n$ of $p$ is odd and the coefficients of $p$ are real, then $p$ has at least one real root. (You may use the fundamental theorem of algebra, if needed.) Conclude also that if $p$ is of degree four and has real coefficients, then $p$ can be factored as the product of two quadratic polynomials with real coefficients. (Does this follow “directly” from the argument described in lecture?)

3. Solve exercises 54-56 from Chapter 3 of the book.

4. Show directly that if $a,b,c$ are real numbers, then at least one of the solutions of $x^3+ax^2+bx+c=0$ is a real number. What I mean is that, rather than appealing to problem 2, you want to look at the solutions obtained by Cardano’s method as described in lecture, and argue directly from the formulas so obtained that at least one of the solutions must be real. Be careful, since your argument should not give you that all three roots are real, since this is not true in general.

5. Show directly that a quartic with complex coefficients admits only 4 roots. What I mean is that, rather than appealing to problem 1, you want to look at the solutions obtained by Ferrari’s method as described in lecture, and argue directly that they only produce 4 roots, even though, in principle, they produce 24 (since they involve solving a cubic and then taking a square root to obtain parameters from which four solutions are then found).

## 305 -3. Complex numbers.

February 9, 2009

Mathematicians first approached complex numbers cautiously. Although it was clear that they were useful in solving certain problems at least formally (for example, they are needed to even make sense of the formulas we found in the previous lectures) what was not clear was that they made sense. Perhaps indiscriminate use of them would lead to contradictions.

Gauß solved this problem by realizing that one can define ${\mathbb C}$ and its operations in terms of ${\mathbb R}$ and its operations. As long as we are willing to accept that ${\mathbb R}$ makes sense, then no contradictions will come up from the use of complex numbers.

## 580 -Cardinal arithmetic (3)

February 9, 2009

It is easy to solve negatively the question immediately following Homework problem 5 on lecture II.1. I asked whether if $X$ is Dedekind-finite but ${\mathcal P}(X)$ is Dedekind-infinite, then it followed that there is an infinite Dedekind-finite set $Y$ such that ${\mathcal P}(Y)\preceq X$.

To exhibit a counterexample, it is enough to know that it is consistent to have an infinite Dedekind finite set $X$ that is the countable union of finite sets (in fact, sets of size 2). Notice that $\omega$ is a surjective image of $X,$ so ${\mathcal P}(X)$ is Dedekind-infinite. Suppose that ${\mathcal P}(Y)\preceq X.$ Then certainly $Y\preceq X,$ so $Y$ is a countable union of finite sets $Y_n.$ If $Y$ is infinite then $Y_n\ne\emptyset$ for infinitely many values of $n.$ But then $\omega$ is also a surjective image of $Y$, so $\omega$ (and in fact $P(\omega)$) injects into ${\mathcal P}(Y)$ and therefore into $X,$ contradiction.

At the end of last lecture we showed Theorem 10, a general result that allows us to compute products $\kappa^\lambda$ for infinite cardinals $\kappa,\lambda,$ namely:

Let $\kappa$ and $\lambda$ be infinite cardinals. Let $\tau=\sup_{\rho<\kappa}|\rho|^\lambda.$ Then $\displaystyle \kappa^\lambda=\left\{\begin{array}{cl} 2^\lambda & \mbox{if }\kappa\le 2^\lambda,\\ \kappa\cdot\tau & \mbox{if }\lambda<{\rm cf}(\kappa),\\ \tau & \begin{array}{l}\mbox{if }{\rm cf}(\kappa)\le\lambda,2^\lambda<\kappa,\mbox{ and }\\ \rho\mapsto|\rho|^\lambda\mbox{ is eventually constant below }\kappa,\end{array}\\ \kappa^{{\rm cf}(\kappa)} & \mbox{otherwise.}\end{array}\right.$