← Principles of Mathematical Analysis

Chapter 1: The Real and Complex Number Systems

Walter Rudin

Exercise 2

Prove that there is no rational number whose square is $12$.

Exercise 4

Let $E$ be a nonempty subset of an ordered set; suppose $\alpha$ is a lower bound of $E$, and $\beta$ is an upper bound of $E$. Prove that $\alpha \leq \beta$.

Exercise 6

Fix $b > 1$.

  1. If $m, n, p, q$ are integers, $n > 0$, $q > 0$, and $r = m/n = p/q$, prove that $(b^m)^{1/n} = (b^p)^{1/q}$. Hence it makes sense to define $b^r = (b^m)^{1/n}$.
  2. Prove that $b^{r+s} = b^r b^s$ if $r$ and $s$ are rational.
  3. If $x$ is real, define $B(x)$ to be the set of all numbers $b^t$, where $t$ is rational and $t \leq x$. Prove that $b^r = \sup B(r)$ when $r$ is rational. Hence it makes sense to define $b^x = \sup B(x)$ for every real $x$.
  4. Prove that $b^{x+y} = b^x b^y$ for all real $x, y$.

Exercise 8

Prove that no order can be defined in the complex field that turns it into an ordered field.

Exercise 10

Suppose $z = a + bi$, $w = u + vi$, and $$a = \left(\frac{|w| + u}{2}\right)^{1/2}, \quad b = \left(\frac{|w| - u}{2}\right)^{1/2}.$$ Prove that $z^2 = w$ if $v \geq 0$, and that $(\bar{z})^2 = w$ if $v \leq 0$. Conclude that every complex number has two complex square roots.

Exercise 12

If $z_1, z_2, \ldots, z_n$ are complex, prove that $$|z_1 + z_2 + \cdots + z_n| \leq |z_1| + |z_2| + \cdots + |z_n|.$$

Exercise 14

If $z$ is a complex number such that $|z| = 1$, that is, such that $z\bar{z} = 1$, compute $$|1 + z|^2 + |1 - z|^2.$$

Exercise 16

Suppose $k \geq 3$, $\mathbf{x}, \mathbf{y} \in \mathbb{R}^k$, $|\mathbf{x} - \mathbf{y}| = d > 0$, and $r > 0$. Prove:

  1. If $2r > d$, there are infinitely many $\mathbf{z} \in \mathbb{R}^k$ such that $|\mathbf{z} - \mathbf{x}| = |\mathbf{z} - \mathbf{y}| = r$.
  2. If $2r = d$, there is exactly one such $\mathbf{z}$.
  3. If $2r < d$, there is no such $\mathbf{z}$.

Exercise 18

If $k \geq 2$ and $\mathbf{x} \in \mathbb{R}^k$, prove that there exists $\mathbf{y} \in \mathbb{R}^k$ such that $\mathbf{y} \neq \mathbf{0}$ but $\mathbf{x} \cdot \mathbf{y} = 0$. Is this also true if $k = 1$?