A manifold is a structure

This theorem guarantees the existence of fixed points in complete metric spaces under a contraction mapping.

The REINFORCE Algorithm

Normed Vector Spaces

Often in mathematics, it is useful to be able to compute an approximation to a given function \(f(x)\).
One way of doing this is to find some \(a\) for which \(f(x)\) is easy to compute. We can then calculate
the approximate value of \(f(x)\) when \(x\) is near \(a\). We will call this approximation \(f_{a}(x)\).
More concretely, we might say \(x\) *is near* \(a\) when \(\left|x - a\right| < \varepsilon\) for some
threshold value of \(\varepsilon\) our choosing. One way of settling upon such an \(\varepsilon\) might be
by determining an acceptable upper bound \(U\) on the error of our approximation \(\left|f(x) - f_{a}(x)\right|\) and
then finding the greatest \(\varepsilon\) such that:
\[\left|f(x) - f_{a}(x)\right| < U \quad \forall x \in (a - \varepsilon,\ a + \varepsilon)\]

## Solutions to Rudin's Principles of Mathematical Analysis

Exercises from Chapter 1 of the book.

**Exercise 1:** *If \(r\) is rational (\(r \neq 0\)) and \(x\) is irrational,
prove that \(r + x\) and \(rx\) are irrational.*

Suppose \(r \in \mathbb{Q}\) and \(x \in \mathbb{I}\) but \(r+x,\ rx \in \mathbb{Q}\). Then we have \(r +x = \frac{a}{b}\) for some \(a,b \in \mathbb{Z}\). This implies that \(x = \frac{a}{b} - r\). Since \(r \in \mathbb{Q}\), \(r = \frac{c}{d}\) for some \(c, d \in \mathbb{Z}\). Therefore \[x = \frac{a}{b} - \frac{c}{d} = \frac{ad-bc}{bd}\] which implies \(x \in \mathbb{Q}\): a contradiction. Therefore, \(r + x\) is irrational. Now, consider \(rx\). If \(rx \in \mathbb{Q}\), then

\begin{align}
rx &= \frac{a}{b}\ \text{for}\ a,b \in \mathbb{Z}

\implies x &= \frac{a}{b} \cdot \frac{d}{c}

\implies x &\in \mathbb{Q}
\end{align}

which is a contradiction. Therefore, \(rx\) is irrational.