Introduction to Smooth Manifolds -- Some Solutions
The definitive graduate-level textbook on differential topology.

The definitive graduate-level textbook on differential topology.
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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.