Exercises §3 — Ross, Elementary Analysis, 2nd Edition
Exercise 3.3. Prove \((-a)(-b) = ab\) for all \(a, b \in \mathbb{F}\).
Proof. Recall \(a + (-a) = (-a) + a = 0\).
The book proved \((-a)b = -(ab)\). Swapping the roles of \(a\) and \(b\) (identical proof) gives \(a(-b) = -(ab)\). Replacing \(a\) with \((-a)\) in this identity,
\[ (-a)(-b) = -\bigl((-a)b\bigr) = -\bigl(-(ab)\bigr) = ab. \qquad\square \]
Also prove: if \(ac = bc\) and \(c \neq 0\), then \(a = b\).
Proof. Since \(c \neq 0\), the inverse \(c^{-1}\) exists. Then
\[ \begin{aligned} ac = bc &\Rightarrow (ac)c^{-1} = (bc)c^{-1} \\ &\Rightarrow a(cc^{-1}) = b(cc^{-1}) \\ &\Rightarrow a(1) = b(1) \\ &\Rightarrow a = b. \qquad\square \end{aligned} \]
Exercise 3.4. Prove \(0 < 1\).
Proof. We use the previously proven fact \(0 \le a^2\) for all \(a \in \mathbb{F}\). Taking \(a = 1\),
\[ 0 \le 1^2 = 1 \cdot 1 = 1. \]
Since \(\mathbb{F}\) is a field it has more than one element, so \(0 \neq 1\). Hence \(0 < 1\). \(\square\)
Also prove: if \(0 < a < b\), then \(0 < b^{-1} < a^{-1}\).
Proof. Suppose \(0 < a < b\) with \(a, b \in \mathbb{F}\). By a previous exercise, \(0 < a \Rightarrow 0 < a^{-1}\) and \(0 < b \Rightarrow 0 < b^{-1}\).
Multiplying \(a < b\) by \(a^{-1} > 0\):
\[ a\,a^{-1} < b\,a^{-1} \;\Rightarrow\; 1 < b\,a^{-1}. \]
Multiplying by \(b^{-1} > 0\):
\[ b^{-1}(1) < b^{-1}(b\,a^{-1}) \;\Rightarrow\; b^{-1} < a^{-1}. \]
Together with \(0 < b^{-1}\), this gives \(0 < b^{-1} < a^{-1}\). \(\square\)
Exercise 3.5. (a) Show \(|b| \le a\) if and only if \(-a \le b \le a\).
Proof.
(\(\Rightarrow\)) Suppose \(|b| \le a\). Note \(0 \le |b|\), hence \(0 \le a\), and therefore \(-a \le 0\).
Case \(0 \le b\): then \(|b| = b\), so \(|b| \le a \Rightarrow b \le a\). Since \(-a \le 0 \le b\), we get \(-a \le b \le a\).
Case \(b \le 0\): then \(|b| = -b\), so \(|b| \le a \Rightarrow -b \le a \Rightarrow -a \le b\). Since \(b \le 0 \le a\), we get \(-a \le b \le a\).
(\(\Leftarrow\)) Suppose \(-a \le b \le a\).
Case \(0 \le b\): then \(|b| = b\). Since \(b \le a\), we have \(|b| \le a\).
Case \(b \le 0\): then \(|b| = -b\). Since \(-a \le b \Rightarrow -b \le a\), we have \(|b| \le a\). \(\square\)
Exercise 3.5. (b) Prove \(\bigl|\,|a| - |b|\,\bigr| \le |a - b|\).
Proof. Note \(|a|, |b|, |a-b| \ge 0\).
Case \(|b| \le |a|\): then \(|a| - |b| \ge 0\), so \(\bigl|\,|a| - |b|\,\bigr| = |a| - |b|\). By the triangle inequality,
\[ |a| = \bigl|(a - b) + b\bigr| \le |a - b| + |b| \;\Rightarrow\; |a| - |b| \le |a - b|, \]
so \(\bigl|\,|a| - |b|\,\bigr| \le |a - b|\).
Case \(|a| \le |b|\): then \(|a| - |b| \le 0\), so \(\bigl|\,|a| - |b|\,\bigr| = -(|a| - |b|) = |b| - |a|\). By the triangle inequality,
\[ |b| = \bigl|(b - a) + a\bigr| \le |b - a| + |a| = |a - b| + |a| \;\Rightarrow\; |b| - |a| \le |a - b|, \]
so \(\bigl|\,|a| - |b|\,\bigr| \le |a - b|\).
In both cases \(\bigl|\,|a| - |b|\,\bigr| \le |a - b|\). \(\square\)
Exercise 3.6. (a) Prove \(|a + b + c| \le |a| + |b| + |c|\) for all \(a, b, c \in \mathbb{R}\).
Proof. By the triangle inequality, \(|(a + b) + c| \le |a + b| + |c|\), and again \(|a + b| \le |a| + |b|\). Hence
\[ |a + b + c| \le |a + b| + |c| \le |a| + |b| + |c|. \qquad\square \]
Exercise 3.6. (b) Prove \(|a_1 + a_2 + \cdots + a_n| \le |a_1| + |a_2| + \cdots + |a_n|\) for \(n\) numbers \(a_1, a_2, \dots, a_n\), \(n \in \mathbb{N}\).
Proof. By induction on \(n\).
Base. Trivial for \(n = 1\). The case \(n = 2\) is the triangle inequality, already proven.
Inductive Step. Assume the inequality holds for \(n - 1\); we prove it for \(n\), where \(n \ge 3\). By the triangle inequality,
\[ \bigl|a_1 + a_2 + \cdots + a_{n-1} + a_n\bigr| = \bigl|(a_1 + a_2 + \cdots + a_{n-1}) + a_n\bigr| \le \bigl|a_1 + a_2 + \cdots + a_{n-1}\bigr| + |a_n|. \]
By the inductive hypothesis, \(\bigl|a_1 + a_2 + \cdots + a_{n-1}\bigr| \le |a_1| + |a_2| + \cdots + |a_{n-1}|\). Therefore
\[ \bigl|a_1 + a_2 + \cdots + a_n\bigr| \le |a_1| + |a_2| + \cdots + |a_n|. \qquad\square \]