The group of complex p-power roots of unity is a proper quotient of itself
Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 3.3 Exercise 3.3.8 Let $p$ be a prime and let $G$ be the group of $p-$power roots of 1 in…
Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 3.3 Exercise 3.3.8 Let $p$ be a prime and let $G$ be the group of $p-$power roots of 1 in…
Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 3.3 Exercise 3.3.5 Let $QD_{16} = \langle \sigma, \tau \rangle$ be the quasidihedral group of order 16 described in…
Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 3.3 Exercise 3.3.4 Let $A$ and $B$ be groups, with $C \leq A$ and $D \leq B$ normal. Prove…
Chapter 1 Riemann Integration §1A Review: Riemann Integral §1B Riemann Integral Is Not Good Enough Chapter 2 Measures §2A Outer Measure on R (#1) (#2) (#3) (#4) (#5) (#6) §2B…
Chapter 2 Linear Algebra #2.1, #2.2, #2.3, #2.4, #2.5, #2.6, #2.7, #2.8, #2.9, #2.10, #2.11, #2.12, #2.13, #2.14, #2.15, #2.16, #2.17, #2.18, #2.19, #2.20 Chapter 3 Analytic Geometry #3.1, #3.2,…
Chapter 1. Basic Notions Vector spaces #1.1, #1.2, #1.3, #1.4, #1.5, #1.6, #1.7, #1.8 Linear combinations, bases #2.1, #2.2, #2.3, #2.4, #2.5, #2.6 Linear Transformations. Matrix–vector multiplication #3.1, #3.2, #3.3, #3.4, #3.5, #3.6, #3.7 Linear transformations as…
Exercise 7.1 Solution: (a) Note that\[\sqrt{x+3}-\sqrt{c+3}=\frac{(x+3)-(c+3)}{\sqrt{x+3}+\sqrt{c+3}}=\frac{x-c}{\sqrt{x+3}+\sqrt{c+3}}.\]We have\begin{align*}f'(c)&=\lim_{x\to c}\frac{f(x)-f(c)}{x-c}\\ &=\lim_{x\to c}\frac{\sqrt{x+3}-\sqrt{c+3}}{x-c}\\ &=\lim_{x\to c}\frac{\sqrt{x+3}-\sqrt{c+3}}{x-c}\\&=\lim_{x\to c}\frac{1}{x-c}\frac{x-c}{\sqrt{x+3}+\sqrt{c+3}} \\ &=\lim_{x\to c}\frac{1}{\sqrt{x+3}+\sqrt{c+3}}\\&=\frac{1}{2\sqrt{c+3}}. \end{align*}(b) Note that\[\frac{1}{x}-\frac{1}{c}=\frac{c-x}{xc}.\]We have\begin{align*}g'(c)&=\lim_{x\to c}\frac{g(x)-g(c)}{x-c}\\ &=\lim_{x\to c}\frac{\frac{1}{x}-\frac{1}{c}}{x-c}\\ &= \lim_{x\to c}\frac{\frac{c-x}{xc}}{x-c}\\ &=\lim_{x\to c}\frac{-1}{xc}=-\frac{1}{c^2}.\end{align*}(c) Note that…
Exercise 6.5 Solution: If $a=0$, for any $\e >0$ and any $x_0\in\ R$, and any $\delta >0$, we have\[|f(x)-f(x_0)|=|b-b|=0 < \e\] for all $|x-x_0|<\delta$. Hence $f$ is continuous at $x_0$.…
Exercise 5.1 Solution: (a) Closed. As it has no limit points, $\mathbb Z$ is closed. Clearly, it is not open. $\mathbb Z$ is not compact since $\mathbb Z$ is not…
Exercise 4.1 Solution: (1) Converges conditionally. It is convergent by Alternating Series Test. But $\sum_{k=1}^\infty\dfrac{1}{\sqrt{k}}$ is divergent by Proposition 4.16. (2) Converges absolutely. By Geometric Series Test and the fact…