Exercise 1.3.1 The matrix of coefficients is $$\left[\begin{array}{cc}1-i&-i\\2&1-i\end{array}\right].$$Row reducing $$\rightarrow \left[\begin{array}{cc}2&1-i\\1-i&-i\end{array}\right]\rightarrow\left[\begin{array}{cc}2&1-i\\0&0\end{array}\right] $$Thus $2x_1+(1-i)x_2=0$. Thus for any $x_2\in\mathbb C$, $(\frac12(i-1)x_2,x_2)$ is a solution and these are all solutions. Exercise 1.3.2 We…
Exercise 1.2.1: Verify that the set of complex numbers described in Example 4 is a subfield of $\mathbb C$. Solution: Let $F=\{x+y\sqrt{2}\mid x,y\in\mb Q\}$. Then we must show six things:…
Chapter 1. Linear Equations 1.1 Fields (no exercises) 1.2 Systems of Linear Equations (#1) (#2) (#3) (#4) (#5) (#6) (#7) (#8) 1.3 Matrices and Elementary Row Operations (#1) (#2) (#3)…
Exercise 4.4.1 (a) It is already shown in Example 4.3.5. (b) Let $x_n=n+\dfrac{1}{n}$ and $y_n=n$, then we have $$ |x_n-y_n|\to 0. $$On the other hand, we have \begin{align*} |f(x_n)-f(y_n)|=&\ \left|\Big(n+\dfrac{1}{n}\Big)^3-n^3\right|\\…
Exercise 4.3.1 See Understanding Analysis Instructors’ Solution Manual Exercise 4.3.1. Exercise 4.3.2 Solution: (a) Let $f(x)\equiv 0$ (constant function). (b) Let $f(x)=x$. Please check that this function is not onetinuous.…
Exercise 4.2.1 See Understanding Analysis Instructors’ Solution Manual Exercise 4.2.5 Exercise 4.2.2 (a) We would like $|(5x-6)-9|<1$, that is $|5(x-3)|<1$. Hence we need\[|x-3|<\frac{1}{5}.\]Therefore, the largest possible $\delta$ is $\dfrac{1}{5}$. (b)…
Exercise 2.8.1 See Understanding Analysis Instructors’ Solution Manual Exercise 2.8.1 Exercise 2.8.2 See Understanding Analysis Instructors’ Solution Manual Exercise 2.8.2 Exercise 2.8.3 See Understanding Analysis Instructors’ Solution Manual Exercise 2.8.3…
Exercise 2.7.1 See Understanding Analysis Instructors’ Solution Manual Exercise 2.7.1 Exercise 2.7.2 (a) Converges. Since $\sum\limits_{n=0}^\infty \dfrac{1}{2^n}$ converges, see Example 2.7.5 (Geometric Series), and $\dfrac{1}{2^n+n}\leqslant \dfrac{1}{2^n}$, by Theorem 2.7.4 (Comparison…
Exercise 2.6.1 Let $\varepsilon > 0$ be arbitrary. Since $(x_n)$ converges to $x$, there exits $N\in\mathbf N$ such that \begin{equation}\label{eq2.6.1.1}|x_n-x|<\frac{\varepsilon}{2}\end{equation} for all $n\geqslant N$. Whenever $n,m\geqslant N$, we have\begin{align*}|x_n-x_m|\leqslant &~|x_n-x|+|x-x_m|\\…
Exercise 2.5.1 (a) Impossible by Theorem 2.5.5. (b) Consider the sequence given by\[a_n=\frac{1+(-1)^n+\frac{1}{n}}{2}.\]Then we have\[a_{2n}=1+\frac{1}{4n},~a_{2n-1}=\frac{1}{2(2n-1)}.\]It is clear that $(a_{2n})$ converges to $1$ and $(a_{2n-1})$ converges to $0$. Moreover, it is…