Merge branch 'main' into BabyRudinCh01Ex02

.github/workflows/blank.yml

@@ -1,11 +1,16 @@
 name: Build LaTeX document
-on: [push, workflow_dispatch]
+on: [push, pull_request, workflow_dispatch]
 jobs:
   build_latex:
     runs-on: ubuntu-latest
     steps:
       - name: Set up Git repository
         uses: actions/checkout@v3
+      - name: Compile DoneRight LaTeX document
+        uses: xu-cheng/latex-action@v2
+        with:
+          working_directory: Books/DoneRight
+          root_file: DoneRight.tex
       - name: Compile BabyRudin LaTeX document
         uses: xu-cheng/latex-action@v2
         with:
@@ -45,6 +50,7 @@ jobs:
         with:
           name: PDF
           path: |
+            Books/DoneRight/DoneRight.pdf
             Books/BabyRudin/BabyRudin.pdf
             Neukirch.pdf
             Notes.pdf

.gitignore

@@ -5,3 +5,4 @@
 *.log
 *.out
 *.pdf
+*.synctex.gz

Books/BabyRudin/BabyRudin.tex

@@ -14,5 +14,7 @@
 \section*{Ch01 - The Real and Complex Number Systems}
 \input{Chapter01/ex01.tex}
 \input{Chapter01/ex02.tex}
+\input{Chapter01/ex08.cecilia.tex}
+\input{Chapter01/ex08.gapry.tex}
 
 \end{document}

Books/BabyRudin/Chapter01/ex08.cecilia.tex

@@ -0,0 +1,29 @@
+\subsection*{Exercise 08 (Cecilia)}
+After definition 1.26, Rudin identifies the complex number $ (a, 0) $ as the real number $ a $ on the basis that arithmetic of $ (a, 0) $ and $ a $ are compatible. However, that does not say anything about the ordering of the complex numbers. 
+In fact, suppose an ordering exist, it might be the case that the ordering of $ (a, 0) $ is not compatible with the ordering of $ a $, so identification could be confusing. In the sequel, we will not use the identification, and use the ordered pair representation instead.
+
+Suppose (for contradiction) that an ordering of complex number can be defined.
+
+First, we show that $ (-1, 0) > (0, 0) $ is false. Suppose $ (-1, 0) > (0, 0) $, we have:
+
+\begin{eqnarray*}
+             (-1, 0) &>& (0, 0)          \\
+    (-1, 0) + (1, 0) &>& (0, 0) + (1, 0) \\
+              (0, 0) &>& (1, 0)
+\end{eqnarray*}
+
+\begin{eqnarray*}
+           (-1, 0) &>& (0, 0)        \\
+    (-1, 0)(-1, 0) &>& (0, 0)(-1, 0) \\
+            (1, 0) &>& (0, 0)
+\end{eqnarray*}
+
+Apparently these two facts contradict each other. Therefore, $ (-1, 0) > (0, 0) $ is false.
+
+With that result, suppose the complex number can be assigned an order so that it becomes an ordered field. Then we know that either $ i > (0, 0) $ or $ i < (0, 0) $.
+
+Suppose $ i > (0, 0) $, then $ i^2 > (0, 0) $, which means $ (-1, 0) > (0, 0) $, which is false.
+
+Otherwise suppose $ i < (0, 0) $, then $ (0, 0) = i - i < -i $, which means $ (0, 0) < (-i)^2 $, which means $ (0, 0) < (-1, 0) $, which is false.
+
+Therefore we have a contradiction and it is impossible to impose an order on the complex numbers so that it becomes an ordered field.

Books/BabyRudin/Chapter01/ex08.gapry.tex

@@ -0,0 +1,11 @@
+\subsection*{Exercise 08 (Gapry)}
+Let $c_1 = i = (0,\ 1)$ are the ordered pair in the complex field $\mathbb{C}$
+
+\begin{enumerate}
+    \item{$1 > 0$ (by 1.18d) }
+    \item{$-1 < 0$  (by 1.18a applied on (1))}
+    \item{$c_1^2 = i^2 = -1 \implies i \neq 0 \implies i^2 > 0 \implies -1 > 0$ (by 1.18d)}
+\end{enumerate}
+
+Obviously, 2. and 3. are contradictory.
+

Books/DoneRight/Chapter01/1A/ex01.tex

@@ -0,0 +1,17 @@
+\subsubsection*{Exercise 01}
+
+\begin{flushleft}
+Let $\alpha$ and $\beta$ be two complex numbers such that $\alpha = p + iq$ and $\beta  = m + in$, where $p,\ q,\ m,\ n \in \mathbb{R}$. Then, we can express the sum of $\alpha$ and $\beta$ as follows:
+\begin{align*}
+\alpha + \beta &= (p + iq) + (m + in) \\
+               &= (p + m) + i(q + n)
+\end{align*}
+
+Similarly, the sum of $\beta$ and $\alpha$ is:
+\begin{align*}
+\beta  + \alpha &= (m + in) + (p + iq) \\
+                &= (m + p) + i(n + q)
+\end{align*}
+
+Since addition is commutative in the set of real numbers $\mathbb{R}$, we have $(p + m) = (m + p)$ and $(q + n) = (n + q)$. Therefore, we can conclude that addition is also commutative in the set of complex numbers $\mathbb{C}$.
+\end{flushleft}

Books/DoneRight/Chapter01/1A/ex02.tex

@@ -0,0 +1,16 @@
+\subsubsection*{Exercise 02}
+
+\begin{flushleft}
+Let $\alpha$, $\beta$, and $\lambda$ be three complex numbers such that $\alpha = p + iq$, $\beta  = m + in$, and $\lambda = u + iv$, where $p,\ q,\ m,\ n,\ u,\ v \in \mathbb{R}$.
+
+Then, we can express the sum of $\alpha$, $\beta$, and $\lambda$ as follows:
+\begin{align*}
+(\alpha + \beta) + \lambda &= ((p + iq) + (m + in)) + (u + iv) \\
+                           &= (p + m) + i(q + n) + (u + iv)    \\
+                           &= (p + m + u) + i(q + n + v)       \\
+\alpha + (\beta + \lambda) &= (p + iq) + ((m + in) + (u + iv)) \\
+                           &= (p + iq) + (m + u) + i(n + v)    \\
+                           &= (p + m + u) + i(q + n + v)
+\end{align*}
+Since the two expressions are equal, we can conclude that addition is associative in the set of complex numbers $\mathbb{C}$.
+\end{flushleft}

Books/DoneRight/Chapter01/1A/ex03.tex

@@ -0,0 +1,18 @@
+\subsubsection*{Exercise 03}
+
+\begin{flushleft}
+Let $\alpha$, $\beta$, and $\lambda$ be three complex numbers such that $\alpha = p + iq$, $\beta  = m + in$, and $\lambda = u + iv$, where $p,\ q,\ m,\ n,\ u,\ v \in \mathbb{R}$. Then, we can express the product of $\alpha$, $\beta$, and $\lambda$ as follows:
+\begin{align*}
+(\alpha \cdot \beta) \cdot \lambda &= ((p + iq) \cdot (m + in)) \cdot (u + iv)               \\
+                                   &= ((pm - qn) + i(pn + mq)) \cdot (u + iv)                \\
+                                   &= ((pm - qn)u - v(pn + mq)) + i((pm - qn)v + u(pn + mq)) \\
+                                   &= (pmu - qnu - vpn - vmq) + i(pmv - qnv + upn + umq)     \\
+\alpha \cdot (\beta \cdot \lambda) &= (p + iq) \cdot ((m + in) \cdot (u + iv))               \\
+                                   &= (p + iq) \cdot ((mu - nv) + i(nu + mv))                \\
+                                   &= p(mu - nv) - q(nu + mv) + i(p(nu + mv) + q(mu - nv))   \\
+                                   &= (pmu - pnv - qnu - qmv) + i(pnu + pmv + qmu - qnv)     \\
+                                   &= (pmu - vpn - qnu - vmq) + i(upn + pmv + umq - qnv)     \\
+                                   &= (pmu - qnu - vpn - vmq) + i(pmv - qnv + upn + umq)     
+\end{align*}
+Since the two expressions are equal, we can conclude that multiplication is associative in the set of complex numbers $\mathbb{C}$.
+\end{flushleft}

Books/DoneRight/Chapter01/1A/ex04.tex

@@ -0,0 +1,20 @@
+\subsubsection*{Exercise 04}
+
+\begin{flushleft}
+Let $\alpha$, $\beta$, and $\lambda$ be three complex numbers such that $\alpha = p + iq$, $\beta  = m + in$, and $\lambda = u + iv$, where $p,\ q,\ m,\ n,\ u,\ v \in \mathbb{R}$. Then, we can express the product of $\alpha$, $\beta$, and $\lambda$ as follows:
+\begin{align*}
+\lambda \cdot (\alpha + \beta) &= (u + iv) \cdot (p + iq + m + in) \\
+                               &= (u + iv) \cdot ((p + m) + i(q + n)) \\
+                               &= u(p + m) + iu(q + n) + iv(p + m) - v(q + n) \\
+                               &= (u(p + m) - v(q + n)) + i(u(q + n) + v(p + m)) \\
+                               &= Left \\ 
+\lambda \cdot \alpha + \lambda  \cdot \beta &= (u + iv) \cdot (p + iq) + (u + iv) \cdot (m + in) \\
+                                            &= (up + iqu + ivp - vp) + (um + inu + ivm - vn) \\
+                                            &= (up - vp + um - vn) + i(qu + vp + nu + vm) \\
+                                            &= (up + um - vp - vn) + i(qu + nu + vp + vm) \\
+                                            &= (u(p + m) - v(q + n)) + i(u(q + n) + v(p + m)) \\
+                                            &= Right
+\end{align*}
+Since Left = Right, we can conclude that $\lambda \cdot (\alpha + \beta) = \lambda \cdot \alpha + \lambda  \cdot \beta$.
+\end{flushleft}
+

Books/DoneRight/DoneRight.tex

@@ -0,0 +1,23 @@
+\documentclass{article}
+
+\usepackage{amsmath}
+\usepackage{amssymb}
+\usepackage{amsfonts}
+\usepackage{hyperref}
+
+\title{Solution to Exercises on the \\Linear Algebra Done Right, Fourth Edition}
+\author{Gapry}
+
+\begin{document}
+\maketitle
+
+\section*{Chapter 1 - Vector Spaces}
+
+\subsection*{1A}
+\input{Chapter01/1A/ex01.tex}
+\input{Chapter01/1A/ex02.tex}
+\input{Chapter01/1A/ex03.tex}
+\input{Chapter01/1A/ex04.tex}
+
+\end{document}
+