BabyRudin: Solution: Ex03 in Ch01: Update

Books/BabyRudin/Chapter01/ex03.gapry.tex

@@ -1,30 +1,37 @@
 \subsection*{Exercise 03 (Gapry)}
 
 \subsubsection*{1.15 Proposition (a)}
+As we know $x \neq 0$
 \begin{flalign*}
-                        xy &= xz                                      &\\
-\implies \frac{1}{x}{x}{y} &= \frac{1}{x}{x}{z} \text{ (by 1.12(M5))} &\\
-\implies       1 \cdot {y} &= 1 \cdot {z}       \text{ (by 1.12(M4))} &\\
-\implies               {y} &= {z} 
+                        xy &= xz                                       &\\
+\implies \frac{1}{x}{x}{y} &= \frac{1}{x}{x}{z} \text{ (by 1.12 (M5))} &\\
+\implies       1 \cdot {y} &= 1 \cdot {z}       \text{ (by 1.12 (M5))} &\\
+\implies               {y} &= {z}               \text{ (by 1.12 (M4))}
 \end{flalign*}
 
 \subsubsection*{1.15 Proposition (b)}
+As we know $x \neq 0$
 \begin{flalign*}
-         x \cdot 1 &= x \text{ (by 1.12(M4))}               &\\
-\implies x \cdot y &= x \text{ (use $y$ to substitute $1$)} &\\
-\implies         y &= 1
+                        xy &= x                                     &\\
+\implies \frac{1}{x}{x}{y} &= \frac{1}{x}{x} \text{ (by 1.12 (M5))} &\\
+\implies       1 \cdot {y} &= 1              \text{ (by 1.12 (M5))} &\\
+\implies               {y} &= 1              \text{ (by 1.12 (M4))}
 \end{flalign*}
 
 \subsubsection*{1.15 Proposition (c)}
+As we know $x \neq 0$
 \begin{flalign*}
-         x \cdot \frac{1}{x} &= 1 \text{ (by 1.12(M5))}                         &\\
-\implies           x \cdot y &= 1 \text{ (use $y$ to substitute $\frac{1}{x}$)} &\\
-\implies                   y &= \frac{1}{x}
+                        xy &= 1                                  &\\
+\implies \frac{1}{x}{x}{y} &= \frac{1}{x} \text{ (by 1.12 (M5))} &\\
+\implies       1 \cdot {y} &= \frac{1}{x} \text{ (by 1.12 (M5))} &\\
+\implies               {y} &= \frac{1}{x} \text{ (by 1.12 (M4))}
 \end{flalign*}
 
 \subsubsection*{1.15 Proposition (d)}
+We know that $x \in F$ and $x \neq 0$. According to 1.12, there must exist an inverse element $\frac{1}{x} \in F$ such that $x \cdot \frac{1}{x} = 1$. Similarly, since $\frac{1}{x} \in F$, there must exist an inverse element $\frac{1}{\frac{1}{x}} \in F$ such that $\frac{1}{x} \cdot \frac{1}{\frac{1}{x}} = 1$.
 \begin{flalign*}
-\frac{1}{\frac{1}{x}} &= \frac{1 \cdot x}{\frac{1}{x} \cdot x} 
-                       = \frac{x}{1} 
-                       = x \text{ (by 1.12(M5))} &
+        \frac{1}{x} \cdot \frac{1}{\frac{1}{x}} &= 1         \text{ (by 1.12 (M5))}     &\\
+x \cdot \frac{1}{x} \cdot \frac{1}{\frac{1}{x}} &= x \cdot 1 \text{ (by 1.12 (M5))}     &\\
+                  1 \cdot \frac{1}{\frac{1}{x}} &= x         \text{ (by 1.12 (M4, M5))} &\\
+                          \frac{1}{\frac{1}{x}} &= x         \text{ (by 1.12 (M4))}     
 \end{flalign*}