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BabyRudin: Solution: Ex03 in Ch01
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gapry committed Jan 23, 2024
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\subsection*{Exercise 03}

\subsubsection*{1.15 Proposition (a)}
\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}
\end{flalign*}

\subsubsection*{1.15 Proposition (b)}
\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
\end{flalign*}

\subsubsection*{1.15 Proposition (c)}
\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}
\end{flalign*}

\subsubsection*{1.15 Proposition (d)}
\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))} &
\end{flalign*}

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