Minor edits

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Mark 2024-01-30 22:10:59 -08:00
parent b122a150ff
commit 82064f890c
Signed by: Mark
GPG Key ID: C6D63995FE72FD80

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@ -51,13 +51,6 @@ An ordered field must satisfy the following properties:
\definition{}
An ordered field that contains $\mathbb{R}$ is called a \textit{nonarchimedian extension} of $\mathbb{R}$.
\vfill
\pagebreak
\problem{}
Show that each of the following is true in any ordered field.
\begin{enumerate}
@ -68,15 +61,14 @@ Show that each of the following is true in any ordered field.
\end{enumerate}
\begin{solution}
\textbf{Part A:}
We know that $x^{-1} \times (x^{-1})^{-1} = 1$ \par
Thus $x \times (x^{-1} \times (x^{-1})^{-1}) = x \times 1 = x$ \par
We can rewrite this as $(x \times x^{-1}) \times (x^{-1})^{-1} = x$ \par
When then becomes $1 \times (x^{-1})^{-1} = x$ \par
And thus $(x^{-1})^{-1} = x$
\end{solution}
%\begin{solution}
% \textbf{Part A:}
% We know that $x^{-1} \times (x^{-1})^{-1} = 1$ \par
% Thus $x \times (x^{-1} \times (x^{-1})^{-1}) = x \times 1 = x$ \par
% We can rewrite this as $(x \times x^{-1}) \times (x^{-1})^{-1} = x$ \par
% When then becomes $1 \times (x^{-1})^{-1} = x$ \par
% And thus $(x^{-1})^{-1} = x$
%\end{solution}
\vfill