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Mark 2024-01-30 22:10:59 -08:00
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Advanced/Nonstandard Analysis/parts/1 extensions.tex
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@ -51,13 +51,6 @@ An ordered field must satisfy the following properties:
\definition{} \definition{}
An ordered field that contains $\mathbb{R}$ is called a \textit{nonarchimedian extension} of $\mathbb{R}$. An ordered field that contains $\mathbb{R}$ is called a \textit{nonarchimedian extension} of $\mathbb{R}$.
\vfill
\pagebreak
\problem{} \problem{}
Show that each of the following is true in any ordered field. Show that each of the following is true in any ordered field.
\begin{enumerate} \begin{enumerate}
@ -68,15 +61,14 @@ Show that each of the following is true in any ordered field.
\end{enumerate} \end{enumerate}
\begin{solution} %\begin{solution}
\textbf{Part A:} % \textbf{Part A:}
% We know that $x^{-1} \times (x^{-1})^{-1} = 1$ \par
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
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
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
When then becomes $1 \times (x^{-1})^{-1} = x$ \par % And thus $(x^{-1})^{-1} = x$
And thus $(x^{-1})^{-1} = x$ %\end{solution}
\end{solution}
\vfill \vfill