Minor edits
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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 $\mathbb{R}$. | ||||
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| \vfill | ||||
| \pagebreak | ||||
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| \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 | ||||
|  | ||||
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