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@ -32,13 +32,18 @@
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}
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% This handout is pretty difficult. Make sure you can all solve all the problems yourself,
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% and remember that each SECTION was a two-hour lesson with a smart class.
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% From experience, the following holds:
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% supremum is a better lesson than dual numbers, which is better than extensions.
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\begin{document}
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\maketitle
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\input{parts/1 extensions}
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\input{parts/2 dual}
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\input{parts/x supremum}
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\input{parts/supremum}
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\input{parts/dual}
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\input{parts/extensions}
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\end{document}
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@ -21,8 +21,7 @@ In the problems below, let $\varepsilon$ a positive infinitesimal so that $\vare
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\note{Note that $\varepsilon \neq 0$.}
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\definition{}
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The set of \textit{dual numbers} is a nonarchimedian extension of $\mathbb{R}$ \par
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that consists of elements of the form $a + b\varepsilon$, where $a, b \in \mathbb{R}$.
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The set of \textit{dual numbers} consists of elements of the form $a + b\varepsilon$, where $a, b \in \mathbb{R}$.
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\problem{}
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Compute $(a + b\varepsilon) \times (c + d\varepsilon)$.
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@ -33,11 +32,7 @@ Compute $(a + b\varepsilon) \times (c + d\varepsilon)$.
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\definition{}
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Let $f(x)$ be an algebraic function $\mathbb{R} \to \mathbb{R}$. \par
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(that is, a function we can write using the operators $+-\times\div$, powers, and roots) \par
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\note[Note]{
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Why this condition? These are the only operations we have in an ordered field! \\
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Powers, roots, and division aren't directly available, but are fairly easy to define.
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}
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(that is, a function we can write using the operators $+-\times\div$ and integer powers) \par
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\vspace{2mm}
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@ -13,7 +13,7 @@
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% If you edit this, please give credit!
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% Quality handouts take time to make.
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\section{Nonarchimedian Extensions}
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\section{Extensions of $\mathbb{R}$}
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\definition{}
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An \textit{ordered field} consists of a set $S$, the operations $+$ and $\times$, and the relation $<$. \par
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@ -49,7 +49,16 @@ An ordered field must satisfy the following properties:
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\end{itemize}
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\definition{}
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An ordered field that contains $\mathbb{R}$ is called a \textit{nonarchimedian extension} of $\mathbb{R}$.
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An ordered field that contains $\mathbb{R}$ is called an \textit{extension} of $\mathbb{R}$.
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\definition{}
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The \textit{Archimedian property} states the following: \par
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For all positive $x, y$, there exists an $n$ so that $nx \geq y$.
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\theorem{}
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All extensions of $\mathbb{R}$ are nonarchemedian. \par
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Proving this is difficult.
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\vfill
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\pagebreak
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@ -71,7 +80,8 @@ Which of the following are ordered fields?
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\problem{}
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Show that each of the following is true in any ordered field.
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Show that each of the following is true in any ordered field. \par
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The list of field axioms is provided below, for convenience.
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\begin{enumerate}
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\item if $x \neq 0$ then $(x^{-1})^{-1} = x$
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\item $0 \times x = 0$
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@ -89,8 +99,36 @@ Show that each of the following is true in any ordered field.
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% And thus $(x^{-1})^{-1} = x$
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%\end{solution}
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\vfill
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\begin{itemize}
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\item \textbf{Properties of $+$:}
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\begin{itemize}
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\item Commutativity: $a + b = b + a$
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\item Associativity: $a + (b + c) = (a + b) + c$
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\item Identity: there exists an element $0$ so that $a + 0 = a$ for all $a \in S$
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\item Inverse: for every $a$, there exists a $-a$ so that $a + (-a) = 0$
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\end{itemize}
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\item \textbf{Properties of $\times$:}
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\begin{itemize}
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\item Commutativity
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\item Associativity
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\item Identity (which we label $1$)
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\item For every $a \neq 0$, there exists an inverse $a^{-1}$ so that $aa^{-1} = 1$
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\item Distributivity: $a(b + c) = ab + ac$
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\end{itemize}
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\item \textbf{Properties of $<$:}
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\begin{itemize}
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\item Non-reflexive: $x < x$ is always false
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\item Transitive: $x < y$ and $y < z$ imply $x < z$
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\item Connected: for all $x, y \in S$, either $x < y$, $x > y$, or $x = y$.
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\item If $x < y$ then $x + z < y + z$
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\item If $x < y$ and $z > 0$, then $xz < yz$
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\item $0 < 1$
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\end{itemize}
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\end{itemize}
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\pagebreak
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@ -14,7 +14,7 @@
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% Quality handouts take time to make.
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\section*{Bonus: The supremum \& infimum}
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\section*{The supremum \& the infimum}
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\definition{}
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In this section, we'll define a \say{real number} as a decimal, infinite or finite.
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