\documentclass[11pt,letterpaper,answers]{exam}
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\newcommand{\R}{\mathbb{R}}
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\newcommand{\Q}{\mathbb{Q}}
\newcommand{\Z}{\mathbb{Z}}
\newcommand{\N}{\mathbb{N}}
\newcommand{\F}{\mathcal{F}}
\newcommand{\set}[2]{\left\{ #1 \, \middle| \, #2 \right\} }

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% Change "YOUR NAME HERE" below to your name
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\header{Math 3010}{Final}{YOUR NAME HERE}
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\begin{document}

\noindent \textbf{Instructions:} You may treat this take--home final
just like a homework assignment. You \emph{may} seek outside help from
\emph{any} source including me, the Internet, other students, other
teachers, etc.. Just like homework, if you do not have time to typeset
your answers via $\LaTeX{}$, then submit your answers on paper, but
you will suffer a 10\% penalty. Please submit your answers as a PDF
attachment via email, or printed on paper to my mailbox in JWB 228 by
5:00 pm, Friday 5/2.

\bigskip

\begin{questions}

  \question A field is a set which allows you to perform arithmetic
  with its elements. More precisely, a field is a set, $F$, that
  satisfies three criteria:
  \begin{enumerate}
  \item $F$ is an abelian group under addition $(+)$.

  \item $F - \{ 0 \}$ is an abelian group under mulitiplication
    $(\times )$.

  \item Addition and multiplication are related through the
    distributive law: 
    \[
    a(b+c) = ab + ac.
    \]
  \end{enumerate}

  Some examples of fields are $\Q, \R$ and $\C$. We can enlarge a
  field by \emph{adjoining} (explained in the next problem) a new
  number that is not already in the set. Adjoining is \emph{not} the
  same as unioning the field and the new number.

  For example, $\Q \cup \{\sqrt{2} \}$ is not a field because this new
  set does not contain the additive inverse of this new number,
  $-\sqrt{2}$. But neither is the set $\Q \cup \{ \sqrt{2}, -\sqrt{2}
  \}$ a field. This set does not contain the multiplicative inverses
  of the two new elements. OK, well perhaps if we compute the
  multiplicative inverses and then union them to $\Q$ we will get a
  new field!
  \[
  (\sqrt{2})^{-1} = \dfrac{1}{\sqrt{2}} =
  \dfrac{1}{\sqrt{2}} \cdot \dfrac{\sqrt{2}}{\sqrt{2}} =
  \dfrac{\sqrt{2}}{2}
  \]
  Similarly, we see that $(-\sqrt{2})^{-1}= -\sqrt{2}/2$. Maybe if we
  union all four of these new numbers to $\Q$, then we will get a
  field.
  \[
  S = \Q \cup \left\{ \sqrt{2}, -\sqrt{2}, \frac{\sqrt{2}}{2},
    -\frac{\sqrt{2}}{2} \right\}.
  \]

  \begin{parts}
    
    \part Explain why the process described in the preceding paragraph
    is doomed to fail. That is, explain why the above process is
    unusable in practice for extending a field to a larger field.

    \begin{solution}
      Your answer here...
    \end{solution}

    \part Give a specific example of an expression obtained by
    combining two elements from $S$ (defined above) via one of the
    four operations of arithmetic which is \emph{not} in the set
    $S$. That is show that $S$ is not closed under one of its four
    arithmetic operations.

    \textbf{Hint:} An extremely simple expression will suffice. The
    simpler the better for my sanity's sake.
    
    \begin{solution}
      Your answer here...
    \end{solution}

  \end{parts}

  \newpage

  \question Adjoining a new number to a field is how we create larger
  fields. For example, we can adjoin $\sqrt{2}$ to $\Q$, and we write
  $\Q(\sqrt{2})$ which is pronounced ``Q adjoin root 2'', and defined
  by
  \[
  \Q(\sqrt{2}) = \set{a+b\sqrt{2}}{a,b \in \Q }.
  \]
  Now suppose we wish to extend the field $\Q(\sqrt{2})$ by adjoining
  $\sqrt{3}$. Before we do this we should ask ``Is $\sqrt{3}$ already
  in $\Q(\sqrt{2})$?'' If it is already a member of that set, then
  adjoining it is superfluous.

  Show that $\sqrt{3} \not\in \Q(\sqrt{2})$.

  \textbf{Hint 1:} Use proof by contradiction, that is assume the opposite:
  $\sqrt{3} \in \Q(\sqrt{2})$, and see if this leads to a
  contradiction. Start by assuming,
  \[
  \sqrt{3} = a + b\sqrt{2} \quad \text{for some } a, b \in \Q,
  \]
  then square both sides, to get:
  \begin{align*}
    3 &= (a + b\sqrt{2})(a + b\sqrt{2}), \\
    3 &= (a^2 + 2b^2) + (2ab)\sqrt{2}, \\
    3 + 0\sqrt{2} &= (a^2 + 2b^2) + (2ab)\sqrt{2}.
  \end{align*}
  That last equation decomposes into two equations. Show that these
  two equations lead to a contradiction.

  \textbf{Hint 2:} What do you know must be true if the product of two
  rational numbers is 0?

  \begin{solution}
    Your answer here...
  \end{solution}

\end{questions}
\end{document}
