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\begin{center} \textbf{Project on constructing the rational numbers $\mathbb{Q}$.} \end{center}

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Define the set $\mathbb{Q}$ of pairs of integers as follows: $\mathbb{Q} = \mathbb{Z} \times ( \mathbb{Z} - \{0\})$.  Equivalently $(x,y) \in \mathbb{Q}$ if and only if $x, y \in \mathbb{Z}$ and $y \ne0$.  We define the following relation $\sim$ on $\mathbb{Q}$
\begin{eqnarray*}
(a,b) \sim (c,d) & \mbox{if and only if} & ad = bc.
\end{eqnarray*}

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Part I.  Note that you may only use the axioms of the integers and the theorems that we have derived      from them to do the following problems.  Part I is due Monday Mar. 30.

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\noindent
Problem 1. Prove that $\sim$ is an equivalence relation.

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[Note that now we can talk about equivalence classes.]  Define:
\begin{eqnarray*}
[(a,b)] & = & \{ (c,d) | (c,d) \sim (a,b) \}.
\end{eqnarray*}

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\noindent
Problem 2. [Note that the operations on the right in the following are ordinary multiplication and addition on the integers.]

a.) Consider the operation $\cdot_\sim$ defined by
\begin{eqnarray*}
[(a,b)] \cdot_\sim [(c,d)] & = & [(ac, bd)].
\end{eqnarray*}
Show that $\cdot_\sim$ is well defined.

Note, this is with respect to the equivalence relation $\sim$:
In other words: show that if $(a,b) \sim (a',b')$ and $(c,d) \sim (c',d')$ then $(ac,bd) \sim (a'c',b'd')$.  Equivalently, show that if $(a,b) \sim (a',b')$ and $(c,d) \sim (c',d')$ then
$[(ac,bd)] = [(a'c',b'd')]$

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b.) Consider the operation $\oplus$ defined by
\begin{eqnarray*}
[(a,b)] \oplus [(c,d)] & = & [(a+c, b+d)].
\end{eqnarray*}
Show that $\oplus$ is not well defined.

In other words: show that there exist $(a,b) \sim (a',b')$ and $(c,d) \sim (c',d')$ that  $(a+c,b+d) \nsim (a' + c',b' + d')$.  Equivalently, show that there exist $(a,b) \sim (a',b')$ and $(c,d) \sim (c',d')$ so that
$[(a+c,b+d)] \neq [(a'+c',b'+d')]$.

c.) Consider the operation $+_\sim$ defined by
\begin{eqnarray*}
[(a,b)] +_\sim [(c,d)] & = & [(ad+bc, bd)].
\end{eqnarray*}
Show that $+_\sim$ is well defined.

In other words: show that if $(a,b) \sim (a',b')$ and $(c,d) \sim (c',d')$ then $(ad + bc,bd) \sim (a'd' + b'c',b'd')$.  Equivalently, show that if $(a,b) \sim (a',b')$ and $(c,d) \sim (c',d')$ then
$[(ad + bc,bd)] = [(a'd' + b'c',b'd')]$.

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[Big hint: since we are constructing the rationals from the axioms of the integers, you should be aware that we are constructing the usual set of fractions in the form $\frac nm$.  Our goal is to prove from the axioms that our $\mathbb{Q}$ is equivalent to our usual fractions.  So there has to be a one-to-one relationship between $\mathbb{Q}$ and our familiar fractions.  Use that to help answer the problems - then use our knowledge of the integers to prove your responses to the problems posed.]

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