\documentclass{article}
\usepackage{fullpage,amsmath,amssymb}


\begin{document}

\noindent
Math 541 \\
Problem Set 1
\begin{enumerate}


\item[0.1.5.] 
\begin{enumerate}
\item No.  $1/2 = 2/4$, but $f(1/2) = 1$ while $f(2/4) = 2$.
\item Yes.  If $a/b = a'/b'$ then there is a $c$ such that $a' = ca$ and $b' = cb$, so
\[ \frac{a'^2}{b'^2} = \frac{a^2 c^2}{b^2 c^2} = \frac{a^2}{b^2}. \]
\end{enumerate}


\item[0.1.7.] The relation $\sim$ is reflexive: for all $a \in A$, $f(a) = f(a)$, so $a \sim a$.  It is symmetric: if $a \sim b$ then $f(a) = f(b)$, so $f(b) = f(a)$, so $b \sim a$.  It is transitive: if $a \sim b$ and $b \sim c$ then $f(a) = f(b)$ and $f(b) = f(c)$, so $f(a) = f(c)$, so $a \sim c$.  Thus $\sim$ is an equivalence relation.

Every equivalence class of $\sim$ is a fiber of $f$, as follows.  Let $a \in A$.  Then the equivalence class of $a$ equals the fiber over $f(a)$, for $a' \sim a$ iff $f(a') = f(a)$ iff $a' \in f^{-1}(\{f(a)\})$.

Conversely, every fiber of $f$ is an equivalence class of $\sim$.  Let $b \in B$.  Since $f$ is surjective, there is an $a \in A$ such that $f(a) = b$.  The fiber over $b$ is the equivalence class of $a$ as we showed above.


\item[0.2.1.]
\begin{enumerate}
\item The greatest common factor is $1 = 2 \cdot 20 - 3 \cdot 13$.  The least common multiple is 260.
\item The Euclidean Algorithm gives
\begin{align*}
372 &= 5 \cdot 69 + 27 \\
69 &= 2 \cdot 27 + 15 \\
27 &= 1 \cdot 15 + 12 \\
15 &= 1 \cdot 12 + 3 \\
12 &= 4 \cdot 3
\end{align*}
so the greatest common factor is $3$, which we can also find by writing $372 = 2^2 \cdot 3 \cdot 31$ and $69 = 3 \cdot 23$.  The least common multiple is $372 \cdot 69 / 3 = 8556$.  Rewriting our divisions above,
\begin{align*}
27 &= 372 - 5 \cdot 69 \\
15 &= 69 - 2 \cdot 27 = 69 - 2 \cdot (372 - 5 \cdot 69) = -2 \cdot 372 + 11 \cdot 69 \\
12 &= 27 - 15 = (372 - 5 \cdot 69) - (-2 \cdot 372 + 11 \cdot 69) = 3 \cdot 372 - 16 \cdot 69 \\
3 &= 15 - 12 = (-2 \cdot 372 + 11 \cdot 69) - (3 \cdot 372 - 16 \cdot 69) = {\bf -5 \cdot 372 + 27 \cdot 69}.
\end{align*}
\end{enumerate}


\item[0.2.3.] Since $n$ is composite, let $a$ be a positive divisor of $n$ different from $1$ and $n$, and let $b = n/a$.  Then $a < n$, so $n \nmid a$, and $a > 1$, so $b = n/a < n$, so $n \nmid b$, but $ab = n$, so in particular $n \mid ab$.


\item[0.2.4.]
\[ ax + by
= a\left( x_0 + \frac{b}{d} t \right) + b \left( y_0 - \frac{a}{d}t \right)
= a x_0 + \frac{ab}{d}t + b y_0 - \frac{ab}{d}t
= a x_0 + b y_0
= N. \]


\end{enumerate}
\end{document}