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\newtheorem{conjecture}{Conjecture}
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\lhead{MATH 311-02}
\chead{Problem Set \#3}
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\rfoot{\bf due February 18}
\begin{document}
\begin{enumerate}
\item \textbf{(15 points)} \emph{Below are two proofs relating to
divisibility of products.}
\begin{enumerate}
\item \textbf{(7 points)} \emph{Determine the sufficient (and if
possible necessary) conditions on an integer $n$ for the
following statement to be true: ``For integers $a$ and $b$, if
$n\nmid ab$, then $n\nmid a$ and $n\nmid b$.''. Then state the
condition on $n$ and the resulting impliccation as a proposition
and prove it.}
% \mid and \nmid are a divisibility bar and a negated divisibility
% bar, respectively.
% Below, for your edification, a construction of a proposition and
% its proof in LaTeX. Obviously these are just filler text at
% present, and you must determine exactly what conditions on $n$
% to include in the proposition, and what text to include in the
% proof.
% These environments ("proposition" and "proof") are given by
% the amsthm package; "proof" is explicitly defined by it, while
% "proposition" we defined ourselves way back at the beginning of
% the document, by using a \newtheorem environment.
\begin{proposition}
% As written this is wrong. I threw the word "cromulent" in as
% an example of a restriction on $n$ beyond merely being an
% integer. As far as I know, there is no such thing as a
% "cromulent" integer, so try to figure out a exactly what
% restrictions on $n$ are appropriate instead.
Let $n$ be a cromulent integer. If $a$ and $b$ are integers such
that $n\nmid ab$, then $n\nmid a$ and $n\nmid b$.
\end{proposition}
\begin{proof}
We restate the premise here. Then we argue by steps from our
premise. Lorem ipsum dolor sit amet, consectetur adipisicing
elit, sed do eiusmod tempor incididunt ut labore et dolore magna
aliqua. Ut enim ad minim veniam, quis nostrud exercitation
ullamco laboris nisi ut aliquip ex ea commodo consequat. Duis
aute irure dolor in reprehenderit in voluptate velit esse cillum
dolore eu fugiat nulla pariatur. Excepteur sint occaecat
cupidatat non proident, sunt in culpa qui officia deserunt
mollit anim id est laborum. Finally we reach our conclusion.
\end{proof}
\item \textbf{(8 points)} \emph{Determine the sufficient (and if
possible necessary) conditions on an integer $n$ for the
following statement to be true: ``For integers $a$ and $b$, if
$n\mid ab$, then either $n\mid a$ or $n\mid b$.''. Then state the
condition on $n$ and the resulting impliccation as a proposition
and prove it.}
\end{enumerate}
\item \textbf{(12 points)} \emph{In class we saw a proof that $\sqrt{2}$ is irrational. Here we will explore variations on it.}
\begin{enumerate}
\item \textbf{(8 points)} \emph{Modify the proof of $\sqrt2$'s
irrationality to produce a proof that for integer $n$, $\sqrt n$
is irrational if $n$ is not exactly the square of some
integer.}
% \sqrt{...} puts its argument under a radical; Here it's
% sometimes used with a single-character argument. Beware of such
% forms as, e.g. \sqrt16; that wouldn't typeset right, and you
% probably mean \sqrt{16}.
\item \textbf{(4 points)} \emph{Explain why this proof couldn't be
modified to prove that $\sqrt 4$ is irrational. Don't merely
point out that $\sqrt 4$ is rational; exhibit why the specific
proof of the irrationality of $\sqrt 2$ doesn't work when the 2
is replaced with a 4.}
\end{enumerate}
\item \textbf{(7 points)} \emph{Let $a,b\in\mathbb Z$. Prove that if
$a^2+2b^2\equiv 0\pmod{3}$, then either $a$ and $b$ are both
congruent to zero modulo 3 or neither is congruent to zero modulo
3.}
% \equiv is a three-lined equals-sign, as is commonly used to assert
% congruence (in some modulus). \pmod{...} puts a parenthesised
% "(mod ...)" in the expression. You can explore and compare to
% "\bmod ...", which is the binary operator "mod", common in
% computer science pseudocode.
\item \textbf{(6 points)} \emph{Let $A$, $B$, and $C$ be sets. Prove
that $(A-B)\cup(A-C)=A-(B\cap C)$.}
% You only need to do the following if you want to -- which is why
% it's commented out in the basic template, so that it doesn't show
% up unless you uncomment it and work on it.
% \item \textbf{(4 point bonus)} \emph{Prove (possibly following the
% structure of the similar proof in class) that if every two
% vertices of a regular 17-sided polygon are joined with segments
% colored red, green, or blue, then regardless of how the segments
% are colored, some three vertices are joined by three edges of the
% same color.}
\end{enumerate}
\end{document}