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\title{MAT 108: Problem Set 2}
\author{(ADD NAME)}
\date{Due 1/24/23 at 11:59 pm on Canvas}
\begin{document}
\maketitle
\paragraph{Reminders:}
\begin{itemize}
\item Put your name at the top!
\item You will receive feedback on PS1 by next Tuesday, 1/24. PS1 revisions are due Friday, 1/27 at 11:59 pm.
\item Dr.~Zhang will be away all of next week (no instructor office hours). I will be in Germany, where the time is 9 hours ahead of California. You may email me your questions, and I will respond once daily as usual. The TA will still hold his office hours.
\end{itemize}
\paragraph{Another reminder} Figuring out how to prove something should feel like doing a puzzle. Writing down and expressing your proof should feel like you're trying to write in a new language. Trust the process!
\paragraph{How much detail is needed?} In PS2, you no longer need to cite the axioms / propositions from Chapter 1. For example, it's clear to the audience (e.g.\ your classmates) that $-0 = 0$.
On the other hand, in Chapter 2, we defined the natural numbers $\N$ using a set of axioms that are not very obvious to your peers. You should cite these axioms as you use them.
%%%%
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\exercise
\textbf{Prove that $1 \in \N$ via a proof by contradiction. Then, deduce that that if $n \in \N$, then $n+1 \in \N$. }
\begin{remark}
The phrase \emph{deduce that} here is indicating that the second statement follows quite immediately from the first.
\end{remark}
\solution
% INSERT YOUR SOLUTION HERE
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\exercise
\begin{definition}
Let $m, n \in \Z$. If $m - n \in \N$, then we say $n$ \emph{is less than} $m$, and write $n < m$. We also say $m$ \emph{is greater than} $n$, and write $m > n$.
\end{definition}
\textbf{Prove that there exists no integer $x$ such that $0 < x < 1$. }
\solution
% INSERT YOUR SOLUTION HERE
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\exercise
Use induction to prove that for any $n \in \N$, the following formula holds:
\[
1 + 2 + 3 + \ldots + (n-1) + n = \frac{n(n+1)}{2}.
\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\exercise
\begin{definition}
Let $p \in \N$. If the only $k\in \N$ such that $k \divides p$ are $k = 1$ and $k=p$, then $p$ is \emph{prime}.
\end{definition}
\textbf{Prove that there are infinitely many prime numbers. }
\begin{remark}
We haven't rigorously discussed the term \emph{infinite} just yet. We will discuss cardinality in detail in a few weeks. For now, \emph{infinite} means \emph{not finite}.
You should begin by assuming there are finitely many prime numbers, so that you can label them $p_1, p_2, p_3, \ldots, p_n$ for some finite $n\in\N$. Then try to derive a contradiction.
\end{remark}
You may use the following Lemma without proof:
\begin{lemma}
Let $p$ be a prime number, and let $m \in \N$. If $p$ divides $m$, then $p$ does not divide $m+1$.
\end{lemma}
(Proving this lemma will be easier once we've developed the language of \emph{modular arithmetic} later on in the course.)
\solution
% INSERT YOUR SOLUTION HERE
\end{document}