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\author{Math 300 -- Spring 2019 (Hachtman)}
\title{Assignment 5: Primes and density}
\date{Due Monday, April 8}

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\begin{enumerate}
  \item A natural number $n$ is a \emph{divisor} of $m$ if $kn = m$ for some natural number $k$.  A number $p \in \mathbb{N}$ is \emph{prime} if it has exactly two divisors: 1 and itself.

\fbox{Prove that the set of primes is infinite.}

You may want to use the outline here:
    \begin{itemize}
      \item Show that if $n \neq 1$ and $n$ divides $m$, then $n$ does not divide $m+1$.
      \item Show every natural number $m>1$ has at least one prime divisor.
      \item Suppose that the set of primes is finite.  Let $m$ be the product of all of them, and consider $m+1$ to obtain a contradiction.
    \end{itemize}


\item Recall we defined the density of $A \subseteq \mathbb{N}$ to be the value of the limit
  \[
   \rho(A) := \lim_{n\to \infty} \frac{|A \cap [0,n]|}{n+1}.
  \]
However, the limit does not always exist!  

\fbox{\parbox{4in}{Define a set $A$ where the limit in the definition of density $\rho(A)$ does not exist, and prove it.}}

Hints:  
\begin{itemize}
\item Your goal is to define $A$.  For any increasing sequence $\langle n_k \rangle_{k \in \mathbb{N}}$ of natural numbers, you can define $A$ one piece at a time, by deciding what $A \cap [n_k,n_k+1)$ will be. 

\item Suppose that \[\frac{|A \cap [0,n]|}{n+1}\] is very very small.  Show there is a number $m$ so that \[\frac{|(A \cap [0,n]) \cup [n+1,m]|}{m+1}\] is very very close to 1.  
\end{itemize}
\end{enumerate}

Write up your solutions in {\LaTeX} and submit them by email as usual.
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