homework_problems_week_12.tex

\activitytitle{Homework problems, week 12}{Due on (put date here).}

Write up solutions of each of the problems below.
They are designed to be straightforward problems.
The goal is to come as close to perfection in your solutions as you can.
\begin{itemize} \itemsep 1pt
\item Do not take shortcuts.
\item If you need to show that something is true for all $n$, or for all $x,y$, start the proof with ``Let \ldots''
\item If you need cases, explain what the cases are and why they cover all the possibilities.
\item If you are doing a proof by contradiction, start that part by saying ``Assume \ldots''
\item If you are doing a proof by contrapositive, tell what $P$ and $Q$ are, and that you will be showing that $\lnot Q$ implies $\lnot P$.
\item Take small steps in each proof, and explain each step.
\item Follow good form.
\item If your proof started with ``Let \ldots'' it will probably end by saying ``We made no further assumption \ldots''
\end{itemize}
Here are the problems to do.  You can write them in your notebook or on separate paper.
\blist{0.1in}
\item Show that if $n$ is an integer and $7n$ is odd, then $n$ is odd.
{\bf Hint:} Be clear what facts you are using about even and odd numbers.

\item Without consulting your book or your notes, prove that $\sqrt{2}$ is irrational.
I mean it.  
Do this from memory.
You should be able to write a very nice proof, with no missing steps.

\item Let $x$ and $y$ be real numbers, and suppose that the product $xy$ is irrational.
Show that either $x$ or $y$ (or both) must be irrational.
{\bf Hint:} You can do this.  Be patient, think about it.

\item Let $A = \{2k+1 : k \in \Z\}$ and let $B = \{ 2m-11 : m \in \Z \}$.
Show that $A = B$ by showing containment both ways.
{\bf Hint:}  Use good form!

\item Let $A = \{ (x,y) \in \R^2 : y = 5x/7 - 2/7 \}$ and $B = \{ (x,y) \in \R^2 : 5x - 7y = 2 \}$.
Show that $A = B$ by showing containment both ways.

\item Let $A = \{ m \in \Z : m = 15k$ for some $k \in \Z \}$, let $B = \{ m \in \Z : m = 35j$ for some $j \in \Z \}$, and let $C = \{ m \in \Z : m = 105n$ for some $n \in \Z \}$.
Show that $A \cap B = C$ by showing containment both ways.
One direction is easier than the other.
Label one of them ``the easy direction'' and the other ``the hard direction''.
{\bf Hint:} Yes, we worked on a problem just like this in class.
Don't go back and find it, work through this one on your own.
{\bf Another hint:}  In the hard direction, you should come to something like $3k = 7j$ where $j$ and $k$ are integers.
You will need to conclude that $j$ is a multiple of 3.
If you are up for the challenge, show this using the division algorithm.
Don't use any ideas about prime factorization.
\elist

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