thanks

kayla ]]>

i was wondering if you could give me some help on what to write for the paper. and also can i just send it to you through google docs? ]]>

Theorem 2 requires two halfs. For each half, you can do whatever proof technique (direct, contradiction, or contrapositive) makes sense to you. You can do each half differently. This one is probably the hardest of the four theorems (in my opinion) despite looking easy on the surface.

Try weak induction for Theorem 3. This problem is one of my favorites. To get some intution, take the picture that I drew for 5 points and ask yourself how many additional lines you would need to add if you added a 6th point. And then how many new lines if you added a 7th? How many new lines would you need to add to a circle with $k$ points and all the lines if you added one more point? Once you set this one up correctly, it is very short.

For Theorem 4, use strong induction and mimick the proofs we've done involving the Fibonacci sequence (except this isn't the Fibonacci sequence). If you know how to phrase the first sentence of the inductive step correctly, there aren't really any tricks. Don't get bogged down in the indices. The recurrence relation says: each term is equal to 5 times the previous term minus 6 times the terms before that. For example, $a_{11}=5a_{10}-6a_{9}$. Despite appearances, this one is quite friendly.

]]>Thank you,

Saddened Exams Students (Ian & Tabby)

http://jeromyanglim.blogspot.com/2010/10/getting-started-with-writing.html

]]>For number two, I am not sure how I am supposed to come up with a conjecture. I do not think I am supposed to randomly make something up for a

I feel like a pain, sorry for all the questions. ]]>

First, let's recall the statement that we are trying to prove:

If $x$ and $y$ are odd integers, then $xy$ is odd.

To form the contrapositive, we need to negate the hypothesis and conclusion and reverse the order of the implication. The negation of "$xy$ is odd" is "$xy$ is even." However, the negation of "$x$ and $y$ are odd integers" is a bit trickier. Before negating, notice that the original hypothesis is equivalent to "$x$ is an odd integer **and** $y$ is an odd integer." The negation of this statement is "$x$ is an even integer **or** $y$ is an even integer". Notice that the "and" in the original statement becomes an "or". The "or" in this case is an inclusive "or", which means one or the other or both. This is an application of a rule of logic called DeMorgan's Law, which you can read more about here. One can quickly write down a truth table to convince themselves this must be the case. In addition, it matches our use of everyday language. Think of some statements that involve "and" and ask yourself what it would mean if the statement was false. (There are other cases to consider, but this should be enough to convince that our model is accurate.) OK, in summary, the contrapositive is:

If $xy$ is even, then $x$ is an even integer or $y$ is an even integer.

To prove the contrapositive directly, you would start your proof with something like: "Assume that $x$ and $y$ are integers such that $xy$ is even." Your goal is to conclude that $x$ is an even integer or $y$ is an even integer.

If you wanted to finish this up by using a proof by contradiction, there wouldn't be anything wrong with that. In this case, you would write something like: "For sake of a contradiction, assume that it is not the case that either $x$ is an even integer or $y$ is an even integer." This implies that both $x$ and $y$ are odd. Now, do what we have done in lots of problems. What can you conclude?

Another approach could be to first prove as a lemma that if $p$ is prime and divides $xy$ where $x,y\in\mathbb{Z}$, then $p$ must divide either $x$ or $y$. A special case of this is when $p=2$. By the way, if you remove the requirement that the divisor be prime, it is no longer true. Y'all provided a counterexample to this already on a previous homework assignment.

If people ask, I'll provide more hints on the other problems (which are not nearly as difficult as the one discussed above).

]]>First on number 1 part c, how do we begin the proof when there is nothing said about "divides" in the "P" but there is in the "Q"?

Second, on 1 part d, does the square root effect the outcome somehow?

Third, I am not sure how to fill out the "If Q, then P" and "If not P, then not Q" portions of the truth table.

Finally, for number 3 do you want a mathematical example or just a real life example in which the inverse and converse are not neccessarily true?

Sorry for all the questions.

Kayla

First, recall that you can always use `Lurch` from a computer on campus by following the instructions that were on the original Circle-Dot handout. In case, you don't have the Circle-Dot handout handy, you can get to it here.

If you want to install `Lurch` on your own computer, it is easy to do. Follow these instructions:

- Go to http://lurch.sourceforge.net/
- Click on "Downloads" in the menu at the top of the page.
- Click on the big arrow next to "Download LurchLite."
- The download should automatically begin.
- Once the download has completed, double-click the file (if it doesn't open for you automatically).
- If you are using a Mac, drag the application to your Applications folder.
- If you are using a PC with MS Windows, then do whatever you usually do with applications.
- Once
`Lurch`(`LurchLite`technically) is installed, you can just double-click the application to get started.

If you have any trouble, post your questions here.

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