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One of the most important concepts we got to last time was the power set, and you should know what that is, and how to form them. Do you?
Winners receive an extra Get-Out-Of-Quiz-Free card -- congratulations!
We'll do logos on Tuesday, and projects on Thursday. If you have any conflicts on those days, let me know in advance. You are expected to attend to support your classmates. Please be on time.
I'll want your digital materials in advance. Please get those to me by Monday evening (logos), and by Wednesday evening (projects). Anything that we'll want to display, so that I can have it ready to go.
If you want to go early in the presentations, let me know. Otherwise, your presentations will be ordered randomly (although I'll have prepared the order in advance, with the materials lined up).
You'll have three minutes for your presentation. You should talk about the logo/project itself, your motivation or what it means to you, and about the particular mathematics involved. You should definitely have three minutes of material -- make me cut you off!
At the end of class, you'll be asked to give me some positive comments about three of the presentations that resonated with you. These will be shared (anonymously) with the presenters at a later date.
We were ready to show that the power set of the natural numbers is bigger than the natural numbers, via a "proof by contradiction": we assume something impossible (i.e. that the natural numbers and its power set can be put into one-to-one correspondence), and then proceed to show that that's impossible.
This means that, although the natural numbers $1,2,3,\ldots$ is infinite as a set, there's a bigger set (the power set of the natural numbers).
Imagine not: then we can set up a one-to-one correspondence between the natural numbers and its power set,
\(n \in N\) | \(ss(n) \in P(N)\) |
1 | \(N\) |
2 | {} |
3 | odds |
4 | primes |
5 | {1,2} |
6 | evens |
7 | Fibonaccis |
... | ... |
We can find a subset of \(N\) that's not on the right hand side! That is, we can show that it's not really one-to-one, because there's a subset that's not at the dance -- that had no partner, and so is sitting quietly and sadly at home.
We construct this sad set as follows:
We will call \(ss(n)\) the partner subset corresponding to a natural number \(n\); and we construct the set \(A\), the lonely set, by the following rule: for each natural number \(n\), we either add \(n\) to \(A\) or not, depending on whether \(n \in ss(n)\):
If \(n \in ss(n)\), then \(n \notin A\)
If \(n \notin ss(n)\), then \(n \in A\)
And by this means, we can see that \(A\) is different from every \(ss(n)\) that's gone to the dance, and so \(A\) says to itself sadly, "Well, I guess tonight I'll be dancing by myself!"
That symbol that you've been familiar with for all your lives, $\infty$: you thought it stood for a single thing; but it stands for a whole collection of monstrously big things, all too big to really think about properly. (Well, Cantor did!:)
"I love you more than the power set of your set of infinite love."
Amen!
Question of the day:
\[ \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\ldots=1 \]
Let's make ourselves a copy of our old friend, but let's leave a little room at the top. Skip four rows before starting in with the "1".
I'm going to try to convince you what I think that it should add up to1.
(Your homework is to have Vi Hart help you to see the invisible....)
But a great mathematician named Jacob Bernoulli long ago made them visible...
but only from one side!:) Let's put them on both sides.
What are some options?
Symmetry is a fundamental property of the triangle, and it seems like a useful thing to preserve. That leads us to just one choice for those two entries above 1 -- and once we make that choice, the entire row above the first one is complete.
And once again, that first choice is going to determine it all....