- Your quiz this week will be over infinity.
One of the 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?
- Art contest winners announced tomorrow! (Your professor forgot the list at home...:(
- Reminder: we're a week from our project and logo demonstrations (the last week of
class).
We'll do logos on Monday, and projects on Tuesday. If you have any conflicts on those days, let me know in advance.
- Here's a logo from Pompeii....
- We had a quiz:
- Median: 6.5
- Key
- We wrapped up (mostly) our demonstration that there are an
infinite number of bigger and bigger infinities.
We showed 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.
- We didn't quite get to this last time, so let me begin with the
punch line:
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,
and we enthusiastically claim that we've somehow managed to find a natural number partner for every subset of the natural numbers. Except....\(n \in N\) \(ss(n) \in P(N)\) 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 (a tango!) -- that had no partner, and so is sitting quietly and sadly at home, perhaps even crying quietly....
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!"
- And the power set of the natural numbers is a set, having a power
set of its own: \(P(P(N))\). And that power set is bigger yet,
and so on, and on and on, and on, and on, and on, and so on
forever, forever, Hallelujah, Hallelujah!
-
Upshot: infinity comes in various sizes! Infinitely
many, in fact. Crazy sounding, I know -- but it's true.
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!:)
- That's too crazy, it seems; but, nonetheless, mathematicians
know that it's true, because we can prove it! The key to
determining if two sets have the same size is the one-to-one
correspondence -- and that is key for infinite sets, as well.
- So, in the end, here's what you say on the playground,
when you want to love your friend more:
"I love you more than the power set of your set of infinite love."
Amen!
Question of the day:
- Can we add up an infinite number of things and get a sensible
answer? The answer is sometimes yes, as we saw when discussing
the Dichotomy of Zeno:
\[ \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\ldots=1 \]
- The reason for us to consider this problem seriously is contained
in Pascal's triangle.
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".
- Now for what seems like a little diversion: we're going to start
by figuring out how to add up a succession of alternating plus
ones and minus ones:
\[
S_1 = 1-1+1-1+1-1+1-1+\ldots
\]
(What do you think that should add up to?)
I'm going to try to convince you what I think that it should add up to1.
- Now what does this have to do with Pascal's triangle? Let's first
add all the invisible zeros....
(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.
- Having added all the infinite zeros, we're going to
figure out how to continue Pascal's triangle in the
backwards direction! We know how to go forward,
but how do you go backward? It turns out that we have
infinitely many options, which can be good or bad.
What are some options?
- What might guide us as we try to figure out what two numbers
add to 1?
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.
- Recall what the entries of Pascal's triangle add up to:
powers of 2. That pattern should continue. But we've
been adding finite sums down below. We're now in the
presence of an infinite sum -- or rather two infinite
sums!
\[
\frac{S_1}{2} = \frac{1}{2}-\frac{1}{2}+\frac{1}{2}-\frac{1}{2}+\frac{1}{2}-\frac{1}{2}+\ldots
\]
These have terms half the size of the terms of \(S_1\), so this
sum should be half of \(S_1\), which makes it \(\frac{1}{4}\);
add two of those together and you get \(\frac{1}{2}\), which is
exactly
\[
\frac{1}{2}=2^{-1}
\]
Perfect! Exactly what we thought it should be.
- Once again we're facing an infinitude of options. How does the
next row look?
And once again, that first choice is going to determine it all....
- Let's now focus on the sum
\[
S_2 = 1-2+3-4+5-6+7-8+\ldots
\]
A similiar trick allows us to calculate what that sum "should
be" -- which is exactly what we expect it to be!
- One more: let's focus on
\[
S_3 = 1-3+6-10+15-21+28+\ldots
\]
Let's try the same trick again, and we'll see that it just
keeps on working.
- But now to the big story:
\[
S = 1+2+3+4+5+6+7+8+\ldots
\]
What should that add up to? I think that you're going to be
surprised....!