# Partitions are everywhere

We talked about equivalence relations and partitions as one of the last units in my FOM class. Equivalence relations are basically classes of relations that various things can fit into. Described in a less vague ways, think about elements of sets being balls, and think about equivalence classes being buckets that these balls can be thrown into. You can sort the balls by color, size, or any other way, and that’s where the relation part of the definition comes in. Partitions come in with the fact that balls are in different buckets. Partitions separate elements in some way.

I pretty much understood this concept in the class, but there’s something really weird about the partition part. It shows up everywhere in my life.

Like in my India class (Religions and Cultures of India, but I called it my India class), we learned about the British occupation of India, and the partition of India. I didn’t even have to pay attention in that class to know that that was about splitting up India! The partition of India was basically about splitting up India based on the equivalence class of religions.

The reason I’m bringing this up now is because I was helping my mom at our church set up for an event, and she asked me to help her with these little dividers that the preschool uses to separate the hall into different sections. Guess what they’re called?! Partitions.

# I’m a big deal

I should’ve blogged about this a while ago, but whatever I’m doing it now.

BEST NEWS EVER: Julia and I won best blogs in our FOM class! While this may not seem like a huge accomplishment, it actually is. I got a pint glass out of it, filled with candy. What’s on the pint glass? The proof of Cantor’s Theorem! The mathy part of me is drooling right now.

FYI, Cantor’s Theorem states that the cardinality of the power set of a set A, is greater than the cardinality of A. This theorem is true for infinite and finite sets.

We were very excited about our pint glasses, to say the least.

# Scavenger Hunt: Done, Thank God.

Amanda and I just finished the scavenger hunt, big thanks to everyone who helped us (which was a lot of people). We’ll post all our clues later!

If anything, this scavenger hunt made me realize that I cannot do math by myself, especially under a time limit.

And that math majors are super nice and always willing to help, even if you ask them a random calc II question in the middle of the green bean.

# Procrastination

Okay so I have this Comparative Economic Systems final in 13 days. It’s 25% of my grade, and cumulative. I basically have to know everything about capitalism, socialism, and the economic framework and history of 7 countries to the point where I could compare and contrast any aspect of them with any other country.

So instead of starting to study the insurmountable amount of information I’m required to know, or finding more clues for the scavenger hunt, I’m just going to prove that thing we talked about in class on Monday.

Using the definition of limit given, prove:

lim (x→0) x² = 0

lim(x→17) x² = 17²
The definition of limit as given before in class is: (∀ε>0)(∃δ>0) such that if 0<|x-a|<δ then |f(x) – L|<ε.
We’ll start with lim (x→0) x² = 0. First, let’s define some terms/match them up with the definition of a limit.
f(x) =
a = 0
L = 0.
So based on the definition, if 0<|x|<δ, then |x²|<ε. Since the definition says that this definition must work for all ε>0, we just need to choose one δ that makes this statement true. If δ ≥0, this statement works, since x is approaching 0.
0<|x|<1, then |0-0|<ε, which is true because ε>0.
For lim(x→17) x² = 17²,
f(x) =
a = 17
L = 17².
Based on the definition, if 0<|x-17|<δ then |x² – 17² |<ε.
Adding 17 to both sides, we get that if -17<x<17, then |x² – 17² |<ε, which is true, because ε must be greater than 0. No matter what you put in for x, because of the absolute values, cannot be less than 0.
That’s really all I’ve got. I thought these would be much easier to prove since we’re given a definition to plug things into, but I don’t think I proved either of these. I’m ignoring the fact that I need to assume the hypothesis is true and focus on the conclusion, and I’ve lost sight of what needs to really be proved in this statement, since we’ve got δ and ε to prove.
Dang it. And I thought I was being productive.

# Clue #1

So the first clue has something to do with marbles and distributions. Amanda figured out that it has something to do with Pascal’s triangle, where Pascal’s marble run distributes the marbles in a bell shaped curve.

Also when I googled Pascal’s Marble Run this picture came up:

Is this what we’re looking for??

# How to Brighten Someone’s Day: FOM Edition.

Because who doesn’t want to compliment someone/pick someone up while showing off their knowledge of mathy math?

• “I like you more than the power set of the power set of the real numbers.”
• “My mood tends to turn into a strictly increasing function when I’m around you.”
• “We relate better than the bijection relating the integers to the naturals.”
• “I’ve been working on the induction of my life for a while now, are you my k+1?”
• “I’m hoping the intersection between the two of us isn’t the empty set.”
• “I’m contrapositive you’re the one for me.”

# “Casual Thinking and Hardcore Drinking”

The above quote is one of the many reasons I love FOM/Casey. Anyways, Casey wanted us to work on proving a theorem as part of our homework this weekend, and I am 75% I figured it out last night.

Cantor-Bernstein-Schroeder Theorem:

If ∃ injective functions f : A → B and g : B → A, then ∃ a bijection b: A → B.

Okay, so first of all we’re assuming that those two injections exist, and we’re trying to prove that those two injections mean that a bijection exists.

For f : A → to exist, there must be at least as many outputs as there are inputs so that f(a)=f(b) ⇒ a=b can be true. If there were more inputs than there were outputs, every input would not have a unique output, and would therefore not be an injection. So, based on this |A| ≥ |B|.

But, there exists another injection  g:B → A, which means that |B|≥|A| based on the same reasoning above.

For both of these functions to be injective, we need to combine |A|≥|B| and |B|≥|A|, which means that only way both injections could be true is if |A| = |B|.

For a function to be bijective in between two sets, they have to satisfy onto and one-to-one. For one-to-one, as said before, there needs to be at least as many inputs as outputs. There can be more outputs not mapped to with injectivity. But with onto, all outputs must be mapped to, so there must be at least as many inputs for all outputs. This means that for a bijection to exist, |A| must equal |B|.

Hey! We just proved that it does if there are two injections in existence!

I think this proves this, right? Word. Happy Saturday ya’ll.

# Models – Micro ∩ FOM Edition

Blah blah blah, yes I talk about Micro a lot.

But these two classes have been overlapping like crazy to me lately. They’re both using ambiguous models that tend to confuse the crap out of me.

In micro, we’re talking about oligopolies. Oligopolies focus on the most important firms in a market. It’s pretty much in between perfect competition (with a lot of firms fighting for consumer money) and monopoly (with one big guy making production decisions), but it’s categorized more towards the monopoly end of the spectrum.

Like FOM, Micro uses a lot of models for more of a self-reference tool to see how economics works in theory, only to apply it to real world situations later. The three models we’re talking about right now are the Cournot Model, the Bertrand Model, and the Stakelberg Model

## Kickin’ It With Cournot

With every economic (or other any field really) model, you’ve got to assume some things first. This model keeps it simple, and only keeps up with two firms in the market, which is called a duopoly. You’ve also got to assume that each extra unit of output produced costs the companies nothing (Marginal Cost = 0; but the model can compensate for positive, constant marginal coast if need be). The last assumption is stretching a little, but it’s crucial to make this model work. We need to imagine that each firm assumes that the other (it’s rival) will continue producing at it’s current level.

Basically this model shows how each firm will react to it’s rival’s output, through two reaction functions. Through brutal algebra, we can take a total demand function and marginal cost, and tell you how much each firm will produce, what price they’ll charge, and even how much money they’ll probably make out of it.

This graph shows the stable equilibrium of a Cournot duopoly. At point E, there is no need to react any differently and revise output, because both firms are acting according to the other’s assumptions. This equilibrium is called the Nash equilibrium, named after a really cool (but crazy) math/econ dude. If you have not seen A Beautiful Mind, you need to.

## Ballin’ Bertrand Model

The Bertrand Model is much simpler, and makes most of the same assumptions the Cournot model makes, but imagines that firms are reacting off of announced prices.

Say Firm 1 announces their price on a product. Firm 2 has three options. They can charge a higher price, and sell nothing. They can charge the same price, and split the market with Firm 1. OR, they can charge a lower price, and capture the whole market. Both firms obviously want to make the most amount of money, so both of them strive to undercut prices just low enough to make some money, but to be lower than their rivals price (think price matching at grocery stores).

This actually works against both of them. Both firms will end up cutting prices so low, that price will be set equal to the amount it took to make that one extra unit (MC), and the firms will split the market.

Silly firms.

## Stackelberg Swag

The Stackelberg Model is probably the most realistic of all three models. It considers two models, but takes into consideration that one firm is more experienced than the other. We treat the follower’s output level as given, and substitute the follower’s reaction function into the market demand equation to find the leader’s output.

This makes sense intuitively, because the follower firm is really the only firm doing the reacting. The leader firm is the one making the big decisions.

The take away message from these three models is very simple: the more competition you have in a given market, the more output and lower prices you’ll get. Competition is a good thing, people.

Anyways, these models all seem simple enough, but when put to the real-life application test, they seem to break apart. You can’t just assume that companies will act rationally.

I’m just thinking about all of the high-school algebra I have to work through to make any three of these models work for my Micro final.

# Math ≠ Ambiguity (Right?)

We had a huge discussion in FOM today that brought to light some of the concerns I’ve been having with numbers lately. We talked about Euclidian Axioms, and how basically, that’s just one model that we use to reference numbers and the world around us. And get this. Basically, the type of math we use is still a working model. Some brilliant, John Nash-type could wake up tomorrow morning and find a contradiction in the math we’ve all grown to know and love.

Our best defense against it? Whenever a contradiction comes up, we go back to the axioms, and make sure there are no errors.

No offense, but that sounds a lot to me like bringing a knife to a gun fight.

This also made me think about what we’ve been proving lately in class. We went over one horse theorem in class, with a proof that proved that all horses are the same color. One subtle detail made the proof fall apart. We ended up pulling it apart because they intersected two sets – and why would you do that?!

But then I’m thinking back to another complicated set Casey showed us in class, where he did some random mathy thing (forgive me for not remembering), and said something basically to the effect of, it works, and took someone a really long time to figure out that this proof works this way, so yeah.

I see differences in these two situations obviously, but they are sort of similar to me as well. What makes math work? Does anyone really know?

I think more than ever I’m realizing that math is 20% number crunching, 5% knowing what you’re doing, and 75% acting like you know what you’re doing.

This is sort of eye-opening for me, because I’ve loved math all this time because of the lack of ambiguity.

FOM right now is just making me really philosophical. I don’t think I can handle this much longer, man.

# Recap/Nostalgia

As per the second to last Monday of the semester, I’m thinking over the past 12 or so weeks, and wondering if I’m any smarter at this math stuff. Casey did say we’d learn how to prove things in his class, but am I there yet?

I’m not going to lie, I really thought I would take this class, and somewhere right in the middle of the semester we’d come up with this universal proof writing formula that could be used for all mathy things, everywhere.

What I’m realizing ties back to the quotes we blogged about at the beginning of the semester (corny I know). Math isn’t all formulas. It requires common sense, and the sense to throw that common sense out the window when necessary (ie thinking about infinity). Math is what you make it, and for me it’s still fun. Here’s what I’ve learned so far:

• What a hypothesis is. I may have learned about this in all science classes since 3rd grade, but now I finally get it. Really all you need to know about the hypothesis, is that you always assume it true. And if it’s false, your work there is done. (ie if a unicorn runs through the library, I’ll ace college. neither is going to happen, so why worry about it. in math terms, we call it a vacuously true statement).
• What a conclusion is, and how easy it is to make a proof into proving a very, very small part of a statement. Like with implication statements. You basically assume 75% of the statement is true and just prove the last conclusion. The key here is distinguishing identifying the conclusion and having something be easy to prove. Just because you only have to prove 25% of a statement, doesn’t mean it’s easy.
• Counterexamples are the holy grail of FOM. Very rare, hard to find, but freaking awesome when found.
• Implications are straightforward, sometimes. At least you know when you see a little ⇒, you know what you’re dealing with.
• Crazy letters can help too. Knowing what ∀ and ∃ mean is usually about 33% of the battle.
• Proving sets are equal requires two things: patience, and proving that each is a subset of the other one.
• Functions are not just functions. If you’re a fan of the good old calculus function, run far way from FOM right now.
• Injectivity, surjectivity, and bijectivity will help you in ways unimaginable at the beginning of the semester. Just learn what they mean and never forget it.
• Sometimes, when you can’t figure out a statement, it helps to turn it around, a lot. (ie converse, contrapositive, etc.)
• Induction is probably the easiest concept (which isn’t really even that simple). Just think in terms of k, and prove that every other k after that acts the same way. Or, start with k, and prove that all k’s before that are true. Sound good?
• In the words of Casey Douglas “And when it all comes down to it, all you’re doing is wiggling”. Wiggles are equivalence classes, which are partitions. There is two much notation for all three of these things, so once you understand any of it, stick with one and chug through with it.
• When you’re proving an equivalence relation, just prove three things: transivity, reflexivity, and symmetry.
• There are so many different kinds of infinity. Knowing this should make you feel very small (so eat the extra donut if you want to).
• Just because there are different kinds of infinity, does not mean you can leave infinity be. Match infinite sets up to each other using functions and keep on provin’.
• Modulo is not spanish for anything, at least not in FOM.
• And finally, math is only fun when you give yourself time to think about how really awesome it is.

Maybe FOM is making us smarter, even though it feels pretty confusing mostly all of the time.