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. IMG_6192 IMG_6194

Class Review: Conditional Statements – the most confusing thing you will ever think about (not really)

So a conditional statement is basically a formula to set up a hypothesis and a conclusion. The basic formula is

IF (hypothesis), THEN (conclusion).

These statements do not need to be true to be considered a conditional statement. For example, the statement If Casey likes teaching FOM, then we will all get A’s, is a conditional statement, even though it is (unfortunately) not true.

How do we know then if these statements are true or false? Simple.

To make a conditional statement true:

  • Any hypothesis that is true should be followed by any conclusion that is true. So, a statement like If I like pancakes, then my name is Kate. They don’t necessarily have to be connected in any way, both parts just must be true. It seems weird, but logical causality is different from straight-forward causality. 
  • Any hypothesis that is false that is followed by any sort of conclusion. So, a statement like If the Earth is flat, I’ll legally change my name to Beyoncé. It doesn’t matter what the conclusion is, because the condition of this statement will never be proven true, and therefore the conclusion will never be used. This condition is vacuously true. 

To prove a conditional statement false:

  • The hypothesis must be true, while the conclusion proves false. For example, a statement like If I am in a FOM class, I am a math major. The conclusion is false because I am a math minor, but I am in a FOM class. Therefore the hypothesis is true, but not the conclusion. 

With this knowledge, I want to talk about the problem we were given in class today that I’m still trying to figure out. The problem is:

You see four cards and are told that each one of them has a letter on one side and a number on the other. You see the four cards as pictured here:

A 3 T 8

Which of the four cards must you turn over to verify the claim:

“On the opposite side of every vowel is an even number”

First, we’ll make it a conditional statement. If a card has a vowel on one side, then there will be an even number on the other side. 

So to prove this right or wrong, we must focus on the conclusion, not the hypothesis. If we were to focus on the hypothesis and prove that wrong, we would be focusing on a whole other statement, not the one at hand. We don’t need to worry about what happens when the card doesn’t have a vowel on one side, we’re focusing on what happens when it does, so we’ll make the condition that the hypothesis is true.

When the hypothesis is true, the statement is true if the conclusion is true, and false if the statement is false. Casey also mentioned in class that it is much easier to prove something false than true. To prove something false, you only need to bring up one instance that proves it false, but to prove something unquestionably true, you would need to flip all the cards.

Or would we? Let’s say we focus on only vowels. T would then be irrelevant to the statement all together because it is not mentioned in the hypothesis. To prove this statement right then, we would have to flip both the 8 card and the A card, to see that there is a vowel on the other side of the 8 card and an even number on the other side of the A card.

To prove this statement wrong, we would need to flip A to see if there is an odd number, and flip 3 to see if there is a vowel on the other side. If either of these cases happened, the statement would be false.

It seems to me that A is the crucial card. Either way, no matter if you are trying to prove it true or prove it false, you need to flip A to see what is on the other side.

Whew, that was confusing. I hope it makes any sort of sense.


Class Review: Sets

On Friday we talked about sets in class.


A set is just a collection of things, mathematical or otherwise. I could have a set of numbers, symbols, ice cream cones, pandas, whatever.

Say I have a set of: {7, 42, #, h, @, $} Each of the characters in this set is called an element. Since this set also has numbers, symbols, and letters in it, we can also call it a set of mixed types.

Some Common Sets

N = all natural numbers

Z = all integers

Q = all rational numbers

R = all real numbers

These are not the only sets in existence, just the most common.

Another common set we talked about was the null or empty set, which is exactly what you think it is. The set with nothing in it. It’s written as : {   }.

The set {∅} does not mean the same thing. That set is not empty, because it has another set inside of it: the null set.

Set Relationships

Sets can be related to each other as well. For example, all natural numbers are integers. Therefore, the natural numbers are a subset of integers. Similarly, all integers are rational, so integers are a subset of rationals. All rational numbers are real, so rational numbers are a subset of real numbers.

Then we bring complex numbers into the mix. Complex numbers are all real numbers..plus the unreal numbers. This means that we can say that real numbers are a subset of complex numbers, as all real numbers fall into the category of complex. We can write out what complex numbers are made of as:

{a + (i)b: a,b \in \!\, \mathbb{R} \!\,}, so when we’re talking about a real number, b=0.


A union of two sets means:

\cup \!\, B = {x: x \in \!\, A or x\in \!\,B}

Which basically means that this new set contains elements of A or B, it doesn’t matter where they come from.

An intersection means:

\cap \!\,B = {x: x \in \!\, A and x\in \!\,B}

Which basically means that this new set only contains elements that are in both A and B.

When you subtract two sets, the new set becomes:

A – B = {x: x \in \!\, A and x∉B}, which basically just a mathy version of realizing that when you subtract two sets, the new set is all of the elements in A, with all elements of B taken away.

You can compare more complicated sets as well. Say we compare 2×2 matrices and all real numbers.

M2x2 \cap \!\, \mathbb{R} \!\, = ∅, because the two sets have nothing in common, so their intersection set contains nothing. The 2×2 matrices may contain real numbers, but the sets are in two different formats, and can therefore not be compared.


If B \subseteq \!\, A, then B \cap \!\, A = B.

If B \subseteq \!\, A, then B \cup \!\, A = A.

When you think about it, these two theorems make sense.

If B is a subset of A, that means that all elements in B are elements in A as well, so when only looking at elements in common with both A and B, that is simply all of B.

Since the same holds true for B being a subset of A, and we compare the two sets so that we use all elements part of A and B, this set would just be A, because A is comprised of all of B’s elements plus some unique elements only in A.


The cool thing about intersection and union is the fact that you can perform both an infinite amount of times.

For example, if you wanted to compare these sets:

[-1 , 1] \cap \!\, [-1/2 , 1/2] \cap \!\, [-1/4 , 1/4]….(to infinity and beyond)

You would look at what all of these sets that are getting infinitely smaller, and see what elements they share. In these case, this intersection would equal the set {0}, since the sets are getting smaller and smaller.