(Semi) Related, Non-FOMy, Econ-ish Math

So today in class, Casey split us up in groups with post-it notes of different numbers, and post-it’s with a function and domain. The thought was that function people had to find the max and min of the function to find their group members. It was at that point that I had a That’s So Raven flashback.

Yesterday, in Microeconomics, we started talking about where supply for a good comes from. Guess where it comes from?

Inputs (like capital and labor) are put into the production function, to get outputs (aka goods). Which is exactly what we’re talking about in FOM. Go figure.

The reason the whole “max” concept with functions made me think of this is because the production function takes inputs, and gives you the maximum amount of outputs you can get from those inputs.

Weird how all my classes are connecting like this.

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Onto/Clarifying Confusing Things/Triangles/Etc.

Today’s class was a relief. I figured out what an image set actually is, why image sets are not always equal to the codomain, why 4 = 2, and what happens when A and B are different sizes! I’m feeling pretty nice about it right now.

What An Image Set Is..Actually/Why B≠Im(f)

An image set is the set of outputs that a function can equal. The codomain may or may not be equal the image set. For example, f: ℝ→ℝ, where f = x². The image set of f [or Im(f)], is all reals greater than or equal to zero. The codomain is bigger than the image set, only because x² can never spit out negative numbers. Therefore, this function is not onto. BUT. If you change the codomain to ℝ≥0, the set is onto, because the image set and codomain are equal. When these two are equal, it means that every possible output that a given function could spit out is “accounted for” in a sense by the codomain.

Let’s try one more example to see if I got it. f: {1,2,3} → {5,6,11}, where f = 3x+2. The image set can be found by plugging in the inputs, and seeing if any of the outputs match up with anything in the codomain. The outputs for plugging in 1 and 3 are included in the codomain, so the Im(f) = {1,3}. This means that this function is also not onto, because with the inputs available (1,2 and 3), you can’t get the output of 6. It’s possible to get 6 by plugging in 4/3, but that is not an input choice in the domain. Therefore, all of the elements in the codomain are not accounted for.

4 = 2

Casey started out class by proving this to us. Take a triangle with sides 1, 1, and √2 (by pythagorean theorem). The hypotenuse is √2, so if you were to drive a car along that length, you would go for √2. But if you decided to go in a criss-cross manner, starting at the top corner of the triangle, going down .5 units, then traveling over .5 to touch the hypotenuse, then down .5 units, then over (touching the bottom of the triangle) .5 units again to get to the final destination of the bottom corner of the triangle.

Drawing this would be way more beneficial, but blogs suck at that feature, so just go with it. Adding up all the .5 distances, you go for 2 units to get to the same point as you would have if you had just driven on the hypotenuse. Do it again with more turns, and only going .25 before each turn, it’s a more confusing route, but you still go for 2 units. When you take the limit of these lengths as the number of turns you make approaches infinity, you get 2.

BUT, when you look at the line that starts approaching, it’s √2. Therefore, √2 = 2, making 4 = 2.

 When you think about it, if you’re taking a limit, there will always be really small triangles with all of their hypotenuses touching that larger hypotenuse of √2. If you find the hypotenuse of each of the .5 triangles, each one has a √2/2 hypotenuse, which, multiplied by two, equals √2. When you find the hypotenuse of a triangle with 1/4 side lengths, the hypotenuse is √2/4, and four of those hypotenuses equals: √2.

I think that’s starting to put a hole in Casey’s argument that 2 = 4. The limit of the drunken turning path may be two, but that drunken path’s hypotenuse sum will always equal √2.

When sets are different sizes.

We started on a worksheet in class today, and working over the first problem, I figured something out. If the domain is smaller than the codomain, out of the many possible functions between the two sets, none of them will be onto, because you will always have elements from the codomain left over (at least if you want that equation to pass the horizontal line test). If the domain is larger than the codomain, out of all the possible functions between the two sets, none of them will be one-to-one, because each input must go to some output, forcing inputs to double, tripe, or more-up on outputs.

So boom, that’s my conjecture.

Happy Monday!

FOM Book: 3.2 (Functions)

This section is really confusing to me, because it’s taking a subject I thought I pretty much knew most things about, and turned it on it’s head. So feel free to pull any examples or definitions I explain wrong apart so I actually know what I’m saying. Here we go.

Defining Functions the FOMy Way

Functions are basically input-output machines when you think about it. You stick some number, symbol, or otherwise into an f(x), and something else (or the same thing) pops out. In FOMy speak, we have to introduce some other features. That is, the set A and set B. We’re using these two sets as destinations to start from (A) and a place to end up (B). Since we can think of these outputs and inputs as ordered pairs (the first number/thing/symbol in the pair being what you put in, and the second number/symbol/thing in the pair being what you get out), it’ll be helpful to think of these ordered inputs/outputs of a Cartesian Product of our destinations. It makes sense. Cartesian products basically takes every possibility from ordering A before B (order is important in function and Cartesian products), which is what happens to a function too! 

A x B probably includes more ordered pairs than what the function actually takes and puts out, though, it just makes sense that the possibilities of the function should be introduced and included in A x B since A x B includes non-function pairs. Remember that functions’ inputs can only go to exactly one output. A x B includes ordered pairs where the input number is paired with multiple outputs. Therefore, the set of input/output ordered pairs we are dealing with does not equal A x B, but is a subset of it. Okay, now we’ll introduce the real FOMy stuff:

For any sets A and B, a function from A to B is a set S ⊆ A x B which satisfies the following conditions:

A. (∀ u ∈ A) (∃ v ∈ B) ((u,v) ∈ S), and
B. (∀ u ∈ A) (∀ v,w ∈ B) ([(u,v) ∈ S and (u,w) ∈ S] ⇒ v=w)
A is called the domain of the function, and the set B is the codomain of the function.
Let’s unpack all that. Basically all A. is saying is that for every term “u” that’s a part of the input set A, there’s another term “v” that’s a part of the output set B, such that the input/output pair (u,v) is an element of the set S, which is a subset of the Cartesian product of both of the sets that “u” an “v” are from.
All B. is saying is that for all “u” terms that are part of the A set, and for all terms “v” and “w” that are a part of the set B, when both the input/output ordered pairs (u,v) and (u,w) are elements of A x B, then “v” has to equal “w”.  This conclusion follows from A, and the need for a function to pass the vertical line test. Because the inputs are the same in (u, v) and (u,w) and because all inputs must have only one output, “v” has to be equal to “w”.
Another way to word this: For set A and B, “f: A→B” means: f is a function from A to B. If f is a function, then “f(u) = v” means: (u,v) ∈ f. 

Image Sets

Definition: For a function f: A→B, the image set of f [written “Im(f)” or “f  (A)”] is {f(t): t ∈ A}.

In simpler terms, if a function is going from set A to set B, an image set of this would just include all elements of B which are images (outputs) of an input from set A. You can think of a function going from A to B as going from A to Im(f).

This may seem a little confusing, so here’s an example.  If A = {1,2,3}, B = {1,2,3} and S = {(1,2),(2,3),(3,2)}. The image set for this function is {2,3}.

If f is a function from A to B , and u and v are elements of A and B respectively, f(u) = v AND:

  • v is the image set of u under f
  • v is the value of f at u
  • u is a preimage of v under f
  • u is mapped to v

Surjective

A function is surjective (aka onto aka epic) if every element of B matches up with some element of A. Put the technical way:

For any function going from A →B (∀ v∈B) (∃ u∈A) (f(u) = v)

For example. If |A| = 4, and |B| = 5, surjectivity from B to A would be possible, but not from A to B. This is because from A to B, all elements from A could go to a unique element from B, but one element from B would not have an input from A to attach to, since all elements from A can only go to one element of B. When going from B to A, it is possible for two inputs to go to the same output into B, so every output in B will be accounted for by all elements in the set A.

Put in a very simple term in a graphical sense, x² is not surjective if B equals ℝ, because no matter what x you put into x², you can never have x² equal ALL reals, ie the negatives. Because f(u) does not equal all “v”s possible, it is not surjective.

Injective

A function is injective (aka one-to-one aka monic) if the function passes the vertical line test, which also means that every output has only one corresponding input. In FOMy terms:

For a function f: A→B, (∀u,w ∈A) (f(u)=f(w) ⇒ u=w)

For example, take x² again. This function is not one-to-one, because f(-1²) = f(1²), but -1≠1. x is one-to-one though, because every input has only one corresponding output: itself.

Relating Image Sets and Injectivity/Surjectivity

A function who’s image set is equal to it’s codomain is surjective.

Question: How can we relate image sets and injectivity?

Bijective

A function is bijective (aka one-to-one correspondence) when it is both injective and surjective. Basically, a function has to have all outputs accounted for, and every input has to have exactly one unique output.

Also, if a function is bijective, it makes sense that if the function goes from A to B, A and B must have the same amount of elements. If a function is surjective, A should be larger than or equal to B. If a function is one to one, each input needs one unique output, so B should be larger than or equal to B. For a function to be both, the only option is for A to be equal to B.

For example, function x³ from {1,2,3,4} →{1,8,27,64} is bijective, as all outputs from B have a corresponding input from A, and all inputs have one unique corresponding output.

Composition

For functions f: A→B and g: B→C, the composition of f and g (written g∘f) is the subset of A x C, given by:

g∘f = {(u,w) ∈ AxC: g(f(u)) = w}; (u,w) ∈ g ∘ f means w= g(f(u))

or, in other words:

g∘f = {(u,w) ∈ AxC: (∃ v ∈ B)((u,v) ∈ f and (v,w) ∈ g)}
So basically, in this format, we find the image set of f under u first, then the image set of f(u) under g.

I’m pretty confused about the concepts of image sets and composition still (because it uses the image set concept), so I’m definitely bringing it up in class on Monday. Anybody else confused?

Negation – Edward Sharpe Edition

If you haven’t heard the song “Home” by Edward Sharpe and the Magnetic Zeroes already, do it RIGHT MEOW. It’s adorable.

Let’s see if we can negate some of the lyrics. I’m going to pick out some of the easiest ones to do (because I’m lazy).

Home is whenever I’m with you. 

Since the statement is saying that every time they’re with this person, they’re home, it’s a universal statement. So the negation of this would be an existential statement:

Home exists at a time when I am not with you.

Girl, I’ve never loved one like you. 

So, to me, this is a negative universal statement. That is, instead of saying that they always love this girl, they have never loved anyone else like this girl. Therefore, the negation would say something about how they actually have loved someone like this girl before (what a letdown):

Girl, there exists a person that I have loved exactly like you. (This sounds like a really badly worded break up)

Man oh man, you’re my best friend, I scream it to the nothingness. 

For the sake of the educational purposes of this post, I’m going to assume this is an existential claim. That is, there exists a time that they had this girl as their best friend, and in that same time they screamed it to the nothingness. (Doesn’t sound as romantic when explained). So, to negate this, we need to make a universal statement that says they’ve never been best friends, and that they don’t scream into nothingnesses. So:

For all elements in my set of best friends, you are not included, and for every instance I have screamed, I have not done so into nothingness. 

With that said, lyrics are really hard to dissect this way because there’s a lot of ambiguity in trying to pick out what the free variable and what the bounded variable would be. For example, I could have switched the last statement and said: For every time I have screamed into nothingness, I have not screamed that you are my best friend. 

And that, my friends, is why I am horrible at diagramming sentences, bad at grammar, and not an english major.

POW: Gauss is in the House!

This Problem of the Week was pretty cool, and it will be even cooler if I did it right! The question is asking what would happen if you had some arbitrary number of coins, n, in a bag that each had a unique value (1,2,3….all the way up to n), how many charities could you donate to if you wanted to donate with the same money.

First you can start by modeling the problem with 10 coins. So you have 10 coins worth, 1,2,3,4,5,6,7,8,9, and 10 dollars. You can set them up into pairs, with the largest coin and the smallest coin, then the second biggest and the second smallest, and so on. That way, every pair will be the same. You get 10+1, 9+2, 8+3, 7+4, and 6+5: 5 pairs of $11!

And then all of a sudden, I remembered good old Gauss.

Gauss was a german mathematician. Legend has it, that when he was little, one of his especially cruel teachers told him to count up all of the numbers between one and one hundred. Being the clever little squirt that he was (even at that young of age), he did the same thing we just did, but on a bigger scale. He matched up 100 and 1, 99 and 2, and so on. Since he split them into pairs, he realized that there would be 100/2 pairs, or 50 pairs of 101. Therefore, 101 x 50 is 5,050, and you now have the sum of the first 100 hundred digits with pretty much no mathematics needed. Pretty cool right? (Right.)

It gets cooler. We’ll move on to n number of coins..

Okay, so we can still set up pairs. And since there are n coins, we know that there will be n/2 pairs. The least amount a coin is worth would be 1 dollar, so it (n)+1, (n – 1) +2, (n – 2) +3, (n – 3) + 4, and so on. Notice that the bigger numbers become smaller be subtracting 1, then two, then three, and so on. So when you start adding 1, then 2, then 3, you get n + 1 for every pair.

So, all in all, for coins, you’ll be able to donate to n/2 charities, and donate n + 1 dollars to each charity!

(It even works for 10! 10/2 = 5 pairs, and 10 + 1 = $11 per charity.)

Problem of the Week: Officially Solved.

FOM Book: 3.1

I started out this chapter with new hope of learning new FOMy things, and got in return a section full of math hieroglyphics.

Hopefully this run through of what these symbols means will help us all out in using them in FOM (keyword hopefully). There’s actually a point to these symbols too, I asked Casey. Using these symbols instead of writing out statements in everyday language:

  1. Makes it impossible for your non-math major friends to read your homework 
  2. Could be used as a really lame party trick if you want to try to convince your friends you know Egyptians hieroglyphics.
  3. Takes out any ambiguity that English statements provide, by defining each term in the statement. 

The last one is the most important.

Okayyyy, here we go.

  • Universal Quantifier: ∀: “for all”/”for every”, basically a conditional statement “in disguise”
  • Existential Quantifier: ∃: “there exists”
  • Free variable: a not bound variable. This (these) variable(s) are integral parts of the statement. This free variable depends on the quantified statement before it. For example, (∃ x ∈ ℝ) (x² = u). The variable u is free, because it depends on the definition of x, and is not quantified by ∀ or ∃. If a statement has at least one free variable, it is considered an open sentence.
  • Bounded variable: also known as a dummy variable, or quantified variable. Basically a variable that is not integral to the explanation of a statement. For example, if you have (∀x∈ℕ)(x + 2 ≥ 2), you can explain the statement without using the term x. You could say “for any natural number, if you add two to that number, the answer will be greater than 2.” X is really just a place holder. Another way to think about it: x is just a placeholder in a definition, and x is quantified by the ∀. If there are only bounded variables in a statement, the whole statement becomes either true of false.
  • Predicate: Basically the statement after your bounded variable is quantified. The predicate is the meat of the statement that connects the bounded variable and shows it’s relationship to the free variable. For example, (∃ x ∈ ℝ) (x² = u). The predicate is  (x² = u).
  • Open Sentence: For an open sentence p(x) and a set S, “(∀ x ∈ S) (p(x)) means that every element of S makes p(x) true. (∃ x ∈ S) (p(x)) means that there is an element in existence in the set S that makes p(x) true. 
  • Implication: the symbol : ⇒. Usually seen in between a hypothesis and a conclusion. P ⇒ Q.
  • Negation: the symbol: ~. Basically turning a statement false. The negation of a conditional statement/universal statement is finding a counterexample. The negation of an existential statement is to find a contradicting true universal statement. For example, say I state that there’s a person somewhere in the library right now eating Frito’s loudly. The negation of this statement would be to say that for every person in the library right now, no one is eating Frito’s loudly (but that could still mean someone is quietly munching somewhere somehow). NOTICE, that the negation of an existential statement is a universal statement, and the negation of a universal statement is an existential statement. Coincidence? I think not.

Those are the highlights of this section that we went over in class as well. Hope they find you well!

Have a jammin’ weekend ya’ll.

Get Your Hands Dirty: #2

Alright, so I’m going easy on myself and doing a short blog post for this Wednesday, because I have an exam for another class to worry about, AND I SEE MUMFORD & SONS TONIGHT.

So I’m going to go through GYHD #2 right quickly!

GYHD #2:

Find the following GCDs:

GCD(10,-18)

GCD(7,12)

GCD(24,0)

GCD(8,-16)

First of all, remember that GCDs are Greatest Common Divisor for each of the two number listed, which basically means that the number found to be the GCD is the greatest number found in common for (x,y) that divides both of them equally. So, the greatest divisor that divides both 10 and -18, for example, is 2. The divisors of 10 are 1, 2, 5, and 10. The divisors of -18 are 1, 2, 3, 6, 9, and 18 (with any combination of those numbers where one is positive and one is negative). The greatest number in common from these sets is 2! Simple enough!

SO:

  • GCD (7,12):

Divisors of 7: 1,7

Divisors of 12: 1,2,3,4,6,12

GCD = 1

  • GCD (24,0)

Divisors of 0: any number! (any number dividing 0 is zero!)

Divisors of 24: 1,2,3,4,6,8,12,24

GCD = 24

  • GCD (8, -16)

Divisors of 8: 1,2,4,8

Divisors of -16: 1,2,4,8,16 (any combo where one of these numbers is positive and one is negative)

GCD = 8

 

Happy Wednesday ya’ll!

 

Contradictions. They Happen.

Proving by contradiction is actually really cool. Instead of proving by cases, or proving directly, you basically take the statement, try and work out what would happen if the negative of this statement were true, and then BAM, you’ve got a proof. It usually takes the form of “If this were true, THE WORLD COULD END.” Okay, not really, but it’s along those lines.

Say we take the statement: A number cannot both be odd and even. To prove by contradiction, we would start out and say “Okay, what would happen if a number was odd and even? What would it look like?” To that you can define what it is to be odd and what it is to be even, in which lies a contradiction!

An odd number is: m ∈ ℤ, such that x = 2m + 1

An even number is: m ∈ ℤ, such that x = 2m.

The contradiction is as follows. x cannot equal both 2m, and 2m + 1 at the same time, because, when they are set equal to each other (replacing x with 2m), 2m ≠ 2m +1.

 

That was almost a cheat of a contradiction proof though. They are usually never this easy, but at least you get the point now!

Happy Monday!

FOM Book: 2.3 – 2.4

Let’s Explore the Integers, and Then Prove Things About Them!

Multiples and Divisors of Integers

  • Definition 2.4: For n ∈ ℤ, the set of multiples of n is: {n · t: t ∈ ℤ}; usually abbreviated as nℤ. Each element in this set is called a multiple of n. 

Example: 5ℤ = {….,-15, -10, -5, 0, 5, 10, 15….} Notice negative numbers are included!

  • Definition 2.5: For x,y ∈ ℤ, x ≠ 0, and t being some integer, x · t = y means that “x is a divisor and a factor of y”, which is usually written as “x|y”. *Zero is not a divisor of any integer. Includes negative numbers. 

Example: 10|-30. 10 is a divisor of -30, because 10 x -30 = 30. *10|-30 is not a number, though. It’s a statement about the                                        relationship between 10 and -30. It is not a fraction or an equation.

  • Definition 2.6: For ∈ ℕ, where t > 1, “t is a prime” means that t is a positive integer that only has divisors of t and 1. If t has more divisors, then t is a composite number. 

*1 is not a prime or composite number, it’s just a chill no-name.

  • Definition 2.7: For x ∈ ℤ: x is even means that there is an integer t such that x = 2 · t. x is odd when there is an integer y such that x = 2  · y+1.

Exploration 1: Properties of Divisibility: what general statements can we make about divisibility?

  • If a|b and a|c, b and c may or may not be divisors of one another. 
  • If a|c and b|c, a and b may or may not be divisors of one another.
  • Prime numbers can be a divisor (x), but not divided (y).
  • If a|b and c|d, I don’t think we can make any conclusions of the relationship between the two, other than the fact that b and d must be composite, and that a and c can be composite or prime. For example 2|6 and 5|20. We cannot make any general connection between the two statements that would apply to all divisibility statements, because what they might have in common may not apply to other statements, such as 3|9.

Exploration 2: Odd and Even Numbers: What general results can you state about odd and even numbers?

  • odd + odd = even
  • odd + even = odd
  • even + even = even
  • odd – odd = even
  • odd – even = odd
  • even – odd = odd
  • even – even = even
  • odd x odd = odd
  • odd x even = even
  • even x even = even
  • odd/odd = odd
  • even/even = even
  • odd/even = fraction?
  • even/odd = fraction?
  • odd^rasied to any power = odd
  • even^raised to any power = even

Linear Diophantine Equations

Linear Diophantine Equations look like this:

ax + by = c.

A, b, and c are input into the equation as integers, and only integers can be input for x and y when trying to solve the equation. This whole topic just plays around with inputing different integers into this equation, and seeing when x and y are integers, and when x and y must be something else to solve the equation.

Exploration 3: Linear Diophantine Equations: For what integers a, b and c will the equation ax + by = c have integral solutions for x and y?

  • If a,b and c, all equal 1, x and y can equal 0 and 1. 
  • If a, b share a common factor that is not shared with c, you can’t solve the equation with integers.
  • If c is negative, the bigger out of ax and by must be negative, meaning either only x or only a must be negative if ax is bigger, and b or y must only be negative if by is bigger.
  • If c is positive, the bigger of ax and by must be positive, meaning that a and x can both be negative or both positive if ax is bigger, and the same for by if by is bigger.

Basic Facts About Divisors

  • Theorem 2.9: For integers a,b and c, if a|b and b|c, then a|c. 

Proof:

We are proving that a|c. Since a|b, there is an integer, x, such that: ax = b. Since b|c, there is and integer, y, such that: by = c. We need to show that there is an integer, az = c. Define z = xy. From this, we can assume z is an integer, since it is the product of two integers. Since b = ax, (ax)y = c, or a(xy) = c. This shows that z = xy is a solution in az = c, proving the proof!

  • Corollary 2.10: Any multiple of an even number is even.
  • Theorem 2.11: If a ∈ ℤ, than the set aℤ is closed under both addition and subtraction.
  • Theorem 2.12: For a,b and c ∈ ℤ, if a and b are even, and c is not even, then the equation ax+by=c has no integral solution for x and y.

Common Divisors

  • Definition 2.8: For x,y ∈ ℤ, “t is a common divisor of x and y” means: t|x and t|y.
  • Definition 2.9: For x,y ∈ ℤ “t is the greatest common divisor of x and y” means that t is a common divisor of x and y, and every common divisor of x and y is less than or equal to t. (Written as: GCD(x,y))
  • Definition 2.10: For x,y ∈ ℤ, “x and y are relatively prime” means: GCD(x,y) = 1.
  • Corollary 2.14: Suppose a,b,c ∈ ℤ with a and b not both 0, and d = GCD(a,b). If d is not a divisor of c, then the equation ax + by = c has no integer solutions for x and y.

Odd and Even Integers

  • Theorem 2.15: If u ∈ ℤ, then 2 · u ≠ 1.
  • Theorem 2.16: For x ∈ ℤ, if x is odd, then x is not even. And integer cannot be both odd and even.
  • Definition 2.11: For x,y  ∈ ℤ, “t is a common multiple of x and y” means that t is both a multiple of x and y.

 

I’m pretty sure that’s more than anyone ever really wanted to know about even numbers, odd numbers, multiples, and divisors. Now there’s the task of applying all these definitions and theorems to proofs – GYHD Problems, here I come! (But not right now, because I still have FOM Homework, and it’s Sunday. So I’ll think about them eventually.)