Well, kind of done. I know where the last clue is but someone didn’t put it back from earlier today… Laura and I finished it though so YAY

I just wanted to make a quick post about the POW. I am 99.9999% sure that the set S, such that |N|<|S|<|R|, does not exist. Why though?

Cantor’s continuum hypothesis says,

There is no set whose cardinality is between that of the integers and that of the reals.

The set R is uncountably infinite, whereas the set Z is countably infinite. Since |N| = |Z| (so that it applies directly to our POW), we can also say that there is no set whose cardinality is between that of the naturals and that of the reals, because the set S would have to be between a countable and an uncountable set, and no such set exists.

Even though Casey wasn’t in class on Friday, he left us with a confusing set to define. And even though I didn’t give all that much thought to it, because I REALLY wanted bubble tea, I am thoroughly confused by these sets.

We let set S = {all positive integers that can be defined using twelve words or less}. This means S^I = {all positive integers that cannot be expressed in twelve words or less}. We also were told that S^I ≠ Ø because there are a finite amount of English words, but an infinite amount of positive integers.

Next, we let x = “the smallest positive integer not definable in fewer than twelve words.”

Here’s where I got caught up. This means that x itself is being defined in fewer than twelve words. Does that mean that it belongs in S, instead of in S^I as we described it? And couldn’t we technically describe any positive integer as “a positive integer that is defined using twelve words or less” since that definition itself uses less than twelve words? I’m so confused. So confused. Why do you do this to us Casey?!

In our book, countable is defined this way:

For a set S, “S is countable” means: S is finite or countably infinite. A set which is not countable is called uncountable.

The book then goes on to prove that R is uncountable, and I thought the proof was really interesting. Casey’s had talked about some sets having a more infinite number of elements that other sets, but I wasn’t sure how this would be proven. Here is a simplified proof of what the book shows.

Prove: R is uncountable.

We want to use a proof by contradiction. Suppose R is countable. Since R is infinite, this means R must be equivalent to N. This means there must be a bijective function f:R → N.

The proof then goes on to show that for f to be bijective, for all x, x would have to be an element of Im(f). Since x is not an element of Im(f), then f is not bijective, which is a contradiction, and so R is not countable, i.e. R is uncountable.

I was struggling looking for a blog post for today since most of class consisted of kicks, and hugs, and finals. So I Google-d “awesome math theorem” and the very first result was another blog called Algorithm.co.il, and a post titled “10 Awesome Theorems and Results.”

http://www.algorithm.co.il/blogs/computer-science/cryptography/10-awesome-theorems-results/

Most of the post was oriented towards computer science, but one theorem caught my eye: Stokes’ theorem. Stokes’ theorem is just a grand generalization of the fundamental theorem of calculus. Basically, it says that we can reason about what is happening in a specific area because we know what is happening at its perimeter. Stokes’ theorem is a generalization of other theorems such as the divergence theorem and Green’s theorem.

The theorem itself is a bit complicated, but I thought the original blog post was really interesting and I thought I’d share it with you all. 🙂

Enjoy this perfect SPRING day!

Prove (x,y)~(u,v) ↔ x²+y²=u²+v² is an equivalence relation on S.

1)      Reflexive: (∀ x,y ϵ S)((x,y)~(x,y))

Let x,y ϵ R. If x²+y²=x²+y², then (x,y)~(x,y).

2)      Symmetric: (∀ x,y,u,v ϵ S)((x,y)~(u,v) → (u,v)~(x,y)

Suppose (x,y)~(u,v). This means x²+y²=u²+v². We want to know if u²+v²=x²+y². The proof from here is trivial, so if (x,y)~(u,v), then (u,v)~(x,y).

3)      Transitive: (∀ x,y,u,v,w,z ϵ S)((x,y)~(u,v) and (u,v)~(w,z) ↔ (x,y)~(w,z))

Suppose (x,y)~(u,v) and (u,v)~(w,z). This means x²+y²=u²+v² and u²+v²=w²+z².

x²+y²=u²+v²=w²+z²

x²+y²= w²+z² which is what we wanted to prove, so if (x,y)~(u,v) and (u,v)~(w,z), then (x,y)~(w,z).

The second part of this question asks us what S/~ would look like if we were to draw it. I think that we would have a series of circles, centered at the origin, with radius r such that r ϵ R. This is similar to the unit circle.

Let’s talk eigenvectors. This is what we went over today in physics and it’s ALL LINEAR ALGEBRA.

When we are discussing particle physics, we often refer to the Schrödinger equation, which is pretty complicated and icky looking, but can be written in eigenvalue-eigenvector form: Ĥ|ψ> = E|ψ>.

We want to break this into a two-state system, and we will call these states “flavor states.” Why? I don’t know, and neither did Ted. Just go with it. The first one is written |1> = (1/0) and the second |2> = (0/1) (the backslash represent the start of the second row) so that their “dot product” is equal to 0, i.e. <1|2> = 0.

Instead of our usual equation for total energy, energies of this system are given by the linear transformation Ĥ = (h g/g h) where h, g ϵ R in the |1>, |2> basis.

We want to know, if a particle begins in |1> |ξ(t=0)>, what is |ξ(t)>? To find this answer we need to find c, E, and |ψ> in the equation sum(c e^-iEt/h |ψ>.

To find E|ψ> from the equation Ĥ|ψ> = E|ψ>, we solve the matrix equation Ĥ|ψ> = E|ψ>. To do this we set |ψ> equal to (α/β). Now we have (h g/g h) (α/β) = E (α/β). After some simplification we find (h-E g/g h-E) (α/β) = 0. The determinant of this matrix is 0. This means that E has two possible solutions, E = h±g.

To find α, we normalize our equations, which is much more difficult to explain than it is to do, so I’m actually stuck on how to describe this to you. Basically, we’re trying to make the probability of finding either state, |1> or |2>, to be equal to 1.

Next, to find the c’s, we get to use bases! YAY! LINEAR! We end up using (1/1) and (1/-1) as our vectors because they are our vector representations of E = h±g.

MORAL OF THE STORY: Linear and physics use exactly the same methods, so no matter how much I despise reducing matrices 456412 times a day, I guess it’s kind of useful.

(∀x(P(x)⇒Q(x)) ⇒ ∀u(R(u)⇒S(u)))

The outline for this proof is simple:

Assume ∀x(P(x)⇒Q(x)). Prove ∀u(R(u)⇒S(u)), by assuming R(u) is true.

The actual proof is as follows:

Supposed f is a fuction R → R and let g(x) = f(2x).

Prove that is f is increasing, then g is also increasing.

Assume f: R → R and g(x) = f(2x). Assume f is increasing. This means for all x, f^I > 0. By the chain rule, the derivative of f(2x) is 2f^I(2x), and 2f^I(2x) > 0. This means f(2x) is also a strictly increasing function. Since we have already assumed f(2x) = g(x), and we have shown f(2x) is an increasing function, we have also shown that g(x) is increasing.