Paint my testicles blue and call me a moron...

[Kuroneko]

So far so good. So if there's for example 3^x, by definition it would be equal to |{f:X -> {a,b,c}| or for n^x = |{f:X->{x1, x2, x3, ... , xn}| for any cardinal for which |X| = x?

Absolutely.

I don't understand what this means. How can you have a set of functions between one set and another set?

Well, why not? One can have a set of sets, and a function between a domain A and a codomain B is simply a subset of the Cartesian product AxB that satisfies certain properties. In particular, a function f:A→B is a set of ordered pairs (a,b) from AxB such that for every a in A, there is a unique b in B with (a,b) in f. For example, let's have X={0,1} and Y = {a,b,c}. Then, the set of all functions from X to Y is { {(0,a),(1,a)}, {(0,a),(1,b)}, {(0,a),(1,c)}, {(0,b),(1,a)}, {(0,b),(1,b)}, {(0,b),(1,c)}, {(0,c),(1,a)}, {(0,c),(1,b)}, {(0,c),(1,c)} }. Unsuprisingly, there are |Y|^|X| = 9 distinct functions. Note that ordered pairs are themselves typically defined as sets, with (x,y) = {x,{x,y}}.

This is too condensed for me to understand. For a bijection, you need to define a one-to-one and onto function that maps A -> B. How are you achieving this by noting simply that any element of X is in {0,1} or it is not?

That is not what I did; I noted that every element of X is either in a particular subset or not, a conditions which can be made to correspond to one of {0,1}. Pick an arbitrary f:X→{0,1}. Rather trivially, the set Y = {z in X: f(z) = 1}, the set of all elements z of X such that f(z) = 1, is a subset of X that is uniquely determined by f (why must it be unique?). Thus, this mapping between these functions and subsets is injective (one-to-one). The other direction is also easy, since given a Y⊂X, one can define a function f:X→{0,1} by f(z) = 1 if z in Y, and f(z) = 0 if z is not in Y, so every subset has some corresponding "characteristic function" f:X→{0,1}. This mapping is then not only injective but also surjective (onto).

There is something in the supplemental textbook using Schroder-Bernstein Theorem to prove |P(N)| = |R| = c, I'm not going to type it and I don't completely understand it.

I very much doubt that is the case. The Schröder-Bernstein theorem merely states that the cardinals are totally ordered, which is completely independent of the continuum hypothesis. It is likely that either your book did somethign completely different, or it invoked some auxilary axiom to prove |P(N)| = |R|, since a sucessful proof of |P(N)| = |R| from classical set theory alone (it's Zermelo-Fraenkel, by the way) would mean that Gödel and Cohen are fatally wrong and no one has noticed for more than half a century.

A ^ 2 = |{f : 2 -> A}| = |{f : {0,1} -> A}| = |P(N)| = c
2 ^ A = |{f : A -> 2}| = |{f : A -> {0,1}}| = |P(N)| = c

Incorrect on the former, right on the latter. Order often matters quite a lot--and why shouldn't it, since 2^3 is not the same thing as 3^2? Think about what you are doing in the first line. A^2 = |{f:{0,1}→N}|, but what does it mean intuitively? A function f:{0,1}→N can be thought of as an ordered pair of naturals (n,m) [where f(0) = n and f(1) = m; the n is the "0th" coordinate and the m is the "1st" coordinate], so it is really the question of how many points with natural coordinates (or integer, if integers Z is used instead of naturals N) there are on the coordinate plane.

A ^ A = |N|^|N| = |{f : |N| -> |N|| = |N| = A

No. Think about it: c>A>2, thus A^A ≥ 2^A > A. (But is A^A>2^A or A^A=2^A? That is the question.)

(2^A)^A = |{f : A -> 2^A}| = |{f : A -> P(N)| = |{f : A -> R}| = c

Correct answer, far from sufficient proof. This is actually without a doubt the hardest one to prove out of the whole lot, so perhaps it is best to save it for last.

A + A = |N U N| = |N| = A

Yes.

3 ^ A = |{f : A -> 3}| = |{f : A -> {0,1,2}}| = |{f : A -> P(N)}| = |{f : A -> R}| = c

The set {0,1,2} is not equinumerous to P(N); there are undoubtedly more than three subsets of the naturals. This is completely broken.

A * A = |N * N| = |N^2| = |N| = A

Yes, but above you claimed that A^2, which is actually the same cardinal as |N^2|, is equal to c.

(2 ^ A) ^ 2 = |{f : 2 -> 2^2}| = |{f : {0,1} -> R}| = c

Ah, not quite on the second step, but yes, (2^A)^2 = |{f:{0,1}→R}| = c. This is really the statement that there are just as many points on a two-dimensional plane as on a one-dimensional line (since |R|=2^A, we're really dealing with |R|^2). This is true, although counter-intuitive to many; can you prove it? (Assuming this was not already proven in class.)

[Kuroneko]

Incorrect. Numbers are part of the international conspiracy to give stupid people headaches.

This is actually more along the lines of a good side-effect rather than our true objective ... er, nevermind.

They didn't invent math. They preserved it and added to it but they didn't invent it. I think you're going to have to blame a caveman for this invention.

True. But they did invent algebra, which is the bane of high schoolers everywhere, or at least those that waited until high school for algebra. But in a very real sense, the Greeks invented mathematics, since although they were far from the first to use numbers or even geometry, they were to actually investigate them for their own sake. The difference is not dissimilar to the relationship between science and engineering.