Lemma: The square of an odd number is odd.

Proof. Let p = (2n - 1), then p^2 = (2n - 1)(2n - 1) = 4n^2 - 2n -2n + 1 = 4n^2 - 4n + 1 = 2(2n^2 - 2n + 1) - 1 which is of the form 2j - 1 where j = 2n^2 - 2n + 1. Thus, p^2 is odd.

Corollary: If p^2 is even, then p is even.

Proof. Assume p is not even. Then by the prior Lemma, p^2 is not even, a contradiction. Thus, p is even.

Theorem: The root of 2 is Irrational

Proof. Suppose there exists a rational number r such that r^2 = 2. Then by definition, r = (a/b) where a and b are integers. Factor our any common divisors k_j such that a = (k_1 * k_2 * ... * k_n) * p and b = (k_1 * k_2 * ... * k_n) * q, so that a/b = p/q and that p and q do not share common factors. Since p^2/q^2 = 2, then p^2 = 2*q^2 so that p^2 is an even number. By the above corollary, we know that p is even. Now since p and q share no common factors, we know 2 is not a divisor of q, which means that q is odd.

Also, since p is even, there exists an integer m such that p = 2m. Thus, p^2 = 4m^2. Recalling that p^2/q^2 = 2 by hypothesis, and that this means p^2 = 2q^2, we have that 4m^2 = 2q^2, which simplifies to 2m^2 = q^2. This of course means that q^2 is even, and by the above corollary, we have demonstrated that q is even.

Since we can argue from the same premise that q is both even and odd, we know the premise must be at fault. Thus, there is no rational number whose square is 2.

Let f:A->B and g:B->C and let H be a subset of C, then (g(f))^-1(H) = f^-1(g^-1(H)).

Proof. Let x be a member of (g(f))^-1(H), then since H is a subset of C and (g(f)):A -> B, then x is a member of A, and there exists a z in H such that g(f)(x) = z. Thus, f(x) = g^-1(z) and x = f^-1(g^-1(z)), so that x is a member of f^-1(g^-1(H)). This gives (g(f))^-1(H) is a subset of f^-1(g^-1(H)).

Now let x be a member of f^-1(g^-1(H)). By definition, there exists z in H such that z = g(f(x)). Thus, x = (g(f))^-1(z) and since z is a member of H, x is a member of (g(f))^-1(H). This gives us that f^-1(g^-1(H)) is a subset of (g(f))^-1(H).

Since each side is a subset of the other, the two sets are equal; QED

Wanted to try a hand-written proof, image quality is not very good..

Let x be a sequence in R. Then there exists a subsequence y of x such that y is monotone.

Proof. If x is monotone, then construct any increasing sequence j of integers and then y_n = x_(j_n) is also a monotone sequence. If x is not monotone, then there exists a least integer k_1 such that x_(k_1) is neither an upper or a lower bound for x_M where M > k_1. Now consider the sequence of x beyond x_(k_1); if that sequence attains its minimum, let

Not quite right -- got stuck at the end and realized I had written the conjecture down incorrectly, but I wanted to post it anyways because I was pleased that I figured out what I had done wrong. Will take a second stab at it another day.

Let A, B, and C be sets. Then A \ ( B U C) = (A \ B) U (A \ C).

Proof. We show this with containment in both directions. Consider x in A \ (B U C), then x in A, but x not in B U C. This means x not in B and x not in C. Since x in A and x not in B, then x in (A \ B). Since x in A but x not in C, then x in (A \ C). Since x in (A\B) and x in (A\C), then x in (A\B) U (A \C). Thus, A \ (B U C) is a subset of (A\B) U (A\C).

To show the reverse, consider y in (A\B) U (A\C). The either y in A\B or y in A\C or both. If y in A\B then y in A and y not in B. If y in A\C, then y in A and y not in C. Thus, y in A and either y not in B or y not in C or both. ?

Let f:R->R be differentiable with the (n+1)th derivative of f being identically zero, then f is a polynomial of degree at most n.

For notational clarity in ascii, the nth derivative of f will be called "f^(n)", and the value of that function at a point x will be called "f^(n)(x)".

Proof. Since the (n+1)th derivative is identically zero, we know that f', f'', f''', ..., f^(n) all exist on R. Thus, the Taylor Expansion of f can be computed as the sum of a polynomial of degree n and a remainder term. That is, given some u in R,

f(x) = f(u) + f'(u)(x - u) + (1/2!)f''(u)(x - u)^2 + ... + (1/n!)f^(n)(u)(x - u)^n + (1/(n+1)!)f^(n+1)(e)(x - u)^(n+1)

for some e between u and x. However, since f^(n+1) is identically zero, f^(n+1)(e) = 0, so the remainder term in the taylor expasion is also zero. Thus,

f(x) = f(u) + f'(u)(x - u) + (1/2!)f''(u)(x - u)^2 + ... + (1/n!)f^(n)(u)(x - u)^n

Which is to say that f behaves like a polynomial in the variable (x - u) on the entire interval between x and u. Since u was chosen arbitrarily, let it be zero for simplicity's sake. Then,

f(x) = f(0) + f'(0)x + (1/2!)f''(0)x^2 + ... + (1/n!)f^(n)(0)x^n

so that f is a polynomial of degree at most n. QED, bitches.

Suppose that there exists some $N \in \mathbb{Z}$ which has the property $N \geq n \forall n \in \mathbb{Z}$. This means that $N$ is at least as large as any integer.

Now consider that for any $n \in \mathbb{Z}$, we have $(n + 1) \in \mathbb{Z}$ as well. Also, since $0 \lt 1$, we have $(0 + N) \lt (1 + N)$ so that $N \lt (N + 1)$. Thus, we have constructed an integer, $(N + 1)$ which is greater than $N$. Since $N$ was assumed to be the greatest integer, we have shown by contradiction that such a property cannot hold. Thus, there is no largest integer.