All submissions (10)

Then there is a 1:1 mapping between the Natural numbers and the Even numbers.

Proof. Consider the map f(n) := 2n. Then for every integer n, there exists an even integer f(n). Consider two distinct integers j and k, then f(j) = 2j and f(k) = 2k, and since j does not equal k, we know that 2j does not equal 2k. Thus, f is a 1:1 map.

Scrambled Eggs with chopped, pitted dates.

I used some sunmaid dried pitted dates, about 1 date per 2 eggs, and chopped them into smallish bits and put them in the skillet on low for a few minutes. This is to soften them up. If you put them on medium (normally what I scramble eggs at) the sugar in the dates will char and it will taste like eating burnt marshmallows.

Anyways, I started to crack and beat the eggs after I started heating the dates, but I should have given the dates another minute or two because they were not quite soft enough. Threw in a dash of milk and some salt and pepper. Pretty good addition to plain old, boring ass scrambled eggs.

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.

I usually make lamb stew each week, but normally I just throw frozen veggies and frozen lamb into a crockpot and let it run overnight. Decent results, very little effort.

Today, I opted to thaw the lamb and sear it first, and then put the seared lamb into an already rolling stove pot. Still used frozen veggies, but threw in some diced potatoes to thicken it up and also some fresh mushrooms because mushrooms are fucking cool.

Whole thing was done and packed into containers in thirty minutes. Having a bowl and a Stone IPA right now.

Lesson learned, you gotta peel the potatoes. Otherwise some of the potatoes disintegrate in the heat and your are left with some potato-free skins, which does not for an appetizing dinner make.

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.

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