Prove that the crocodile is longer than it is wide.

Lemma 1. The crocodile is longer than it is green: Let’s look at the crocodile. It is long on the top and on the bottom, but it is green only on the top. Therefore, the crocodile is longer than it is green.

Lemma 2. The crocodile is greener than it is wide: Let’s look at the crocodile. It is green along its length and width, but it is wide only along its width. Therefore, the crocodile is greener than it is wide.

From Lemma 1 and Lemma 2 we conclude that the crocodile is longer than it is wide.

How do you hunt elephants?

MATHEMATICIANS hunt elephants by going to Africa, throwing out everything that is not an elephant, and catching one of whatever is left. Experienced mathematicians will attempt to prove the existence of at least one unique elephant before proceeding to step 1 as a subordinate exercise. Professors of mathematics will prove the existence of at least one unique elephant and then leave the detection and capture of an actual elephant as an exercise for their graduate students.

COMPUTER SCIENTISTS hunt elephants by exercising Algorithm A:

1. Go to Africa

2. Start at the Cape of Good Hope.

3. Work northward in an orderly manner, traversing the continent alternately east and west.

4. During each traverse pass:

a) catch each animal seen

b) Compare each animal caught to a known elephant

c) Stop when a match is detected.

Experienced COMPUTER PROGRAMMERS modify Algorithm A by placing a known elephant in Cairo to ensure that the algorithm will terminate. Assembly language programmers prefer to execute Algorithm on their hands and knees.

ENGINEERS hunt elephants by going to Africa, catching gray animals at random, and stopping when any one of them weighs within plus or minus 15 percent of any previously observed elephant.

ECONOMISTS don’t hunt elephants, but they believe that if elephants are paid enough, they will hunt themselves.

STATISTICIANS hunt the 1st animal they see N times, and call it an elephant.

CONSULTANTS don’t hunt elephants, and many have never hunted anything at all, but they can be hired by the hour to advise those people who do. Operations Research Consultants can also measure the correlation of hat size and bullet color to the efficiency of elephant-hunting strategies, if someone else will only identify the elephants.

POLITICIANS don’t hunt elephants, but they will share the elephants you catch with the people who voted for them.

LAWYERS don’t hunt elephants, but they do follow the herds around arguing about who owns the droppings. Software lawyers will claim that they own an entire herd based on the look and feel of one dropping.

VICE PRESIDENTS of engineering, research, and development try hard to hunt elephants, but their staffs are designed to prevent it. When the vice president does get to hunt elephants, the staff will try to ensure that all possible elephants are completely pre-hunted before the vice president gets to see them. If the vice president does see a non-prehunted elephant, the staff will:

1. compliment the vice president’s keen eyesight,

2. enlarge itself to prevent any recurrence.

SENIOR MANAGERS set broad elephant-hunting policy based on the assumption that elephants are just like field mice, but with deeper voices.

QUALITY ASSURANCE INSPECTORS ignore the elephants and look for mistakes the other hunters made when they were packing the jeep.

SALESPEOPLE don’t hunt elephants but spend their time selling elephants they haven’t caught, for delivery two days before the season opens. Software salespeople ship the first thing they catch and write up an invoice for an elephant. Hardware salespeople catch rabbits, paint them gray, and sell them as ”Desktop Elephants”

|Analysis: 1. Differentiate it and put into the refrig. Then integrate it in the refrig. 2. Redefine the measure on the referigerator (or the elephant). 3. Apply the Banach-Tarsky theorem. Number theory: 1. First factorize, second multiply. 2. Use induction. You can always squeeze a bit more in. Algebra: 1. Step 1. Show that the parts of it can be put into the refrig. Step 2. Show that the refrig. is closed under the addition. 2. Take the appropriate universal refrigerator and get a surjection from refrigerator to elephant. Topology: 1. Have it swallow the refrig. and turn inside out. 2. Make a refrig. with the Klein bottle. 3. The elephant is homeomorphic to a smaller elephant. 4. The elephant is compact, so it can be put into a finite collection of refrigerators. That’s usually good enough. 5. The property of being inside the referigerator is hereditary. So, take the elephant’s mother, cremate it, and show that the ashes fit inside the refrigerator. 6. For those who object to method 3 because it’s cruel to animals. Put the elephant’s BABY in the refrigerator. Algebraic topology: Replace the interior of the refrigerator by its universal cover, R^3. Linear algebra: 1. Put just its basis and span it in the refrig. 2. Show that 1% of the elephant will fit inside the refrigerator. By linearity, x% will fit for any x. Affine geometry: There is an affine transformation putting the elephant into the refrigerator. Set theory: 1. It’s very easy! Refrigerator = { elephant } 2) The elephant and the interior of the refrigerator both have cardinality c. Geometry: Declare the following: Axiom 1. An elephant can be put into a refrigerator. Complex analysis: Put the refrig. at the origin and the elephant outside the unit circle. Then get the image under the inversion. Numerical analysis: 1. Put just its trunk and refer the rest to the error term. 2. Work it out using the Pentium. Statistics: 1. Bright statistician. Put its tail as a sample and say ”Done.” 2. Dull statistician. Repeat the experiment pushing the elephant to the refrig. 3. Our NEW study shows that you CAN’T put the elephant in the refrigerator.

Lemma 1. All horses are the same color. (Proof by induction) Proof. It is obvious that one horse is the same color. Let us assume the proposition P(k) that k horses are the same color and use this to imply that k+1 horses are the same color. Given the set of k+1 horses, we remove one horse; then the remaining k horses are the same color, by hypothesis. We remove another horse and replace the first; the k horses, by hypothesis, are again the same color. We repeat this until by exhaustion the k+1 sets of k horses have been shown to be the same color. It follows that since every horse is the same color as every other horse, P(k) entails P(k+1). But since we have shown P(1) to be true, P is true for all succeeding values of k, that is, all horses are the same color.

Two hydrogen atoms walk into a bar. One says, ”I think I’ve lost an electron.”

The other says, ”Are you sure?” The first replies,

”Yes, I’m positive...”

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