HoUM

## Náhodný vtip

Malicka Sara sa pyta otca Izaka: Ocko, dovoli Hospodin poslat Valentinku cloveku, ktory je ineho vierovyznania nez my? A komu by si chcela poslat Valentinku? - pyta sa Izak. Usamovi bin Ladinovi. Bin Ladinovi? A preco? - pyta sa sokovany otec. No - vravi malicka Sara - ked Bin Ladin dostane Valentinku s vyznanim lasky od malickeho zidovskeho dievcatka, pomysli si, ze nie je vsetko vo svete take zle. Zacne mat trosku rad svet. A potom neskor, ked dostane este niekolko Valentinok, pochopi, ze svet je prekrasny, prestane sa skryvat a verejne prizna svoje chyby... Sarocka - vravi dojaty Izak - to su najkrajsie slova, ktore som kedy pocul... Ja viem, ocko. - odvetila malicka Sara - Navyse, ked sa ten spinavec prestane skryvat po jaskyniach, budu ho moct nasi chlapci z Mosadu konecne zastrelit.

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# Stats/Math Jokes

Jokes found: 8543

|Theorem : All numbers are equal to zero.Proof: Suppose that a=b. Thena = ba^2 = aba^2 - b^2 = ab - b^2(a + b)(a - b) = b(a - b)a + b = ba = 0Furthermore if a + b = b, and a = b, then b + b = b, and 2b = b, which mean that 2 = 1.

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|Theorem: 3=4Proof:Suppose:a + b = cThis can also be written as:4a - 3a + 4b - 3b = 4c - 3cAfter reorganizing:4a + 4b - 4c = 3a + 3b - 3cTake the constants out of the brackets:4 * (a+b-c) = 3 * (a+b-c)Remove the same term left and right:4 = 3

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|Theorem: All positive integers are equal.Proof: Sufficient to show that for any two positive integers, A and B, A = B.Further, it is sufficient to show that for all N > 0, if A and B (positive integers) satisfy (MAX(A, B) = N) then A = B.Proceed by induction.If N = 1, then A and B, being positive integers, must both be 1. So A = B.Assume that the theorem is true for some value k. Take A and B with MAX(A, B) = k+1. Then MAX((A-1), (B-1)) = k. And hence (A-1) = (B-1). Consequently, A = B.

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|Theorem: 1\$ = 1c.Proof:And another that gives you a sense of money disappearing.1\$ = 100c= (10c)^2= (0.1\$)^2= 0.01\$= 1cHere \$ means dollars and c means cents. This one is scary in that I have seen PhD's in math who were unable to see what was wrong with this one. Actually I am crossposting this to sci.physics because I think that the latter makes a very nice introduction to the importance of keeping track of your dimensions.

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|Theorem: 1\$ = 10 centProof:We know that \$1 = 100 centsDivide both sides by 100\$ 1/100 = 100/100 cents=> \$ 1/100 = 1 centTake square root both side=> squr(\$1/100) = squr (1 cent)=> \$ 1/10 = 1 cent Multiply both side by 10=> \$1 = 10 cent

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|Theorem: n=n+1Proof:(n+1)^2 = n^2 + 2*n + 1Bring 2n+1 to the left:(n+1)^2 - (2n+1) = n^2Substract n(2n+1) from both sides and factoring, we have:(n+1)^2 - (n+1)(2n+1) = n^2 - n(2n+1)Adding 1/4(2n+1)^2 to both sides yields:(n+1)^2 - (n+1)(2n+1) + 1/4(2n+1)^2 = n^2 - n(2n+1) + 1/4(2n+1)^2This may be written:[ (n+1) - 1/2(2n+1) ]^2 = [ n - 1/2(2n+1) ]^2Taking the square roots of both sides:(n+1) - 1/2(2n+1) = n - 1/2(2n+1)Add 1/2(2n+1) to both sides:n+1 = n

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|Theorem: e=1Proof:2*e = f2^(2*pi*i)e^(2*pi*i) = f^(2*pi*i)e^(2*pi*i) = 1Therefore:2^(2*pi*i) = f^(2*pi*i)2=fThus:e=1

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|Theorem: 1 = 1/2:Proof:We can re-write the infinite series 1/(1*3) + 1/(3*5) + 1/(5*7) + 1/(7*9)+...as 1/2((1/1 - 1/3) + (1/3 - 1/5) + (1/5 - 1/7) + (1/7 - 1/9) + ... ).All terms after 1/1 cancel, so that the sum is 1/2.We can also re-write the series as (1/1 - 2/3) + (2/3 - 3/5) + (3/5 - 4/7)+ (4/7 - 5/9) + ...All terms after 1/1 cancel, so that the sum is 1.Thus 1/2 = 1.

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|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.

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|Theorem: log(-1) = 0Proof:a. log[(-1)^2] = 2 * log(-1)On the other hand:b. log[(-1)^2] = log(1) = 0Combining a) and b) gives:2* log(-1) = 0Divide both sides by 2:log(-1) = 0

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|Theorem: 4 = 5Proof:-20 = -2016 - 36 = 25 - 454^2 - 9*4 = 5^2 - 9*54^2 - 9*4 + 81/4 = 5^2 - 9*5 + 81/4(4 - 9/2)^2 = (5 - 9/2)^24 - 9/2 = 5 - 9/24 = 5

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|Theorem: All numbers are equal.Proof: Choose arbitrary a and b, and let t = a + b. Thena + b = t(a + b)(a - b) = t(a - b)a^2 - b^2 = ta - tba^2 - ta = b^2 - tba^2 - ta + (t^2)/4 = b^2 - tb + (t^2)/4(a - t/2)^2 = (b - t/2)^2a - t/2 = b - t/2a = bSo all numbers are the same, and math is pointless.

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|Theorem: 1 = -1Proof:1 = sqrt(1) = sqrt(-1 * -1) = sqrt(-1) * sqrt(-1) = 1^ = -1Also one can disprove the axiom that things equal to the same thing are equal to each other.1 = sqrt(1)-1 = sqrt(1)Therefore 1 = -1As an alternative method for solving:Theorem: 1 = -1Proof:x=1x^2=xx^2-1=x-1(x+1)(x-1)=(x-1)(x+1)=(x-1)/(x-1)x+1=1x=00=1=> 0/0=1/1=1

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|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.

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|Theorem: 1 + 1 = 2Proof:n(2n - 2) = n(2n - 2)n(2n - 2) - n(2n - 2) = 0(n - n)(2n - 2) = 02n(n - n) - 2(n - n) = 02n - 2 = 02n = 2n + n = 2or setting n = 11 + 1 = 2

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|An engineer, a physicist, and a mathematician are trying to set up a fenced-in area for some sheep, but they have a limited amount of building material. The engineer gets up first and makes a square fence with the material, reasoning that it's a pretty good working solution. ''No no,'' says the physicist, ''there's a better way.'' He takes the fence and makes a circular pen, showing how it encompasses the maximum possible space with the given material.Then the mathematician speaks up: ''No, no, there's an even better way.'' To the others' amusement he proceeds to construct a little tiny fence around himself, then declares:''I define myself to be on the outside.''

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|It is proven that the celebration of birthdays is healthy. Statistics show that those people who celebrate the most birthdays become the oldest. -- S. den Hartog, Ph D. Thesis Universtity of Groningen.

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|Hello, this is probably 438-9012, yes, the house of the famous statistician. I'm probably not at home, or not wanting to answer the phone, most probably the latter, according to my latest calculations. Supposing that the universe doesn't end in the next 30 seconds, the odds of which I'm still trying to calculate, you can leave your name, phone number, and message, and I'll probably phone you back. So far the probability of that is about 0.645. Have a nice day.

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