ProofWiki:Jokes

0.999...=1

 * Q: How many mathematicians does it take to change a lightbulb?
 * A: 0.999999 ...

Banach-Tarski Paradox

 * Q: Give me an anagram of Banach-Tarski.
 * A: Banach-Tarski Banach-Tarski.

Educational Standards
Two captains of industry, Arthur and George, were in a restaurant discussing the state of educational standards, particularly in the field of mathematics. Arthur was convinced they were slipping badly, and that your average college student was completely mathematically illiterate. George, on the other hand, was confident that any student would at least know the basics of calculus.

"I bet you a hundred bucks," said Arthur, "that if you were to ask a random college student a basic question in calculus, he wouldn't understand the question, let alone furnish you with an answer."

"I'll think about that," said George. "Not sure whether to take you up on your bet or not, but I reckon you'd be wrong."

Arthur slipped off to the mens' room at that point, and while he was gone, George called over the waitress Jody. (He knew that was her name because it was written on a badge pinned to her uniform. This appears to be a custom in certain chain diners.)

"I'd like you to help settle a wager between me and my colleague," he said. "When he comes back, I'm going to call you over, and ask you a question, to which you are to answer: one third x cubed."

"Wuntur dex cue?"

"One third x cubed."

"One thurrd ex cuebd."

"That's it, one third x cubed."

"One third ... x cubed."

"That's it, perfect. There's a good tip in it for you."

Arthur returned. George said, "Yes, I think I will take you up on it. A hundred bucks says our waitress can answer such a question. Hey, Jody! What's the indefinite integral of x squared with respect to x?"

"One third x cubed," replied Jody, dutifully.

"You see?" said George, pocketing Arthur's hundred.

As Jody turned away, she called back over her shoulder, "Plus a constant."

George ruefully took Arthur's hundred back out of his pocket and dropped it onto the table.

Knot Theory

 * Student A: "What's your favourite area of mathematics?"
 * Student B: "Knot theory."
 * Student A: "Me neither."


 * : Knot Jokes and Pastimes (attributed to Martin Scharlemann)

Natural Numbers

 * Q: "Why do computer scientists have nine fingers?"
 * A: "Zero, one, two, three, four; five, six, seven, eight, nine."

Circle Geometry
The roundest knight at King Arthur's round table was Sir Cumference.

He acquired his shape from too much pi.

Sufficiently Large
$1+1 = 3$, for sufficiently large values of $1$.

Why? Because $1.4 + 1.4 = 2.8$.

The result follows after rounding to the nearest integer.

Number Bases
Why do mathematicians get Halloween and Christmas confused?

Because $\mathsf{Dec} \ 25$ equals $\mathsf{Oct} \ 31$.


 * : Halloween $=$ Christmas (but it's a considerably older joke than that.)

Binary
There are 10 sorts of people in the world: those who understand binary and those who don't.

1
Only dead people and me understand hexadecimal. So, how many people understand hexadecimal?

deae people.

(And me too. That's deaf people.)

10
There are 10 sorts of people in the world: those who understand hexadecimal, and F the rest.

... and just plain innumeracy
There are three sorts of people in the world: those who can count, and those who can't.

Computer Prayer

 * Our Program which art in Memory,
 * "Hello World!" be Thy Name.
 * Thy Operating System come, Thy Commands be done,
 * at the Printer as it is on the Screen.
 * Give us this day our Data Dump,
 * and forgive us our I/O Errors
 * as we backup those whose Files are faulty.
 * Lead us not into frustration, and deliver us from email
 * for Thine is the Algorithm, the Application, and the Solution,
 * looping forever and ever.


 * Return.

Average Number of Hands
Most people in the world have more than the average number of hands.

Six
The proof that $\mathrm u \left({x \, \mathrm u! \, s}\right)^{-1}= 9$ is left as an exercise for the reader.

Computer Encoding
Link to a ROT26 encoder:

http://www.rot26.org/

All Odd Numbers Are Prime
Proof by inductive argument:


 * $1$, that's prime (well not technically, but that's just mathematical double-talk).
 * $3$, that's prime.
 * $5$, that's prime.
 * $7$, that's prime.
 * $9$, that's prime (although when I measured it, it looked like it might not be - experimental error, ignore that one)
 * $11$, that's prime.
 * $13$, that's prime.

We can extrapolate from there.

Black Friday customer proof:


 * $1$, that's prime, it's obvious it is, no argument there. I said, no argument there.
 * $3$, that's prime.
 * $5$, that's prime.
 * $7$, that's prime.
 * $9$, that's prime YES IT IS -- DON'T ARGUE! YOU ARE STUPID!
 * $11$, that's prime.
 * $13$, that's prime.

Any more slackwit stupid people out there want to argue wit' me?

Logarithms
Hear about the constipated mathematician?

He worked out logs with a pencil.

Axiom of Intuition
The Axiom of Choice is obviously true; the Well Ordering Principle is obviously false; and who can tell about Zorn's Lemma?
 * -- Jerry Bona

Lightbulbs

 * Q: How many Bourbakists does it take to change a lightbulb?


 * A: Changing a lightbulb is a special case of a more general theorem concerning the maintenance and repair of an electrical system. To establish upper and lower bounds for the number of personnel required, we must determine whether the sufficient conditions of Lemma $2.1$ (Availability of personnel) and those of Corollary $2.3.55$ (Motivation of personnel) apply. If these conditions are met, we derive the result by an application of the theorems in Section $3.11.23$. The resulting upper bound is, clearly, a result in an abstract measure space, in the weak-$*$ topology.


 * Q: How many lightbulbs does it take to change a lightbulb?
 * A: One, if it knows its own Gödel number.



Principia Mathematica

 * Lyrics: Colin Fine
 * Music: Burt Bacharach


 * What do you get if you take a set
 * Add an associative operation
 * Give it an identity, make everything invertible?
 * A-a-ah, a-a-a-a-a-a-ah, it's a group
 * A-a-a-a-ah, it's a group.


 * What do you get if you take a group
 * Add an associative operation
 * Make it distributive over the first one?
 * A-a-ah, a-a-a-a-a-a-ah, it's a ring
 * A-a-a-a-ah, it's a ring.


 * Don't tell me it's too hard for you
 * Cos I can prove it so I know it's true
 * Out of a handful of simple axioms
 * I can build mathematics to satisfaction


 * What do you get if you take a ring
 * Make the multiply commutative
 * Give a reciprocal for each non-zero element?
 * A-a-ah, a-a-a-a-a-a-ah, it's a field
 * A-a-a-a-ah, it's a field.
 * (And not a skew field either!)


 * What do you get if you take a field
 * And you take a group with a scalar multiply
 * Make it associative with the field multiply
 * A-a-ah, a-a-a-a-a-a-ah, it's a vector space
 * A-a-a-a-ah, it's a vector space.


 * You can go on like this all day
 * Building structures in this kind of way
 * You end up feeling pretty cocksure
 * When you get into categories with meta-structure!


 * What do you get with some proofs and rules
 * Some axioms to give you a formal system?
 * You try to prove that it's consistent,
 * A-a-ah, a-a-a-a-a-a-ah ...??
 * Godel's Theorem!

In prime mover's defence, he remembers meeting Colin Fine at one or two SF conventions in the 1980's. He recently found this poem, a copy of which he got hold of in approximately 1987, lurking near the bottom of a pile of old magazines.

Medical Conditions

 * Ring epimorphism? I had one of them once. The doctor had to give me suppositories.
 * -- Mrs. Prime.mover.

Imaginary numbers
After eating too much food, the mathematician announced: "$\sqrt {\left({-1 / 64}\right)}$."
 * -- Kit Yates, via Twitter

A constant argument

 * $i$: "Be rational."
 * $\pi$: "Get real!"

Beerlogical
Three logicians walk into a pub.

The barmaid asks: "Are you all having beer?"

The first logician replies: "I don't know."

The second logician replies: "I don't know."

The third logician replies: "Yes."
 * -- Peter Rowlett, via Kit Yates, via Twitter

More beer
$\aleph_0$ mathematicians walk into a bar. The first one orders a pint of beer. The second one orders half a pint of beer. The third orders a quarter of a pint of beer. The fourth one orders an eighth of a pint of beer. After hearing the seventh order, the barman pours two pints, and says, "You guys should know your limits."

Surreal variant
$\aleph_0$ mathematicians walk into a bar. The first orders one beer. The second one orders two beers. The third one orders three beers. The bartender stops them, says: "You guys are idiots," and pours out $-\dfrac 1 {12}$ of a beer.

Culinary
What goes: $3.1415 \text{baa}$?

Shepherd's pi.

What is the volume of a pizza whose radius is $z$ and whose thickness is $a$?

From Volume of Cylinder: $\text{pi} z z a$.

Proof by Contradiction
A mad scientist captures a mathematician and locks him in a room full of cans of food - but no can opener.

Checking on the cell several weeks later, the mathematician is lying dead, but he has written one last message on the dust on the floor:

Theorem
If I cannot open these cans of food, I will die.

Proof
Suppose not.

The Mathematicians' Party
Click on the link in the title of this section to go to the party yourself.

Optimists versus Pessimists
Consider the real interval:
 * $\mathbb I := \left[{a \,.\,.\, b}\right)$

An optimist regards $\mathbb I$ as half-open.

A pessimist regards $\mathbb I$ as half-closed.

Author Prank
I have started covering a book whose authors state:


 * "The verification that $A$ (a Ring of Idempotents, LF) becomes a Boolean ring in this way is an amusing exercise in ring axiomatics.", p.5

It's indeed most enjoyable to watch people write out $\left({x \oplus y}\right) \circ \left({x \oplus y}\right) = x \oplus y$...

Möbius Strip
I had a fight with a Möbius strip (or Möbius band for those of a prurient mentality).

I'm ashamed to say I lost. It was completely one-sided.

Amusing names
Is there an Emma Lehmer Lemma?

Psychiatry
Let $f = \displaystyle \sum_{i \mathop = 0}^n a_i x^i$ be a polynomial.

If the polynomial coefficient $a_n$ of $f$ is $1$, then $f$ is monic

If the polynomial coefficient $a_{n-1}$ of $f$ is $0$, then $f$ is depressed.

So, a polynomial of the form:
 * $f = x^n + a_{n-2} x^{n-2} + \cdots + a_1 x + a_0$

can be called monic depressive?

Computer Programming
There are only two really difficult things in computer programming: cache invalidation, naming things, and off-by-one errors.

Quantum Mechanics
Heisenberg and Schrödinger in a car speeding down the freeway. Predictably, they are stopped by a traffic policeman.

"Do you know how fast you were going?" asked the cop.

"No, but I know exactly where I was, replied Heisenberg.

Deciding to give the car an inspection, the cop opens the trunk.

"Did you know you've got a dead cat in here?" he asks.

"Well, I do now!" replied Schrödinger.

Shopological
A woman sends her logician husband to the shops. "Get me a loaf of bread," she said, "and if they have eggs, get me a dozen."

The husband returns from the shop with twelve loaves of bread.

Yes or no?
A woman has a baby, and the midwife immediately hands it to her logician husband.

"Well?" says the woman, "is it a boy or a girl?"

"Yes," he replies.

The Evils of Drink
Booze and calculus don't mix.

Don't drink and derive.

Linguistics
A visiting Professor of Linguistics was delivering a lecture.

"In the grammars of many languages throughout the world, a double negative expresses a positive. On the other hand there are some languages, such as Russian and the English of Chaucer, in which a double negative remains a negative.  However, there is not a language in the world in which a double positive can express a negative."

A voice from the back of the room piped up: "Yeah, right."


 * -- Taken from the Facebook page of George Takei on 9th March 2014, but has been around for considerably longer than that.

Rhetorical Questions

 * Q: What do you get when you cross a joke with a rhetorical question?


 * A: Quite.

Diophantus Updated
Amanda is $21$ years older than her son John.

In $6$ years from now, Amanda will be $5$ times as old as John.


 * Question: Where is John's father?


 * Solution :

Let $M$ be the age in years of Amanda now.

Let $C$ be the age in years of John now.

Then:

So the child is $-\dfrac 3 4$ years old, that is, $-9$ months.

That is, the child, will be born in $9$ months time.

So, right now, John's father is inside Amanda.

Fractions

 * Q: What goes: two bloody thirds?
 * A: A vulgar fraction.