Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

First, let me make it clear that I do not mean jokes of the "abelian grape" variety. I take my cue from the following passage in A Mathematician's Miscellany by J.E. Littlewood (Methuen 1953, p. 79):

I remembered the Euler formula $\sum n^{-s}=\prod (1-p^{-s})^{-1}$; it was introduced to us at school, as a joke (rightly enough, and in excellent taste).

Without trying to define Littlewood's concept of joke, I venture to guess that another in the same category is the formula

$1+2+3+4+\cdots=-1/12$,

which Ramanujan sent to Hardy in 1913, admitting "If I tell you this you will at once point out to me the lunatic asylum as my goal."

Moving beyond zeta function jokes, I would suggest that the empty set in ZF set theory is another joke in excellent taste. Not only does ZF take the empty set seriously, it uses it to build the whole universe of sets.

Is there an interesting concept lurking here -- a class of mathematical ideas that look like jokes to the outsider, but which turn out to be important? If so, let me know which ones appeal to you.

share|improve this question

closed as no longer relevant by Yemon Choi, Dan Petersen, Andy Putman, Bill Johnson, Mark Sapir Jan 4 '12 at 13:29

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

2  
Don't think it's exactly the same concept but Scott Aaronson has a post on something similar here: scottaaronson.com/blog/?p=392 –  Harry Altman Sep 15 '10 at 18:40
25  
In regard to the empty set being a joke, Frank Harary and Ronald Read wrote a 1974 paper entitled "Is the null graph a pointless concept?". –  Richard Stanley Sep 15 '10 at 22:29
10  
It's amazing to see how many such jokes involve geometric series. –  Thierry Zell Sep 16 '10 at 12:16
11  
I just noticed that there are two puns in "abelian grape variety," as variety can associate with abelian or with grape. Too bad this is the kind of joke you don't mean. –  Gerry Myerson Sep 26 '10 at 6:34
2  
The two most recent answers have been more-or-less duplicates of previous answers. Time to close? –  Gerry Myerson Oct 3 '10 at 12:25

59 Answers 59

There are other divergent series that fit the bill, such as $1-1+1-1+ \cdots = 1/2$. Here's one from formal language theory: Suppose we define a language $L$ recursively by the rule $L = 1 | aL$, meaning that the empty string $1$ is in $L$, and the letter $a$ followed by any element in $L$ is also in $L$. Jokingly, we note that $|$ is akin to addition and concatenation is akin to multiplication, so we can solve for $L$: $1 = L - aL = L (1-a)$, so $$L = {1\over 1-a} = 1|a|aa|aaa|aaaa \ldots,$$ which is the right answer.

The umbral calculus could also be considered an elaborate joke.

share|improve this answer
10  
Nice example. I like the fact that the very name "umbral calculus" admits its shady nature. –  John Stillwell Sep 15 '10 at 20:51
1  
Dear Mariano, I think that "umbral calculus" was coined by Sylvester. Looking on Mathworld confirms this, but (according to Mathworld) the shadows being alluded to are the combinatorial identities obtained, which "shadow" more obvious polynomial or Taylor series idenities. –  Emerton Sep 16 '10 at 0:10
3  
@Mariano: The term "umbra" (shadow) is attributed to Sylvester, and its first occurrence seems to be the following, from 1851 (in his Collected Mathematical Papers vol. 1, p.242): "Each quantity is now represented by two letters; the letters themselves, taken separately, being symbols neither of quantity nor of operation, but mere umbrae or ideal elements of quantitative symbols." –  John Stillwell Sep 16 '10 at 0:40
2  
Dear John, This is more appealing and romantic than the Mathworld explanation, and presumably more accurate, since you are quoting Sylvester directly. I wonder if the Mathworld explanation is just based on speculation rather than primary sources? –  Emerton Sep 16 '10 at 1:52

A typical proof of Kolmogorov's zero-one law has as its punchline 'therefore A is independent of A'. Perhaps not a joke in the sense of Littlewood, but amusing nonetheless.

share|improve this answer

The geometric series expansion of projective space: $\frac{\mathbf{C}^{n+1} - \mathrm{pt}}{\mathbf{C} - \mathrm{pt}} = \mathbf{C}^n + \cdots + \mathbf{C}^1 + \mathrm{pt}$

share|improve this answer
1  
@Aaron: Could you just consider them as (co?)homology classes inside a suitably large embedding space? (Say, $\mathbb{C}^{n+1}$?) –  drvitek Sep 16 '10 at 6:13
2  
Aaron, I find John Baez's perspective here math.ucr.edu/home/baez/week184.html to be quite interesting. –  Mike Skirvin Sep 16 '10 at 13:55
7  
@Aaron: This equation holds in the Grothendieck ring of varieties (sometimes called ``ring of motives of varieties'' or similar). Whenever you have a Zariski locally trivial fibration $X \to Y$ with every fiber isomorphic to $F$, then $[X] = [F] * [Y]$ in the Grothendieck ring. –  Arend Bayer Sep 17 '10 at 14:57
4  
This equation is fantastic! –  Martin Brandenburg Sep 24 '10 at 9:13
1  
@Vivek: that doesn't mean the equation doesn't have content, e.g. it is meaningful to know that in the Grothendieck ring the terms don't all collapse to zero. (At least, I assume that they don't.) –  Qiaochu Yuan Sep 28 '10 at 16:51

If $1-ab$ is invertible for $a$, $b$ in a (noncommutative) ring then so is $1-ba$.

Proof: $$(1-ba)^{-1} = 1+ba +baba+\cdots = 1+b(1+ab+abab+\cdots)a = 1+b(1-ab)^{-1}a,$$
The meaningless infinite series give the right answer (which is hard to guess).

share|improve this answer
2  
There was a "rational" explanation for this sometime ago on MO; maybe someone with better MO-searching fu than I can find it. –  Mariano Suárez-Alvarez Sep 15 '10 at 23:57
10  
4  
If you think of products of a's and b's as regular expressions, with $(1-x)^{-1}$ playing the role of $x^{\ast}$, the result is fairly obvious. This type of reasoning is, to some extent, captured by Kleene algebras. –  Dan Piponi Oct 4 '10 at 17:31

The field with one element seems a good example.

share|improve this answer
8  
e.g. $A_n = \mathbb{P}GL_n(\mathbb{F}_1)$ –  Peter Arndt Sep 16 '10 at 0:00
1  
I've met this one in a presentation where I was told that the Gauss binomial coefficient $ \genfrac{[}{]}{0pt}{}{n}{k}_q $ is the number of $ k $ dimensional subspaces in an $ n $ dimensional projective space over the field $ GF_q $. –  Zsbán Ambrus Sep 16 '10 at 14:13

The yoga of motives, in the parts which are conjectural but inspire the right guesses, is a huge joke!

share|improve this answer

The chain rule, in the form $${dy\over dx}={dy\over du}{du\over dx}$$ is a joke - you just cancel the $du$, top and bottom.

share|improve this answer
2  
This is not a joke, you CAN cancel those du's, if understood properly. Namely, if we have graph of f(x), and a tangent vector v at point (c, f(c)), then dy = dy(v) = projection of v to y-coordinate, dx = dx(v) = projection of v to x-coordinate. Their quotient dy(v)/dx(v) is equal to f'(c). –  mathreader Sep 16 '10 at 9:20
1  
mathreader: indeed you can do that, but it's still a joke in the sense this question asks. –  Zsbán Ambrus Sep 16 '10 at 13:39
3  
Well, if you have two differentials which are dependent in the sense that $dx \wedge dy = 0$ then there is of course some function such that $dy$ = $f dx$. What could $f$ be called but $dy / dx$? Since $\mathbb{R}$ is one-dimensional, the chain rule joke always works. –  Matt Noonan Sep 16 '10 at 14:28
1  
@mathreader: Be very careful. What happens if u'(x) = 0? –  Theo Johnson-Freyd Sep 16 '10 at 21:55

I know you say "moving beyond zeta function jokes", but I'd say the following two zeta-regularizations deserve to be alongside your Ramanujan example: $$\infty!= \sqrt{2\pi}\qquad\qquad\mbox{and }\qquad\qquad \prod_{\mbox{$p$ prime}}p =4\pi^2.$$ One can also entertain beginning calculus students with $\frac{1}{2}!=\frac{1}{2}\sqrt{\pi}$ as a way of introducing the Gamma function.

share|improve this answer

Another example from intro calculus: I once put a question of the form "$y=f(x)^{g(x)}$, find $y'$" on an exam. One student reasoned, if the exponent were a constant, the answer would be $g(x)f(x)^{g(x)-1}f'(x)$, but that's not right; if the base were a constant, the answer would be $g'(x)f(x)^{g(x)}\log f(x)$, but that's not right either; so I'll put them together to get $g(x)f(x)^{g(x)-1}f'(x)+g'(x)f(x)^{g(x)}\log f(x)$. This joke was on me, since that turns out to be correct.

share|improve this answer
12  
In graduate school I graded one of the questions on a multiple section exam in beginning calculus. The correct answer was 4, and many many students got that, but few by any correct route. My conclusion, as the derivative is a limit process, was that the set consisting of the single number 4 is dense in the real line. –  Will Jagy Sep 16 '10 at 0:52
25  
Of course, this is just the multivariate chain rule $\frac{d}{dx}f(u(x),v(x)) = \frac{\partial f}{\partial u}\frac{du}{dx}+\frac{\partial f}{\partial v}\frac{dv}{dx}$. Perhaps your student was motivated by the product rule $(fg)'=f'g+fg'$, which works the same way. –  Ricky Liu Sep 16 '10 at 17:33
9  
It has naturaly physical sense: if $x$ arises several times in our expression, we may consider small peturbations (from the definition of derivative) of $x$ being independent and then sum up. –  Fedor Petrov Sep 19 '10 at 20:09
3  
I noticed this myself a long time ago (for the special case y=x^x), but always thought it was a cute coincidence -- I never realized that it was the multivariate chain rule! –  Harrison Brown Oct 29 '10 at 19:02

Hausdorff dimension. Try showing a Sierpinski triangle to a non-mathematician and explaining that it is a 1.585-dimensional object.

share|improve this answer
2  
And then try the same thing with the sierpinksi tetraeder (dimension = 2) –  Johannes Hahn Sep 16 '10 at 10:43

Gerry's answer reminded me of the "Lucky Larry" series, which is a regular column in The AMATYC Review.

One particularly nice one is

$$\lim_{x\to\infty}\frac{\ln\ln x}{\ln x}$$

where you can cancel the $\ln x$ and wind up with $\ln 1=0$, which turns out correct.

share|improve this answer

The classical Stokes formula $\int_{\partial\Omega}\omega=\int_\Omega d\omega$ is certainly a Littlewood type joke. That is especially true if you learn it after you've spent a few months covering vector calculus, learned rotor, divergence, path and surface integrals of 2 kinds, etc., which is the standard route to follow.

share|improve this answer

$$\dfrac{16}{64} = \dfrac{1\not{6} }{\not{6} 4} = \dfrac{1}{4}$$

$$25^{1/2} = \not25^{1/\not2} = 5^1 = 5$$

$$\sqrt[6]{64} = \sqrt[\not 6]{\not 64} = \sqrt{4}$$

share|improve this answer
3  
For related developments in this important area, see B. Ruekberg, "Simplified Mathematics,". J. Irreproducible Results 35 (1), 1990. –  Nate Eldredge Sep 21 '10 at 18:59
1  
Don't forget that by crossing off the $9$'s we get $\frac {95}{19}=5$. –  Eric Naslund Feb 16 '11 at 21:32

Let "$\int$" denote $\int_0^x$. We want to find the solution to

$$\int f = f-1.$$

We simply "factor out" $f$, getting $1=\left(1-\int\right)f$. Thus, $f=(1-\int)^{-1}1$.

Using the geometric series,

$$f=\left(1+\int+\iint+\iiint+\cdots\right)1=1+\int_0^x1~dx+\int_0^x\int_0^x1~dx+\cdots$$

Thus,

$$f=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\cdots=e^x,$$

as expected. (This was told to me by Steve Miller)

share|improve this answer
6  
I was going to give this example! It is from a classic book by W. W. Sawyer, Prelude to mathematics. What makes this example interesting is that Sawyer described precisely the phenomenon we are discussing here: to a mathematics student, it appears to be a joke, but in fact, this is a standard technique of solving integral equations. Just to appreciate how a good a joke it is, the passage in Sawyer's book I quoted by memory I read over 20 years ago! –  Victor Protsak Sep 16 '10 at 3:07
15  
"Hmmm... And how is it obvious that evaluating ∫ without any function gives x?" Well, doesn't "Let "$\int$" denote $∫_0^x$." clearly mean: "Let "$\int f\;$" denote $∫_0^xf(x) dx$." If so, then $\int = \int _0 ^x dx = x$. –  Dick Palais Sep 16 '10 at 3:40
6  
I think it should read $1 = (1 - \int) f$. Then $f = (1 - \int)^{-1} 1$, which resolves the question about evaluating $\int$ without an input. –  Jonathan Wise Sep 16 '10 at 16:33
1  
@Victor. I read some of those W.W. Sawyer books too, and they may have planted the seed of this question in my mind. Alas, it was 50+ years ago, and I no longer remember what I read in them. –  John Stillwell Sep 16 '10 at 19:20

We owe Paul Dirac two excellent mathematical jokes. I have amended them with a few lesser known variations.

A. Square root of the Laplacian: we want $\Delta$ to be $D^2$ for some first order differential operator (for example, because it is easier to solve first order partial differential equations than second order PDEs). Writing it out,

$$\sum_{k=1}^n \frac{\partial^2}{\partial x_k^2}=\left(\sum_{i=1}^n \gamma_i \frac{\partial}{\partial x_i}\right)\left(\sum_{j=1}^n \gamma_j \frac{\partial}{\partial x_j}\right) = \sum_{i,j}\gamma_i\gamma_j \frac{\partial^2}{\partial x_i x_j}, $$

and equating the coefficients, we get that this is indeed true if

$$D=\sum_{i=1}^n \gamma_i \frac{\partial}{\partial x_i}\quad\text{and}\quad \gamma_i\gamma_j+\gamma_j\gamma_i=\delta_{ij}.$$

It remains to come up with the right $\gamma_i$'s. Dirac realized how to accomplish it with $4\times 4$ matrices when $n=4$; but a neat follow-up joke is to simply define them to be the elements $\gamma_1,\ldots,\gamma_n$ of

$$\mathbb{R}\langle\gamma_1,\ldots,\gamma_n\rangle/(\gamma_i\gamma_j+\gamma_j\gamma_i - \delta_{ij}).$$


Using symmetry considerations, it is easy to conclude that the commutator of the $n$-dimensional Laplace operator $\Delta$ and the multiplication by $r^2=x_1^2+\cdots+x_n^2$ is equal to $aE+b$, where $$E=x_1\frac{\partial}{\partial x_1}+\cdots+x_n\frac{\partial}{\partial x_n}$$ is the Euler vector field. A boring way to confirm this and to determine the coefficients $a$ and $b$ is to expand $[\Delta,r^2]$ and simplify using the commutation relations between $x$'s and $\partial$'s. A more exciting way is to act on $x_1^\lambda$, where $\lambda$ is a formal variable:

$$[\Delta,r^2]x_1^{\lambda}=((\lambda+2)(\lambda+1)+2(n-1)-\lambda(\lambda-1))x_1^{\lambda}=(4\lambda+2n)x_1^{\lambda}.$$

Since $x_1^{\lambda}$ is an eigenvector of the Euler operator $E$ with eigenvalue $\lambda$, we conclude that

$$[\Delta,r^2]=4E+2n.$$


B. Dirac delta function: if we can write

$$g(x)=\int g(y)\delta(x-y)dy$$

then instead of solving an inhomogeneous linear differential equation $Lf=g$ for each $g$, we can solve the equations $Lf=\delta(x-y)$ for each real $y$, where a linear differential operator $L$ acts on the variable $x,$ and combine the answers with different $y$ weighted by $g(y)$. Clearly, there are fewer real numbers than functions, and if $L$ has constant coefficients, using translation invariance the set of right hand sides is further reduced to just one, $\delta(x)$. In this form, the joke goes back to Laplace and Poisson.


What happens if instead of the ordinary geometric series we consider a doubly infinite one? Since

$$z(\cdots + z^{-n-1} + z^{-n} + \cdots + 1 + \cdots + z^n + \cdots)= \cdots + z^{-n} + z^{-n+1} + \cdots + z + \cdots + z^{n+1} + \cdots,$$

the expression in the parenthesis is annihilated by the multiplication by $z-1$, hence it is equal to $\delta(z-1)$. Homogenizing, we get

$$\sum_{n\in\mathbb{Z}}\left(\frac{z}{w}\right)^n=\delta(z-w)$$

This identity plays an important role in conformal field theory and the theory of vertex operator algebras.


Pushing infinite geometric series in a different direction,

$$\cdots + z^{-n-1} + z^{-n} + \cdots + 1=-\frac{z}{1-z} \quad\text{and} \quad 1 + z + \cdots + z^n + \cdots = \frac{1}{1-z},$$

which add up to $1$. This time, the sum of doubly infinite geometric series is zero! Thus the point $0\in\mathbb{Z}$ is the sum of all lattice points on the non-negative half-line and all points on the positive half-line:

$$0=[\ldots,-2,-1,0] + [0,1,2,\ldots] $$

A vast generalization is given by Brion's formula for the generating function for the lattice points in a convex lattice polytope $\Delta\subset\mathbb{R}^N$ with vertices $v\in{\mathbb{Z}}^N$ and closed inner vertex cones $C_v\subset\mathbb{R}^N$:

$$\sum_{P\in \Delta\cap{\mathbb{Z}}^N} z^P = \sum_v\left(\sum_{Q\in C_v\cap{\mathbb{Z}}^N} z^Q\right),$$

where the inner sums in the right hand side need to be interpreted as rational functions in $z_1,\ldots,z_N$.


Another great joke based on infinite series is the Eilenberg swindle, but I am too exhausted by fighting the math preview to do it justice.

share|improve this answer

This isn't a particularly interesting example, but the existence of different sizes of infinity fits your criterion of being something that makes outsiders laugh (as I know from experience) and that is also very important to mathematicians.

The familiar argument that says that if you want an explicit example of $a^b=c$ with a and b irrational and c rational, then one of $a=b=\sqrt{2}$ or $a=\sqrt{2}^{\sqrt{2}}$ and $b=\sqrt{2}$ will work is certainly an argument that makes people laugh. Though the result itself is not very important, the phenomenon it illustrates is quite important.

Added two minutes later: I've just had a look at Scott Aaronson's post and seen that Erik, one of the earlier commenters, chose precisely the same two examples. It was a coincidence -- honest.

share|improve this answer

The Eilenberg swindle was briefly mentioned by Victor Protsak, and something which always struck me as similar (and seemed like black magic to me when I first saw it) is the Pelczynski decomposition method for proving that complemented subspaces of $\ell^p$ are isomorphic to $\ell^p$. I don't have a link to hand but might try to write something out later (or someone else is welcome to edit this).

share|improve this answer

Many infinite, periodic expression involving numbers is a joke, according to non-mathematicians. Three examples: $$0,999999999\cdots=1,$$ $$1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\cdots}}}=\frac{1+\sqrt5}{2},$$ $$\sqrt{-a^2+\sqrt{-a^2+\sqrt{-a^2+\cdots}}}=\frac12(1+\sqrt{1-4a^2}).$$

share|improve this answer

In a probability oral exam, a student is asked to compute the probability that a random number chosen from the interval $[0,1]$ is larger than $2/3$. The students answers $1/3$. The teacher asks him to explain his argument, and he says: well, there are three possibilities: the number is either less than, or bigger than, or equal to $2/3$, so, the probability is $1/3$!

share|improve this answer

Tim Gowers mentioned infinities that may sound like jokes, especially to outsiders. Here is one specific example: you are standing in a room; at every tick of the clock, someone throws in a pair of numbered ping-pong balls: 1 & 2, then 3 & 4, etc... and you only have enough time to throw out one of them before the next tick. If you throw out the one with the largest number, then after $\omega$ ticks of the clock, you are in the room with all the odd-numbered balls, whereas if you always threw out the ball with the smallest number, you would be rid of them all!

And what if the balls are not numbered? A good way to get non-mathematicians thinking about infinity.

share|improve this answer
1  
A similar example is The Gnome and the Pearl of Wisdom: A Fable by Richard Willmott (animations at komal.hu/cikkek/egyeb/torpe/torpe.h.shtml). This is about a sequence of numbered boxes and numbered marbles. The boxes start out empty, then in step $ t $, the gnome puts the stone number $ t $ to box $ 0 $, then resolves the conflict of two stones being in the same box by repeatedly moving the stone with the higher number (in the first variation; lower number in the second variation) to the next box. The question is where are the stones after $\omega$ steps. –  Zsbán Ambrus Sep 16 '10 at 13:53

The fundamental axioms of mathematics are inconsistent if and only if we can prove that they are consistent.

(Because, you know, it follows from "logic." See Second Incompleteness theorem)

share|improve this answer
12  
As Torkel Franzen pointed out in his wonderful book Godel's Theorem: An Incomplete Guide to Its Use and Abuse, if you harbored serious doubts about the consistency of your axioms, why would you be seek a consistency proof in that same setting? –  Thierry Zell Sep 16 '10 at 16:33
13  
But would a naive person be upset to find a proof of consistency in a supposedly rock-solid system? Probably not---but the joke here is that, nevertheless, they should be upset, as it reveals inconsistency. –  Joel David Hamkins Sep 16 '10 at 21:55

Lagrange's equation:

$\frac{d}{dt}( \frac{\partial}{\partial \dot{q}} \mathcal{L} )=\frac{\partial}{\partial q} \mathcal{L}$,

as $\dot{q}=\frac{dq}{dt}$, you can simplify "$dt$".

share|improve this answer

In characteristic $p$, the so-called biologists' rule

$$(a+b)^p = a^p + b^p$$ (which got its name by mathematics students that worked as teaching assistants for "mathematics for biologists") is correct.

share|improve this answer
29  
I've heard this called the "freshman's dream". –  Michael Lugo Sep 16 '10 at 15:31
3  
Yes, in fact Thomas Hungerford's book Algebra calls it that way in Exercise 11 in page 121. And according to it, the terminology is due to V.O. McBrien. –  Adrián Barquero Sep 19 '10 at 15:41

A good joke about infinity is the following. A hotel has rooms $1,2,\dots$. Every room is full when a new guest arrives. The clerk moves the occupant of room $n$ to $n+1$ to make room for the new guest in room 1. An hour later another guest arrives and the clerk repeats the process. 30 minutes later a third guest arrives and the process is repeated. Then 15 minutes, 7.5 minutes, etc., until two hours after the first new guest infinitely many guests have arrived and been accommodated. The clerk is very pleased with himself for dealing with these infinitely many guests, when he notices to his horror that all the rooms are empty! All the guests have mysteriously disappeared!

share|improve this answer
12  
Is this Achilbert and the Tortoise, or what?... –  darij grinberg Sep 16 '10 at 16:37

This was once presented to me as a kind of proof, though I think it works better as a kind of joke:

To compute ${\partial^n\over\partial x^n}(fg)$, split ${\partial\over\partial x}$ into the sum of a piece $D$ that just acts on $f$ (acting as the identitiy on $g$) and a piece $E$ that just acts on $g$ (acting as the identity on $f$) and write ${\partial^n\over\partial x^n}(fg) = (D+E)^n(fg) = \sum_{i=0}^n \binom{n}{i} D^i E^{n-i}(fg) = \sum_{i=0}^n \binom{n}{i} {\partial^i f\over\partial x^i}{\partial^{n-i} g\over\partial x^{n-i}}$.

share|improve this answer

Let $C(x) = \sum_{n \ge 0} \frac{1}{n+1} {2n \choose n} x^n$ be the generating function for the Catalan numbers. Then $C(x) = \frac{1 - \sqrt{1 - 4x}}{2}$. In particular, $C(1) = \frac{1 - \sqrt{-3}}{2}$ is a sixth root of unity - except of course that $C(1)$ does not converge.

Nevertheless, there is a remarkable sense in which $C^7 = C$; see Blass' Seven Trees in One and the generalization in Fiore and Leinster's Objects of Categories as Complex Numbers. As with many results of this type, I learned about this from John Baez.

share|improve this answer
3  
I think the second author ought to be "Leinster," though I would love to take credit since I like that paper very much. :) –  Eric Finster Sep 24 '10 at 16:00

An expansion on Timothy Chow's example of Grandi's series $1 - 1 + 1 - 1 \pm ... = \frac{1}{2}$. It is possible to interpret the left hand side as computing the Euler characteristic of infinite real projective space $\mathbb{R}P^{\infty}$, which is a $K(\mathbb{Z}/2\mathbb{Z}, 1)$ and therefore rightfully has orbifold Euler characteristic $\frac{1}{2}$! I think I learned this example from somewhere on Wikipedia.

share|improve this answer

The Cayley-Hamilton Theorem:

If $A$ is a square matrix with characteristic polynomial $p(\lambda) = \det(A-\lambda I)$, then $p(A) = 0$.

Because you know, you "just plug in."

share|improve this answer
25  
There are some funny proofs of this, too. Here's one that works over any field: First notice that the theorem holds for diagonalizable matrices. Then, adjoin $n^2$ indeterminants to our field and take the algebraic closure. But the $n \times n$ matrix whose entries are those indeterminants is now diagonalizable! Thus, we've proved the Cayley-Hamilton theorem as a polynomial identity over our original field. –  Gene S. Kopp Sep 25 '10 at 17:19
5  
That's the first proof of Cayley-Hamilton that I've actually liked.... –  Brian Lawrence Aug 1 '11 at 9:09

The chain rule "joke" reminded me of a similar notation joke: Radon-Nikodym derivatives.

If $\mu$, $\nu$, $\lambda$ are $\sigma$-finite measures with $\nu \ll \mu \ll \lambda$, and $f \geq 0$ is measurable, then:

$$\int f\ d\nu = \int f \left[\frac{d\nu}{d\mu}\right]\ d\mu$$

and

$$\left[\frac{d\nu}{d\lambda}\right] = \left[\frac{d\nu}{d\mu}\right]\left[\frac{d\mu}{d\lambda}\right]$$

share|improve this answer

Nobody mentioned the third isomorphism theorem yet? If $B$ and $C$ are normal subgroups of $A$ and $C \le B$ then $\frac{A/C}{B/C} \cong \frac{A}{B}$.

share|improve this answer
39  
+1 for the user name. –  Thierry Zell Sep 17 '10 at 13:11
2  
I don't get this one. The proof for integers is very close to the proof for groups (if you are thinking about integers as cardinalities of sets) –  Steven Gubkin Sep 11 '12 at 20:45

Not the answer you're looking for? Browse other questions tagged or ask your own question.