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

I am looking for jokes which involve some serious mathematics. Sometimes, a totally absurd argument is surprisingly convincing and this makes you laugh. I am looking for jokes which make you laugh and think at the same time.

I know that a similar question was closed almost a year ago, but this went too much in the direction "$e^x$ was walking down the street ...". There is also the community wiki Jokes in the sense of Littlewood, but that is more about notational curiosities. In order to motivate you, let me give an example:

The real numbers are countable. Indeed, let $r_1,r_2,r_3,\dots$ be a list of real numbers and suppose that there is a real number missing. Just add it to the list.

If moderators or audience decide to close this question as off-topic or duplicate, I can fully understand. I just thought it could be interesting and entertaining to have this question open for at least some time.

share|improve this question
3  
The question has been closed. Unlike when the first "math jokes" question was asked, there is now a general-purpose SE Q&A math site: math.stackexchange.com. It seems like this question would be more appropriate there. –  Pete L. Clark Nov 21 '10 at 18:53
1  
I think it's fine. "Advanced math jokes" might be an interesting way to present a new concept, provided they aren't simple play on words and whatever. I see it as similar to those questions like "what mistakes do you/your students make that seem correct?" –  Quadrescence Nov 21 '10 at 18:58
add comment

closed as off topic by André Henriques, Willie Wong, Ryan Budney, Andres Caicedo, Pete L. Clark Nov 21 '10 at 18:51

Questions on MathOverflow are expected to relate to research level mathematics within the scope defined by the community. Consider editing the question or leaving comments for improvement if you believe the question can be reworded to fit within the scope. Read more about reopening questions here.If this question can be reworded to fit the rules in the help center, please edit the question.

3 Answers

The first time I ran into the carry operation from grade school addition presented as a non-trivial group cocycle generating part of the group cohomology of $\mathbb Z/10$, it was introduced as a joke embedded completely within mathematics.

Specifically, for those who haven't seen this yet, the carry operation $c(n,m)$ is defined as $c(n,m) = 0$ if $n+m < 10$ and $c(n,m) = 1$ for $n+m ≥ 10$. You can verify the cocycle condition reasonably easily, and then it remains to check there is no endomap $g:\mathbb Z/10\to\mathbb Z/10$ with $c$ as its coboundary.

share|improve this answer
    
More details on that here: math.berkeley.edu/~ericp/latex/pme-talk.pdf –  Quadrescence Nov 21 '10 at 18:59
add comment

Cosgrove's writings in the Mathematical Intelligencer about 20 or 30(?) years ago had lots of puns, many of which would be understood only by mathematicians. E.g. someone was even worse than an unprincipled infiltrator: he was a non-principal ultrafilter. The biographies of Victoria Cross (famous for Cross products and Cross-ratios, and also particular kinds of word puzzles and a certain style of country running), Montmorency Royce Sebastian Carlow (whose "methods" you've heard of), and Karl-Heinz Normal (Normal subgroups, the Normal distribution,....) were of that sort.

share|improve this answer
add comment

This is from my blog, which I interestingly just posted today (at the time of this posting).

Several mathematicians are asked, "how do you put an elephant in a refrigerator?"

Real Analyst: Let $\epsilon\gt0$. Then for all such $\epsilon$, there exists a $\delta\gt0$ such that $$\left|\frac{\mathit{elephant}}{2^n}\right|\lt\epsilon$$ for all $n\gt\delta$. Therefore $$\lim_{n\to\infty} \frac{\mathit{elephant}}{2^n}=0.$$ Since $1/2^n \lt 1/n^2$ for $n\ge 5$, by comparison, we know that $$\sum_{n\ge 1}\frac{\mathit{elephant}}{2^n}$$ converges — in fact, identically to $\mathit{elephant}$. As such, cut the elephant in half, put it in the fridge, and repeat.

Differential Geometer: Differentiate it and put into the refrigerator. Then integrate it in the refrigerator.

Set Theoretic Geometer: Apply the Banach–Tarski theorem to form a refrigerator with more volume.

Measure Theorist: Let $E$ be the subset of $\mathbb{R}^3$ assumed by the elephant and $\Phi\in\mathbb{R}^3$ be that by the fridge. First, construct a partition $e_1,\ldots,e_i$ on $E$ for $1\le i \le N$. Since $\mu(E)=\mu(\Phi)$, and $$\mu(E)=\mu\left(\bigcup_{1\le i \le N}e_i\right)=\sum_{1\le i \le N}\mu(e_i),$$ we can just embed each partition of $E$ in $\Phi$ with no problem.

Number Theorist: You can always squeeze a bit more in. So if, for $i\ge 0$. you can fit $x_i$ in, then you can fit $x_i + x_{i-1}$ in. You can fit in a bit of the elephant $x_n$ for fixed $n$, so just use induction on $i$.

Algebraist: Show that parts of it can be put into the refrigerator. Then show that the refrigerator is closed under addition.

Topologist: The elephant is compact, so it can be put into a finite collection of refrigerators. That’s usually good enough.

Linear Algebraist: Let $F$ mean "put inside fridge". Since $F$ is linear — $F(x+y)=F(x)+F(y)$ — just put 10% of the elephant in, showing that $F\left(\frac{1}{10}\mathit{elephant}\right)$ exists. Then, by linearity, $F(\mathit{elephant})$ does too.

Affine Geometer: There exists an affine transformation $F:\mathbb{R}^3\to\mathbb{R}^3:\vec{p}\mapsto A\vec{p}+\vec{q}$ that will allow the elephant to be put into the refrigerator. Just make sure $\det A\neq 0$ so you can take the elephant back out, and $\det A \gt 0$ so you don't end up with a bloody mess.

Geometer: Create an axiomatic system in which "an elephant can be placed in a refrigerator" is an axiom.

Complex Analyst: Put the refrigerator at the origin and the elephant outside the unit circle. Then get the image under inversion.

Fourier Analyst: Will $\mathcal{F}^{-1}[\mathcal{F}(\mathit{elephant})\cdot\mathcal{F}(\mathit{fridge})]$ do?

Numerical Analyst: Eh, $\mathit{elephant}=\mathit{trunk}+\varepsilon$, and $$\mathrm{fridge}(\mathit{elephant})=\mathrm{fridge}(\mathit{trunk}+\varepsilon)=\mathrm{fridge}(\mathit{trunk})+O(\varepsilon),$$ so just put the trunk in for a good approximation.

Probabilist: Keep trying to push it in in random ways and eventually it will fit.

Combinatorist: Discretize the elephant, partition it, and find a suitable rearrangement.

Statistician: Put its tail in the refrigerator as a sample, and say, "done!"

Logician: I know it's possible, I just can't do it.

Category Theorist: Isn't this just a special case of Yoneda's lemma?

Theoretical Computer Scientist: I can't decide.

Experimental Mathematician: I think it'd be much more interesting to get the refrigerator inside the elephant.

Set Theorist: Force it.

share|improve this answer
2  
A variation on an old joke. See for example math.ucdenver.edu/~wcherowi/mathmajor/archive/catchlion.pdf I think the joke was reprised in Mathematics Made Difficult. –  Todd Trimble Sep 13 '13 at 1:56
    
@ToddTrimble Nice find! –  Quadrescence Sep 15 '13 at 2:56
    
+1 for "everything is just a special case of Yoneda's lemma". –  Yuri Sulyma Sep 24 '13 at 4:14
add comment

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