A Joke

I don’t know the source of this joke. Modified somewhat a lot by me.

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

Real Analyst: Let $\epsilon>0$. Then for all such $\epsilon$, there exists a $\delta>0$ such that \[\left|\frac{\mathit{elephant}}{2^n}\right|<\epsilon\] for all $n>\delta$. Therefore \[\lim_{n\to\infty} \frac{\mathit{elephant}}{2^n}=0.\] Since $1/2^n < 1/n^2$ 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 a bit of the elephant $x_n$ for fixed $n$ in, 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 $\det A > 0$ 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.

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