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 $\epsilon>0$. 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-Tarsky 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.
