Advanced Math Jokes - MathOverflow [closed] most recent 30 from http://mathoverflow.net 2013-05-21T22:46:58Z http://mathoverflow.net/feeds/question/46837 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/46837/advanced-math-jokes Advanced Math Jokes Andreas Thom 2010-11-21T17:49:55Z 2010-11-21T18:54:33Z <p>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. </p> <p>I know that a similar <a href="http://mathoverflow.net/questions/1083/do-good-math-jokes-exist-closed" rel="nofollow">question</a> 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 <a href="http://mathoverflow.net/questions/38856/jokes-in-the-sense-of-littlewood-examples" rel="nofollow">Jokes in the sense of Littlewood</a>, but that is more about notational curiosities. In order to motivate you, let me give an example:</p> <blockquote> <p>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.</p> </blockquote> <p>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.</p> http://mathoverflow.net/questions/46837/advanced-math-jokes/46841#46841 Answer by Mikael Vejdemo-Johansson for Advanced Math Jokes Mikael Vejdemo-Johansson 2010-11-21T18:24:55Z 2010-11-21T18:24:55Z <p>The first time I ran into the <em>carry</em> 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.</p> <p>Specifically, for those who haven't seen this yet, the carry operation $c(n,m)$ is defined as $c(n,m) = 0$ if $n+m &lt; 10$ 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.</p> http://mathoverflow.net/questions/46837/advanced-math-jokes/46845#46845 Answer by Michael Hardy for Advanced Math Jokes Michael Hardy 2010-11-21T18:50:24Z 2010-11-21T18:50:24Z <p>Cosgrove's writings in the <em>Mathematical Intelligencer</em> 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.</p> http://mathoverflow.net/questions/46837/advanced-math-jokes/46847#46847 Answer by Quadrescence for Advanced Math Jokes Quadrescence 2010-11-21T18:54:33Z 2010-11-21T18:54:33Z <p><em>This is from <a href="http://symbo1ics.com/blog/?p=389%20%22my%20blog%22" rel="nofollow">my blog</a>, which I interestingly just posted today (at the time of this posting).</em></p> <p>Several mathematicians are asked, "how do you put an elephant in a refrigerator?"</p> <p><strong>Real Analyst</strong>: Let $\epsilon&gt;0$. Then for all such $\epsilon$, there exists a $\delta&gt;0$ 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.</p> <p><strong>Differential Geometer</strong>: Differentiate it and put into the refrigerator. Then integrate it in the refrigerator.</p> <p><strong>Set Theoretic Geometer</strong>: Apply the Banach-Tarsky theorem to form a refrigerator with more volume.</p> <p><strong>Measure Theorist</strong>: 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.</p> <p><strong>Number Theorist</strong>: 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 $i$.</p> <p><strong>Algebraist</strong>: Show that parts of it can be put into the refrigerator. Then show that the refrigerator is closed under addition.</p> <p><strong>Topologist</strong>: The elephant is compact, so it can be put into a finite collection of refrigerators. That’s usually good enough.</p> <p><strong>Linear Algebraist</strong>: 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.</p> <p><strong>Affine Geometer</strong>: 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.</p> <p><strong>Geometer</strong>: Create an axiomatic system in which "an elephant can be placed in a refrigerator" is an axiom.</p> <p><strong>Complex Analyst</strong>: Put the refrigerator at the origin and the elephant outside the unit circle. Then get the image under inversion.</p> <p><strong>Fourier Analyst</strong>: Will $\mathcal{F}^{-1}[\mathcal{F}(\mathit{elephant})\cdot\mathcal{F}(\mathit{fridge})]$ do?</p> <p><strong>Numerical Analyst</strong>: 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.</p> <p><strong>Probabilist</strong>: Keep trying to push it in in random ways and eventually it will fit.</p> <p><strong>Combinatorist</strong>: Discretize the elephant, partition it, and find a suitable rearrangement.</p> <p><strong>Statistician</strong>: Put its tail in the refrigerator as a sample, and say, “done!”</p> <p><strong>Logician</strong>: I know it's possible, I just can't do it.</p> <p><strong>Category Theorist</strong>: Isn't this just a special case of Yoneda's lemma?</p> <p><strong>Theoretical Computer Scientist</strong>: I can't decide.</p> <p><strong>Experimental Mathematician</strong>: I think it'd be much more interesting to get the refrigerator inside the elephant.</p> <p><strong>Set Theorist</strong>: Force it.</p>