"Slap your forehead" moments- Greatest Hits - MathOverflow [closed] most recent 30 from http://mathoverflow.net 2013-05-24T22:36:45Z http://mathoverflow.net/feeds/question/32225 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/32225/slap-your-forehead-moments-greatest-hits "Slap your forehead" moments- Greatest Hits Dylan Wilson 2010-07-16T21:16:33Z 2010-07-17T11:07:39Z <p>We've all done it: we have a "proof" that is so pretty... but we know it's wrong. Either it proves something we know is false, or the proof doesn't use one of the hypotheses you know it needs to, or we assume something we can't assume... and the whole pretty argument goes away. I had a professor who said that he sometimes convinces himself that some proof works right before he goes to bed, only to wake up and find out that it's total nonsense. He says it's probably some coping mechanism his body has so he can get some sleep!</p> <p>Though the proofs often end up unsalvageable (is that a word?), I think they are an important part of learning mathematics and being creative. I think it was Sir Ken Robinson (not a mathematician, but a good thinker) who said something like "Being wrong isn't always good, but if we don't have the capacity to be wrong we can never be creative." (If anyone knows the precise quote, do fill it in!)</p> <p>So let's hear them- your pretty proofs that turn out to be nonsense. Like the well-written "How not to prove the Poincare Conjecture," these can be well thought out proofs that have a small hidden assumption that makes everything blow-up, or they can be totally silly like the time I thought I proved Brouwer's fixed point theorem using the Baire Category Theorem during my freshman year...</p> <p>Bonus points if anyone's "false" methods (or a close adaptation) ended up working for a different problem later on!</p> http://mathoverflow.net/questions/32225/slap-your-forehead-moments-greatest-hits/32229#32229 Answer by Tobias Kildetoft for "Slap your forehead" moments- Greatest Hits Tobias Kildetoft 2010-07-16T21:23:35Z 2010-07-16T21:23:35Z <p>I did think that Burnside's pq theorem was very simple to prove. You just use that if a finite group has a Hall-subgroup of each possible order then it is solvable. When I finally bothered to look at the actual proof of this fact it of course turned out to use Burnside as basis for induction.</p> http://mathoverflow.net/questions/32225/slap-your-forehead-moments-greatest-hits/32241#32241 Answer by Mark Schwarzmann for "Slap your forehead" moments- Greatest Hits Mark Schwarzmann 2010-07-16T22:43:52Z 2010-07-17T11:07:39Z <p>I was once given a question (in a take home exam) along the following lines: A sequence of complex-valued bounded measurable functions $f_n (x)$ on a probability space $X$ was given, with the functions $f_n$ satisfying some conditions which I won't specify here. The problem was to show that the $L^2 (X)$ limit of the sequence is a certain given function $f(x)$. After some thought, I came up with the following "solution": I defined a sequence of functions $g_n (x,y) \in L^2 (X \times X)$ and a function $g(x,y) \in L^2 (X \times X)$ such that $g_n (x,x) = f_n (x)$ and $g(x,x)=f(x)$ for all $x \in X$, and I proved that $g_n (x,y) \to g(x,y) \;$ in $L^2 (X \times X)$. I was sure that I had, in fact, proved a generalization of the proposition I was given.</p> <p>Embarrassingly, it took me some time to realize my mistake. I tried to salvage my proof using continuity arguments and what not, but in the end I gave up and managed to concoct a different approach, which unfortunately was a lot more complicated and messy.</p> http://mathoverflow.net/questions/32225/slap-your-forehead-moments-greatest-hits/32254#32254 Answer by Zen Harper for "Slap your forehead" moments- Greatest Hits Zen Harper 2010-07-17T03:07:23Z 2010-07-17T03:07:23Z <p>For some problem in Algebraic Topology (presumably related to homotopy groups or similar, with free groups): I thought that two groups $G$, $H$ were isomorphic, because $G \approx A \subseteq H \approx B \subseteq G$, where $\approx$ means "<em>is isomorphic to</em>" and $\subseteq$ means "<em>is a subgroup of</em>".</p> <p>However, I was very shocked when informed that this does NOT imply that $G \approx H$ in general! (<em>I thought this was true, and spent a long time trying to prove it; but I knew that I hadn't succeeded. So I suppose this doesn't qualify, but anyway</em>).</p> http://mathoverflow.net/questions/32225/slap-your-forehead-moments-greatest-hits/32256#32256 Answer by Greg Kuperberg for "Slap your forehead" moments- Greatest Hits Greg Kuperberg 2010-07-17T04:02:02Z 2010-07-17T04:02:02Z <p>Here are four tempting fallacies that I have seen, that are in my opinion are all teachable and interesting:</p> <ol> <li><p>A finite covering space of a disk with finitely many holes, is again a disk with finitely many holes; in particular it is still planar.</p></li> <li><p>If you lengthen all three edges of a triangle, its area increases.</p></li> <li><p>If $F$ is a field with two finite-index subfields $K$ and $L$, then $K \cap L$ also has finite index in $F$.</p></li> <li><p>There are exactly two Lie groups up to isomorphism that are diffeomorphic to a pair of circles.</p></li> </ol>