Theorems with unexpected conclusions - MathOverflow [closed] most recent 30 from http://mathoverflow.net 2013-05-23T00:55:23Z http://mathoverflow.net/feeds/question/18100 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/18100/theorems-with-unexpected-conclusions Theorems with unexpected conclusions Richard Stanley 2010-03-13T21:15:12Z 2012-01-01T03:18:13Z <p>I am interested in theorems with unexpected conclusions. I don't mean an unintuitive result (like the existence of a space-filling curve), but rather a result whose conclusion seems disconnected from the hypotheses. My favorite is the following. Let $f(n)$ be the number of ways to write the nonnegative integer $n$ as a sum of powers of 2, if no power of 2 can be used more than twice. For instance, $f(6)=3$ since we can write 6 as $4+2=4+1+1=2+2+1+1$. We have $(f(0),f(1),\dots) =$ $(1,1,2,1,3,2,3,1,4,3,5,2,5,3,4,\dots)$. The conclusion is that the numbers $f(n)/f(n+1)$ run through all the reduced positive rational numbers exactly once each. See A002487 in the <em>On-Line Encyclopedia of Integer Sequences</em> for more information. What are other nice examples of "unexpected conclusions"?</p> http://mathoverflow.net/questions/18100/theorems-with-unexpected-conclusions/18107#18107 Answer by Joel David Hamkins for Theorems with unexpected conclusions Joel David Hamkins 2010-03-13T21:51:38Z 2010-03-13T21:51:38Z <p>My favorite example of this phenomenon is <a href="http://en.wikipedia.org/wiki/Goodstein%27s_theorem" rel="nofollow">Goodstein's Theorem</a>. </p> <p>Take any positive number a<sub>2</sub>, such as the number 73, and write it in complete base 2, which means write it as a sum of powers of 2, but write the exponents also in this way. </p> <ul> <li>a<sub>2</sub> = 73 = 64 + 8 + 1 = 2<sup>2<sup>2</sup>+2</sup> + 2<sup>2+1</sup> + 1. </li> </ul> <p>Now, obtain a<sub>3</sub> by replacing all 2's with 3's, and subtracting 1. So in this case, </p> <ul> <li>a<sub>3</sub> = 3<sup>3<sup>3</sup>+3</sup> + 3<sup>3+1</sup> + 1 - 1 = 3<sup>3<sup>3</sup>+3</sup> + 3<sup>3+1</sup>.</li> </ul> <p>Similarly, write this in complete base 3, replace 3's with 4's, and substract one, to get </p> <ul> <li>a<sub>4</sub> = 4<sup>4<sup>4</sup>+4</sup> + 4<sup>4+1</sup> - 1 = 4<sup>4<sup>4</sup>+4</sup> + 3 4<sup>4</sup> + 3 4<sup>3</sup> + 3 4<sup>2</sup> + 4 + 3. </li> </ul> <p>And so on. The surprising conclusion is that:</p> <p><b>Goodstein's Theorem.</b> For any initial positive integer a<sub>2</sub>, there is n &gt; 2 for which a<sub>n</sub> = 0. </p> <p>That is, although it seems that the sequence is always growing larger, eventually it hits zero. So our initial impression that this process should proceed to ever larger numbers is simply not correct. The proof of Goodstein's theorem uses transfinite ordinals to measure the complexity of the numbers that arise, and proves that this complexity is strictly descending with each step. Thus, it must hit zero, and the only way this happens is if the number itself is zero. One can see that we had to split up the complexity of the number somewhat in moving from a<sub>3</sub> to a<sub>4</sub>, although even in this case the number did get larger. Eventually, the proof goes, the complexity drops low enough that the base exceeds the number, and from this point on, one is just subtracting one endlessly.</p> <p>This conclusion is very surprising. But this theorem actually packs a one-two punch! Because not only is the theorem itself surprising, but then thee is the following surprise follow-up theorem: </p> <p><b>Theorem.</b> Goodstein's theorem is not provable in the usual Peano Axioms PA of arithmetic. </p> <p>That is, the statement of Goodstein's theorem is independent of PA. It was a statement about finite numbers that is provable in ZFC, but not in PA. </p> http://mathoverflow.net/questions/18100/theorems-with-unexpected-conclusions/18110#18110 Answer by John Stillwell for Theorems with unexpected conclusions John Stillwell 2010-03-13T22:12:52Z 2010-03-14T06:55:47Z <p>Logic/computability theory is quite good at turning up seemingly special processes with unexpectedly universal outcomes. Goodstein's theorem (already mentioned) is one example. Another is the Matiyasevich theorem that polynomials with integer coefficients produce all computably enumerable sets. One way to state this is that each c.e. set is the set of <em>nonnegative</em> values of such a polynomial.</p> http://mathoverflow.net/questions/18100/theorems-with-unexpected-conclusions/18115#18115 Answer by Quimey Vivas for Theorems with unexpected conclusions Quimey Vivas 2010-03-13T22:36:28Z 2010-03-13T22:36:28Z <p>I like <a href="http://en.wikipedia.org/wiki/Sharkovsky%27s_theorem" rel="nofollow">Sharkovskii's theorem</a>. It says that there is an explicit ordering of the natural numbers such that if $f:\mathbb{R}\rightarrow \mathbb{R}$ has a periodic point of least period m and m precedes n in the above ordering, then f has also a periodic point of least period n.</p> http://mathoverflow.net/questions/18100/theorems-with-unexpected-conclusions/18120#18120 Answer by Igor Pak for Theorems with unexpected conclusions Igor Pak 2010-03-13T23:03:08Z 2010-03-13T23:03:08Z <p>I think the question is very personal, in a sense that what is unexpected for one person or from one point of view, can be very straightforward from another. To further complicate the matter, the notion of what is "unexpected" changes over time. Let me give a couple of familiar examples to illustrate these points:</p> <p>1) the evaluation of the <a href="http://en.wikipedia.org/wiki/Chromatic_polynomial" rel="nofollow">chromatic polynomial</a> of a graph at $(-1)$ is equal to the number of acyclic orientation of the graph (up to a sign). When you (R.P. Stanley) published this theorem in 1973, I bet this was considered a remarkably unexpected result - the conclusion had seemingly nothing to do with the assumption. For people outside of combinatorics, it is probably still unexpected. However, these days, with all those numerous reciprocity theorems (many of which, undoubtedly, grew in part out of this result), it is much harder for a combinatorialist to think of it as "unexpected". Curiously, Wikipedia takes a middle course: prior to the statement of the theorem, it adds "perhaps surprisingly", wisely letting us form our own conclusions. </p> <p>2) take the <em>Fibonacci polytope</em> defined as convex hull of 0-1 vectors in $\Bbb R^n$ with no adjacent ones. Then its volume is the number of <a href="http://en.wikipedia.org/wiki/Alternating_permutation" rel="nofollow">alternating permutations</a> divided by $n!$. Again, if one have never seen "combinatorial polytopes" whose volume is expressed in terms of the number of certain permutations, the conclusion is completely unexpected - there is no obvious connection between <a href="http://en.wikipedia.org/wiki/Fibonacci_numbers" rel="nofollow">Fibonacci numbers</a> and alternating permutations. But for those of us who have seen and studied these, this result is straightforward and a very easy exercise. </p> http://mathoverflow.net/questions/18100/theorems-with-unexpected-conclusions/18122#18122 Answer by Qiaochu Yuan for Theorems with unexpected conclusions Qiaochu Yuan 2010-03-13T23:49:40Z 2010-03-14T01:18:44Z <p>I learned this example from Noam Elkies's excellent article <a href="http://www.msri.org/publications/books/Book35/files/elkies.pdf" rel="nofollow">The Klein Quartic in Number Theory</a>. Elkies observes that Siegel's 1968 paper <a href="http://www.springerlink.com/index/M62M55J770X34145.pdf" rel="nofollow">Zum Beweise des Starkschen Satzes</a>, in order to prove its main result, proves what is equivalent to the following.</p> <p><strong>Theorem:</strong> Suppose that the only Fibonacci numbers which are cubes are $0, \pm 1, \pm 8$. Then the set of negative integers $d$ such that $\mathbb{Q}[\sqrt{d}]$ has class number $1$ is <a href="http://en.wikipedia.org/wiki/Stark%E2%80%93Heegner_theorem" rel="nofollow">$\{ -1, -2, -3, -7, -11, -19, -43, -67, -163 \}$</a>.</p> http://mathoverflow.net/questions/18100/theorems-with-unexpected-conclusions/18125#18125 Answer by JosÃ© Figueroa-O'Farrill for Theorems with unexpected conclusions JosÃ© Figueroa-O'Farrill 2010-03-14T02:10:04Z 2010-03-14T06:37:42Z <p>I have always found <a href="http://en.wikipedia.org/wiki/Kuratowski%27s_closure-complement_problem" rel="nofollow">Kuratowski 14-set problem</a> among the most surprising elementary theorems I know. Why 14?! (This was recently discussed in <a href="http://mathoverflow.net/questions/16363/kuratowski-closure-complement-problem-for-other-mathematical-objects" rel="nofollow">this MO question</a>.)</p> http://mathoverflow.net/questions/18100/theorems-with-unexpected-conclusions/18126#18126 Answer by Bjorn Poonen for Theorems with unexpected conclusions Bjorn Poonen 2010-03-14T02:14:33Z 2010-03-14T02:14:33Z <p><strong>Faltings' theorem</strong> (a.k.a. the Mordell conjecture): Given a smooth projective curve $X$ defined by an equation with rational coefficients, if the set of complex points on $X$ is topologically a surface of genus greater than $1$, then there are only finitely many points on the curve with rational coordinates.</p> <p>(Actually it is proved for curves over finite extensions of $\mathbf{Q}$ too.)</p> http://mathoverflow.net/questions/18100/theorems-with-unexpected-conclusions/18128#18128 Answer by Allen Knutson for Theorems with unexpected conclusions Allen Knutson 2010-03-14T02:38:42Z 2010-03-14T02:38:42Z <p>Let G be a group of order p(p+1), with more than one p-Sylow. Then p is either 2 or a Mersenne prime. (Indeed, G exists uniquely for each such p.)</p> <p>One of my own I'm proud of: <a href="http://front.math.ucdavis.edu/0911.4941" rel="nofollow">http://front.math.ucdavis.edu/0911.4941</a></p> <p>Let H be a degree n hypersurface in n-space (yes, same n) over $F_p$. From H we may be able to construct many other subschemes, by decomposing, intersecting components, decomposing again, intersecting again, ...</p> <p>If the number of $F_p$ points on H is not a multiple of p, then all these subschemes are reduced.</p> http://mathoverflow.net/questions/18100/theorems-with-unexpected-conclusions/18130#18130 Answer by Dan Ramras for Theorems with unexpected conclusions Dan Ramras 2010-03-14T02:51:08Z 2010-03-14T02:51:08Z <p>Maybe by now no one thinks of it as counterintuitive, but what about Poincaré Duality? </p> <p>The following formulation (for psuedomanifolds) might fit this question best:</p> <p>If K is a finite simplicial complex satisfying:</p> <ol> <li>each n-1 simplex lies in exactly two n-simplices</li> <li>any two n-simplices are connected by a chain of n-simplices, each intersecting the previous in an n-1 dimensional face</li> <li>each simplex lies in some n-simplex</li> </ol> <p>then the (mod 2) Betti numbers of K in complementary dimensions must be equal!</p> http://mathoverflow.net/questions/18100/theorems-with-unexpected-conclusions/18136#18136 Answer by Guillermo Mantilla for Theorems with unexpected conclusions Guillermo Mantilla 2010-03-14T04:48:59Z 2010-03-14T04:48:59Z <p>It is well known that a group $G$ can't be written as the union of two proper subgroups. On the other hand there are groups that can be written as the union of three proper subgroups, my favorite one the quaternions $Q_8$. Now, I remember the following fact from my undergrad group theory class: if $G$ is a finite group such that $G$ is the union of three proper subgroups then the Klein four group $V_4$ is a quotient of $G$. </p> http://mathoverflow.net/questions/18100/theorems-with-unexpected-conclusions/18138#18138 Answer by Guillermo Mantilla for Theorems with unexpected conclusions Guillermo Mantilla 2010-03-14T04:59:09Z 2010-03-14T04:59:09Z <p>The existence of two non-isomorphic isospectral Riemannian manifolds "we can't hear the Shape of a Drum" can be deduced from the existence of two quasi conjugated subgroups of $PSL_2(7)$</p> http://mathoverflow.net/questions/18100/theorems-with-unexpected-conclusions/18143#18143 Answer by Anonymous for Theorems with unexpected conclusions Anonymous 2010-03-14T05:47:14Z 2010-03-14T05:47:14Z <p>The Taniyama-Shimura conjecture (now proved, by Wiles and others): all elliptic curves over $\mathbb Q$ are modular. It's magical that one can give a "formula" for the numbers of points on the curve modulo $p$ using modular forms.</p> http://mathoverflow.net/questions/18100/theorems-with-unexpected-conclusions/18144#18144 Answer by Douglas Zare for Theorems with unexpected conclusions Douglas Zare 2010-03-14T05:49:57Z 2010-03-14T14:08:12Z <p>If arbitrary products of nonempty sets are nonempty, then you can decompose a unit ball in $\mathbb R^3$ into finitely many pieces and rigidly reassemble then into two balls of radius 1. That is, the axiom of choice implies the Banach-Tarski paradox. </p> <p>Of course, there are plenty of other results which depend on the axiom of choice, and many of them qualify, whether their conclusion seems to violate physical intuition or not. The point is that the conclusion seems nothing like the assumption. </p> http://mathoverflow.net/questions/18100/theorems-with-unexpected-conclusions/18149#18149 Answer by Amira for Theorems with unexpected conclusions Amira 2010-03-14T07:26:30Z 2010-03-14T07:26:30Z <p>One of my personal favorite theorems with an unexpected application is the Atiyah-Singer index theorem. I don't know if the application can be labeled as "real" mathematics, but it is amazing how it works.</p> <p>In the article <em>An SU(2) Anomaly</em>, Edward Witten shows that certain "SU(2) gauge theories" having an odd number of doublets of Dirac fermions are "mathematically inconsistent". In this case, the latter means that all path integrals vanish.</p> <p>That all path integrals vanish is a consequence of the fact that $\pi_4(SU(2)) = \mathbf{Z}/2\mathbf{Z}$. Thus, there is also some homotopy theory involved! </p> <p>For completeness, here is the reference <strong>E. Witten, An SU(2) Anomaly. Phys. Lett. B 117 (1982), pages 324-328.</strong></p> http://mathoverflow.net/questions/18100/theorems-with-unexpected-conclusions/18156#18156 Answer by Jon for Theorems with unexpected conclusions Jon 2010-03-14T09:11:56Z 2010-12-13T20:50:16Z <p>Definition: Let $A$ and $B$ be self-adjoint matrices, with the partial order $A\ge B$ if $A-B$ is positive semidefinite. If $A$ is self-adjoint with spectrum in the interval $[a,b]$ and $f\colon [a,b] \to \mathbb{R}$ is a real-function, define $f(A)$ using the spectral theorem. The function $f$ is called <em>matrix monotone</em> if $A\ge B$ implies $f(A)\ge f(B)$ for all $A,B$ with spectra in the domain $[a,b]$ of $f$.</p> <p><strong>Loewner's theorem</strong>: A function $f\colon [a,b] \to \mathbb{R}$ is matrix monotone iff it has an analytic extension to the upper and lower half-planes so that the each of these half-planes is mapped into itself.</p> http://mathoverflow.net/questions/18100/theorems-with-unexpected-conclusions/18170#18170 Answer by Douglas Zare for Theorems with unexpected conclusions Douglas Zare 2010-03-14T15:06:00Z 2010-03-14T18:48:25Z <p>Reciprocity/duality theorems may give you unexpected results if you don't expect the connections.</p> <p>Dan Ranmas already mentioned Poincare duality. To clarify, Poincare duality is not just abstract nonsense. It fails for non-manifolds like general abstract simlicial complexes. For a [mod $2$] oriented manifold of dimension $d$, the [mod $2$] homology in dimension $k$ is isomorphic to the [mod $2$] homology in dimension $d-k$.</p> <p>Quadratic reciprocity relates whether $p$ is a square mod $q$ with whether $q$ is a square mod $p$. </p> <p><a href="http://en.wikipedia.org/wiki/Weil_reciprocity_for_algebraic_curves" rel="nofollow">Weil reciprocity</a> relates the values of a rational function $f$ at the zeros and poles of $g$ with the values of $g$ at the zeros and poles of $f$. </p> <p><a href="http://en.wikipedia.org/wiki/Stanley%27s_reciprocity_theorem" rel="nofollow">Stanley reciprocity</a> relates a generating function for the lattice points in a convex cone with a generating function for the lattice points in the interior evaluated at reciprocal arguments.</p> http://mathoverflow.net/questions/18100/theorems-with-unexpected-conclusions/18215#18215 Answer by alex for Theorems with unexpected conclusions alex 2010-03-14T22:21:38Z 2010-03-15T01:16:29Z <p>Given a $C^{\infty}$ function $f(x)$, let $\Delta_n$ be the difference between the integral of $f(x)$ and its $n$'th Riemann sum: $$\Delta_n = \int_0^1 f ~dx ~-~ \sum_{i=1}^{n} f(i/n) \frac{1}{n}$$ Clearly, $\Delta_n$ goes to $0$, but at what rate? Its easy to see that its possible for $\Delta_n$ to decay as $\Theta(1/n)$, e.g consider what happens with $f(x)=x$. </p> <p><strong>Theorem:</strong> If $f(x)$ is periodic with period $1$, then $\Delta_n$ decays <a href="http://www-math.mit.edu/~davis/dws08.pdf" rel="nofollow">faster than any polynomial in $n$</a>. </p> http://mathoverflow.net/questions/18100/theorems-with-unexpected-conclusions/18226#18226 Answer by Antun Milas for Theorems with unexpected conclusions Antun Milas 2010-03-15T00:15:56Z 2010-03-15T00:15:56Z <p><a href="http://en.wikipedia.org/wiki/Whitehead%27s_problem" rel="nofollow"> Shelah's solution </a> of </p> <p>Whitehead problem: Is every abelian group A with $Ext^1(A, \mathbb{Z}) = 0$ a free abelian group?</p> http://mathoverflow.net/questions/18100/theorems-with-unexpected-conclusions/18232#18232 Answer by John D. Cook for Theorems with unexpected conclusions John D. Cook 2010-03-15T01:37:46Z 2010-12-13T20:43:52Z <p><a href="http://www.johndcook.com/blog/2009/05/05/a-surprising-theorem-in-complex-variables/" rel="nofollow">Nevanlinna's theorem</a>: </p> <p>Suppose $f(z)$ and $g(z)$ are two functions meromorphic in the plane. Suppose also that there are five distinct numbers $a_1,\ldots,a_5$ such that the solution sets $\lbrace z : f(z) = a_i\rbrace$ and $\lbrace z : g(z) = a_i\rbrace$ are equal. Then either $f(z)$ and $g(z)$ are equal everywhere or they are both constant.</p> http://mathoverflow.net/questions/18100/theorems-with-unexpected-conclusions/21137#21137 Answer by Diego de Estrada for Theorems with unexpected conclusions Diego de Estrada 2010-04-12T18:26:59Z 2010-04-12T18:26:59Z <p>I was very surprised when I first saw that the product of all primes $p$ such that $p-1|2n,$ is the denominator of Bernoulli number $B_{2n}$.</p> http://mathoverflow.net/questions/18100/theorems-with-unexpected-conclusions/21176#21176 Answer by Andrew L for Theorems with unexpected conclusions Andrew L 2010-04-13T02:20:59Z 2012-01-01T03:18:13Z <p>The first time I ever saw Cayley's Fundamental Theorem of Group Theory - i.e. every group is isomorphic to a group of permutations on a nonempty set - I was floored and I knew anything that contained a statement that bizarre that's true was something I wanted to do in life.</p> http://mathoverflow.net/questions/18100/theorems-with-unexpected-conclusions/22427#22427 Answer by Johan for Theorems with unexpected conclusions Johan 2010-04-24T13:37:53Z 2010-04-24T13:37:53Z <p>There is a theorem by Bernstein that I like: </p> <p>If $f$ is a $C^{\infty}$-function on the intervall $I$ such that $f$ and the derivatives of $f$ to every order are non-negative on $I$ then $f$ is analytic. </p> <p>An example would be $e^x$ which satisfies the assumptions and thus is analytic (on the whole real line). </p> http://mathoverflow.net/questions/18100/theorems-with-unexpected-conclusions/22442#22442 Answer by J. H. S. for Theorems with unexpected conclusions J. H. S. 2010-04-24T17:22:25Z 2010-05-08T02:40:34Z <p>The following pearl by Jacobson can under no circumstances be left out from the list:</p> <p>Let $\mathbf{R}$ be a ring with center $\mathrm{Z}$. Let us suppose that you can find $n \in \mathbb{N}_{>1}$ such that $x^{n}-x \in \mathrm{Z}$ for every $x \in \mathbf{R}$. Then $\mathbf{R}$ is a commutative ring.</p> <p>A good place to learn more about results of this kind is Herstein's <em>Noncommutative rings</em>.</p> http://mathoverflow.net/questions/18100/theorems-with-unexpected-conclusions/23882#23882 Answer by Mikhail Bondarko for Theorems with unexpected conclusions Mikhail Bondarko 2010-05-07T18:21:13Z 2010-05-07T18:21:13Z <p>A proper algebraic group is abelian.</p> http://mathoverflow.net/questions/18100/theorems-with-unexpected-conclusions/23925#23925 Answer by Gil Kalai for Theorems with unexpected conclusions Gil Kalai 2010-05-08T10:02:34Z 2010-05-08T10:02:34Z <p>Here are three examples from combinaorics:</p> <p>1) The <a href="http://gilkalai.wordpress.com/2009/05/21/extremal-combinatorics-vi-the-frankl-wilson-theorem/" rel="nofollow">Frankl Wilson' theorem</a> (The paper <a href="http://www.springerlink.com/content/d8571m0786m6xk81/" rel="nofollow">can be found here</a>). This theorem in extremal combinatorics has a large number of amazing applications: Explicit Ramsey constructions, applications in combinatorial geometry; applications regarding Shannon capacity of union of graphs and many more. </p> <p>2) Trotter-Szemeredi The result by Trotter and Szemeredi regarding the maximum number of incidences between points and lines in the plane had remarkable applications including one discovered by Elekes' to the product-sum theorem.</p> <p>3) The mod p product sum theorem by Bourgain-Katz-Tao had many surprising applications in many directions. (One reason for the wide applicability is that when you multiply matrices sums and products mix.)</p> http://mathoverflow.net/questions/18100/theorems-with-unexpected-conclusions/29730#29730 Answer by Ivan Meir for Theorems with unexpected conclusions Ivan Meir 2010-06-27T20:18:19Z 2010-06-27T20:18:19Z <p>How about the Cook-Levin theorem - boolean satisfiability is NP complete. Though the consequence that "if there exists a polynomial time algorithm for boolean satisfiability then all problems in NP can be solved in polynomial time" may fit the bill better!</p> <p>I mean what does boolean satifiability have to do with finding hamiltonians on graphs or finding shortest roots in networks?!</p> <p>Ivan</p> http://mathoverflow.net/questions/18100/theorems-with-unexpected-conclusions/29732#29732 Answer by Qiaochu Yuan for Theorems with unexpected conclusions Qiaochu Yuan 2010-06-27T21:01:22Z 2010-06-27T21:08:29Z <p>Here's one I was reminded of recently. Recall that a <strong>projective plane</strong> is a triple $(P, L, I)$ where $P$ is a set of "points," $L$ is a set of "lines," and $I$ is a subset of $P \times L$ describing the incidence relations which satisfies certain axioms. A finite projective plane always has $n^2 + n + 1$ points for some $n$ which is known as the <strong>order</strong> of the plane. So far, so geometric and combinatorial.</p> <p><strong>Theorem (<a href="http://en.wikipedia.org/wiki/Bruck%E2%80%93Chowla%E2%80%93Ryser_theorem" rel="nofollow">Bruck-Ryser</a>):</strong> If $n \equiv 1, 2 \bmod 4$, then $n$ is a sum of two squares.</p> <p>This is still the only known general criterion for ruling out orders of projective planes! It's conjectured that $n$ must be a power of a prime (examples include the projective planes which occur as $\mathbb{P}^2 \mathbb{F}_q$), but it's not even known whether there exists a projective plane of order $12$. </p> <p>(There is a "theorem with an unexpected proof" in this area as well. For finite projective planes, <a href="http://en.wikipedia.org/wiki/Desargues_theorem" rel="nofollow">Desargues' theorem</a> implies <a href="http://en.wikipedia.org/wiki/Pappus_hexagon_theorem" rel="nofollow">Pappus's theorem</a>, but the only known proof goes through Wedderburn's little theorem!)</p> http://mathoverflow.net/questions/18100/theorems-with-unexpected-conclusions/31959#31959 Answer by Jesse Madnick for Theorems with unexpected conclusions Jesse Madnick 2010-07-15T04:44:36Z 2010-07-15T04:44:36Z <p>If $f\colon [a,b] \to \mathbb{R}$ is increasing, then $f$ is differentiable almost everywhere [w.r.t. Lebesgue measure].</p> <p>(We can further conclude that $f'$ is measurable and $\int_a^b f'(x)\ dx \leq f(b) - f(a)$, but it's the first part that struck me when I learned it.)</p> <p>And sure it makes sense, but knowing how real analysis often is, one might think that there must be some increasing function that fails to be differentiable on a set of positive measure.</p> http://mathoverflow.net/questions/18100/theorems-with-unexpected-conclusions/48701#48701 Answer by Harry Altman for Theorems with unexpected conclusions Harry Altman 2010-12-09T01:14:06Z 2010-12-09T01:14:06Z <p>A theorem of Erdos and Hajnal: Any graph with no 4-cycles is countably colorable.</p> <p>Now, admittedly, this conclusion is less surprising when you state the actual stronger theorem that this is a corollary to: Any graph which is not countably colorable must contain a copy of $K_{\aleph_1,n}$ for every finite n. But in particular it must contain a 4-cycle, which is not only a surprising statement on its own but is also especially surprising considering that given $k$ and any <em>finite</em> $n$ there are finite graphs with girth at least $k$ and chromatic numberat least $n$, and that given $k$ and an arbitrary cardinal $\kappa$ there are graphs with <em>odd</em> girth at least $k$ and chromatic number at least $\kappa$. But, no 4-cycles? Countably colorable!</p> http://mathoverflow.net/questions/18100/theorems-with-unexpected-conclusions/49337#49337 Answer by Michal Oszmaniec for Theorems with unexpected conclusions Michal Oszmaniec 2010-12-14T01:24:04Z 2010-12-14T02:41:52Z <p>Polya's Theorem: Simple random walk on $\mathbb{Z}^d$ is recurrent for $d\leq2$ and transient for $d>2$. </p> <p>There is also a nice connection between this theorem and infinite networks of resistors. It turns out that the resistance of the whole network $\mathbb{Z}^d$ (one puts a unit source in one point and takes away sinks to $\infty$) is finite iff corresppnding random walk is transient :)</p> http://mathoverflow.net/questions/18100/theorems-with-unexpected-conclusions/62784#62784 Answer by Misha Belolipetsky for Theorems with unexpected conclusions Misha Belolipetsky 2011-04-23T22:10:19Z 2011-04-23T22:10:19Z <p>The classical differential geometry results should definitely be mentioned here. Although it may seem not surprising for us, Gauss found his <a href="http://en.wikipedia.org/wiki/Theorema_Egregium" rel="nofollow">Theorema Egregium</a> to be truly remarkable and unexpected. My favorite example is <a href="http://en.wikipedia.org/wiki/Gauss-Bonnet_theorem" rel="nofollow">Gauss-Bonnet Theorem</a>. </p> http://mathoverflow.net/questions/18100/theorems-with-unexpected-conclusions/62789#62789 Answer by anonymous for Theorems with unexpected conclusions anonymous 2011-04-23T23:17:25Z 2011-04-23T23:17:25Z <p>Weil's conjecture (proved by Grothendieck) that the number of points of an algebraic variety over finite fields is dictated by the topology of the same algebraic variety over ${\mathbb C}$ (more precisely its Betti numbers).</p> <p>Baez-Duarte's criterium: If $1$ is in the closure of the subspace of $L^2([1,+\infty[,\frac{dt}{t^2})$ spanned by the {$\frac{t}{n}$} (fractional part), for $n\geq 1$, then Riemann Hypothesis holds.</p> http://mathoverflow.net/questions/18100/theorems-with-unexpected-conclusions/62793#62793 Answer by roy smith for Theorems with unexpected conclusions roy smith 2011-04-24T00:24:16Z 2011-04-24T00:24:16Z <p>In the strict spirit of your question, the hypothesis that 1 = -1 has as conclusion that the moon is made of green cheese.</p> http://mathoverflow.net/questions/18100/theorems-with-unexpected-conclusions/62811#62811 Answer by Stefan Geschke for Theorems with unexpected conclusions Stefan Geschke 2011-04-24T06:56:44Z 2011-04-24T06:56:44Z <p>How about Shelah's truly remarkable ${\aleph_\omega}^{\aleph_0}\leq 2^{\aleph_0}\cdot\aleph_{\omega_4}$ (and variations of it)?</p> <p>After seeing various independence results in set theory it is very surprising that anything of this generality can be proved in ZFC. Hence the disconnect between the assumptions and the outcome is that there are no assumptions (beyond the usual axioms of set theory). And then there is the ever puzzling (open) question "Why the hell is it $\omega_4$?"</p> http://mathoverflow.net/questions/18100/theorems-with-unexpected-conclusions/62825#62825 Answer by tetrapharmakon for Theorems with unexpected conclusions tetrapharmakon 2011-04-24T09:48:05Z 2011-04-24T09:48:05Z <p><a href="http://en.wikipedia.org/wiki/Eckmann-Hilton_argument" rel="nofollow">Eckmann-Hilton argument</a>. I mean, WHY?</p>