Theorems with unexpected conclusions [closed]

Question

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 On-Line Encyclopedia of Integer Sequences for more information. What are other nice examples of "unexpected conclusions"?

Answer 1

My favorite example of this phenomenon is Goodstein's Theorem.

Take any positive number a₂, 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.

• a₂ = 73 = 64 + 8 + 1 = 2²²⁺² + 2²⁺¹ + 1.

Now, obtain a₃ by replacing all 2's with 3's, and subtracting 1. So in this case,

• a₃ = 3³³⁺³ + 3³⁺¹ + 1 - 1 = 3³³⁺³ + 3³⁺¹.

Similarly, write this in complete base 3, replace 3's with 4's, and substract one, to get

• a₄ = 4⁴⁴⁺⁴ + 4⁴⁺¹ - 1 = 4⁴⁴⁺⁴ + 3 4⁴ + 3 4³ + 3 4² + 4 + 3.

And so on. The surprising conclusion is that:

Goodstein's Theorem. For any initial positive integer a₂, there is n > 2 for which a_n = 0.

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 a₃ to a₄, 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.

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:

Theorem. Goodstein's theorem is not provable in the usual Peano Axioms PA of arithmetic.

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.

Answer 2

I learned this example from Noam Elkies's excellent article The Klein Quartic in Number Theory. Elkies observes that Siegel's 1968 paper Zum Beweise des Starkschen Satzes, in order to prove its main result, proves what is equivalent to the following.

Theorem: 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 $\{ -1, -2, -3, -7, -11, -19, -43, -67, -163 \}$.

Answer 3

I like Sharkovskii's theorem. 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.

Answer 4

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 $G$.

Answer 5

I have always found Kuratowski 14-set problem among the most surprising elementary theorems I know. Why 14?! (This was recently discussed in this MO question.)

Answer 6

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.

Answer 7

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.

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.

Answer 8

Faltings' theorem (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.

(Actually it is proved for curves over finite extensions of $\mathbf{Q}$ too.)

Answer 9

Nevanlinna's theorem:

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.

Answer 10

There is a theorem by Bernstein that I like:

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.

An example would be $e^x$ which satisfies the assumptions and thus is analytic (on the whole real line).

Answer 11

How about Shelah's truly remarkable ${\aleph_\omega}^{\aleph_0}\leq 2^{\aleph_0}\cdot\aleph_{\omega_4}$ (and variations of it)?

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 $\omega_4$?"

Answer 12

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:

1) the evaluation of the chromatic polynomial of a graph at $(-1)$ is equal to the number of acyclic orientations 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.

2) take the Fibonacci polytope defined as convex hull of 0-1 vectors in $\Bbb R^n$ with no adjacent ones. Then its volume is the number of alternating permutations 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 Fibonacci numbers and alternating permutations. But for those of us who have seen and studied these, this result is straightforward and a very easy exercise.

Answer 13

Shelah's solution of

Whitehead problem: Is every abelian group A with $Ext^1(A, \mathbb{Z}) = 0$ a free abelian group?

Answer 14

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.)

One of my own I'm proud of: http://front.math.ucdavis.edu/0911.4941

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, ...

If the number of $F_p$ points on H is not a multiple of p, then all these subschemes are reduced.

Answer 15

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 nonnegative values of such a polynomial.

Answer 16

A proper algebraic group is abelian.

Answer 17

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 matrix monotone if $A\ge B$ implies $f(A)\ge f(B)$ for all $A,B$ with spectra in the domain $[a,b]$ of $f$.

Loewner's theorem: 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.

Answer 18

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 $PSL_2(7)$

Answer 19

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 $B_{2n}$.

Answer 20

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).

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.

Answer 21

A theorem of Erdos and Hajnal: Any graph with no 4-cycles is countably colorable.

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 finite $n$ 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 odd girth at least $k$ and chromatic number at least $\kappa$. But, no 4-cycles? Countably colorable!

Answer 22

Here's one I was reminded of recently. Recall that a projective plane 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 order of the plane. So far, so geometric and combinatorial.

Theorem (Bruck-Ryser): If $n \equiv 1, 2 \bmod 4$, then $n$ is a sum of two squares.

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 $12$.

(There is a "theorem with an unexpected proof" in this area as well. For finite projective planes, Desargues' theorem implies Pappus's theorem, but the only known proof goes through Wedderburn's little theorem!)

Answer 23

The following pearl by Jacobson can under no circumstances be left out from the list:

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.

A good place to learn more about results of this kind is Herstein's Noncommutative rings.

Answer 24

Here are three examples from combinaorics:

1) The Frankl Wilson' theorem (The paper can be found here). 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.

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.

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.)

Answer 25

If $f\colon [a,b] \to \mathbb{R}$ is increasing, then $f$ is differentiable almost everywhere [w.r.t. Lebesgue measure].

(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.)

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.

Answer 26

Maybe by now no one thinks of it as counterintuitive, but what about Poincaré Duality?

The following formulation (for psuedomanifolds) might fit this question best:

If K is a finite simplicial complex satisfying:

1. each n-1 simplex lies in exactly two n-simplices
2. any two n-simplices are connected by a chain of n-simplices, each intersecting the previous in an n-1 dimensional face
3. each simplex lies in some n-simplex

then the (mod 2) Betti numbers of K in complementary dimensions must be equal!

Answer 27

Reciprocity/duality theorems may give you unexpected results if you don't expect the connections.

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 $d-k$.

Quadratic reciprocity relates whether $p$ is a square mod $q$ with whether $q$ is a square mod $p$.

Weil reciprocity 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 $f$.

Stanley reciprocity 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.

Answer 28

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!

I mean what does boolean satifiability have to do with finding hamiltonians on graphs or finding shortest roots in networks?!

Ivan

Answer 29

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.

In the article An SU(2) Anomaly, 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.

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!

For completeness, here is the reference E. Witten, An SU(2) Anomaly. Phys. Lett. B 117 (1982), pages 324-328.

Answer 30

Polya's Theorem: Simple random walk on $\mathbb{Z}^d$ is recurrent for $d\leq2$ and transient for $d>2$.

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 :)