Awfully sophisticated proof for simple facts - MathOverflow [closed]

Awfully sophisticated proof for simple facts

2010-10-17T15:16:59Z

It is sometimes the case that one can produce proofs of simple facts that are of disproportionate sophistication which, however, do not involve any circularity. For example, (I think) I gave an example in this M.SE answer (the title of this question comes from Pete's comment there) If I recall correctly, another example is proving Wedderburn's theorem on the commutativity of finite division rings by computing the Brauer group of their centers.

Do you know of other examples of nuking mosquitos like this?

Answer by Boris Bukh for Awfully sophisticated proof for simple facts

2010-10-17T15:41:06Z

There are infinitely many primes because $\zeta(3)=\prod_p \frac{1}{1-p^{-3}}$ is irrational.

Answer by muad for Awfully sophisticated proof for simple facts

2010-10-17T15:44:25Z

Another example from Math Underflow:

We can prove Fermats Last Theorem for $n=3$ by a simple application of Nagell-Lutz (to compute the torsion subgroup) then Mordells Theorem (to see that the group must be $\mathbf{Z}^r \times \mathbf{Z}/3\mathbf{Z}$) then to finish Gross-Zagier-Kolyvagin theorem (which gives $r = 0$) - and that shows it has no nontrivial solutions. I beleive a similar approach works for $n=4$.

Answer by trb456 for Awfully sophisticated proof for simple facts

2010-10-17T16:04:54Z

And of course there is Fürstenberg's topological proof of the infinitude of primes. I love this because it shows that all the mathematical "plumbing" works; i.e that number theory and topology connect up as they should.

Answer by mt for Awfully sophisticated proof for simple facts

2010-10-17T16:44:06Z

Irrationality of $2^{1/n}$ for $n\geq 3$: if $2^{1/n}=p/q$ then $p^n = q^n+q^n$, contradicting Fermat's Last Theorem. Unfortunately FLT is not strong enough to prove $\sqrt{2}$ irrational.

I've forgotten who this one is due to, but it made me laugh. EDIT: Steve Huntsman's link credits it to W. H. Schultz.

Answer by Andres Caicedo for Awfully sophisticated proof for simple facts

2010-10-17T16:49:54Z

A Turing machine is a mathematical formalization of a computer (program). If $y\in(0,1)$, a Turing machine with oracle $y$ has access to the digits of $y$, and can use them during its computations. We say that $x\le_T y$ iff there is a machine with oracle $y$ that allows us to compute the digits of $x\in(0,1)$.

There are only countably many programs, so a simple diagonalization argument shows that there are reals $x$ and $y$ with $x{\not\le}_T y$ and $y{\not\le}_T x$. $(*)$

Being a set theorist, when I first learned of this notion, I couldn't help it but to come up with the following proof of $(*)$:

Again by counting, every $x$ has only countably many $\le_T$-predecessors. So, if CH fails, there are Turing-incomparable reals. By the technique of forcing, we can find a (boolean valued) extension $V'$ of the universe $V$ of sets where CH fails, and so $(*)$ holds in this extension. Shoenfield's absoluteness theorem tells us that $\Sigma^1_2$-statements are absolute between (transitive) models with the same ordinals. The statement $(*)$, "there are Turing-incomparable reals" is $\Sigma^1_1$ (implementing some of the coding machinery of Gödel's proof of the 2nd incompleteness theorem), so Shoenfield's absoluteness applies to it. Working from the point of view of $V'$ and considering $V'$ and $V$, it follows that in $V'$, with Boolean value 1, $(*)$ holds in $V$. It easily follows from this that indeed $(*)$ holds in $V$.

It turns out that Joel Hamkins also found this argument, and he used it in the context of his theory of Infinite time Turing machines, for which the simple diagonalization proof does not apply. So, at least in this case, the insane proof actually was useful at the end.

Answer by Qfwfq for Awfully sophisticated proof for simple facts

2010-10-17T16:56:45Z

The Jordan curve theorem. As far as I know, the "elementary" proof is quite involved, at least with respect to the intuitive plausibility of the statement.

Answer by Barry for Awfully sophisticated proof for simple facts

2010-10-17T16:59:49Z

Carl Linderholm. Mathematics made difficult.

Answer by Johannes Ebert for Awfully sophisticated proof for simple facts

2010-10-17T17:22:40Z

The Gauß-Bonnet theorem and the Riemann-Roch theorem for Riemann surfaces have both reasonably elementary proofs. Of course, they follow from the general Atiyah-Singer index theorem.

Answer by Harun Šiljak for Awfully sophisticated proof for simple facts

2010-10-17T17:32:46Z

A recent example from MO (I found it quite entertaining) - testing primality of one and two digit numbers using Stirling's formula and Wilson's theorem (to make it even more complicated, one has to use some extensions, calculation tricks and high-precision calculations):

http://mathoverflow.net/questions/42393/has-stirlings-formula-ever-been-applied-with-interesting-consequence-to-wilson

Answer by drvitek for Awfully sophisticated proof for simple facts

2010-10-17T17:34:44Z

A number of high school contest problems in number theory reduce to Mihailescu's theorem. (The only perfect powers with a difference of 1 are 8 and 9.)

Answer by Péter Komjáth for Awfully sophisticated proof for simple facts

2010-10-17T18:42:34Z

The number of real functions is $c^c=2^c$ which is bigger than $c$ by Cantor's theorem ($c$ is cardinality continuum). The number of real continuous functions is at most $c^{\aleph_0}=c$ as they can be recovered from restrictions to ${\bf Q}$, and there are $c^{\aleph_0}$ many functions ${\bf Q}\to {\bf R}$. This argument, which requires several minor steps in an introductory set theory class, eventually shows that there exists a discontinuous real function.

Answer by Owen Sizemore for Awfully sophisticated proof for simple facts

2010-10-17T19:04:52Z

The proof that the reduced $C^*$-algebra of the free group has no projections has the nice corollary that the circle is connected.

Answer by Denis Serre for Awfully sophisticated proof for simple facts

2010-10-17T19:07:28Z

I was once flamed because I gave (in my book on Matrices) a short proof of a weak version of Perron-Frobenius' theorem (the spectral radius of a non-negative matrix is an eigenvalue, associated with a non-negative eigenvector), by using Brouwer's fixed point theorem. In my mind, that was to give students an occasion to illustrate the strength of Brouwer's theorem. Of course, there are more elementary proofs of the Perron-Frobenius theorem, even of the stronger version of it.

Answer by Andrej Bauer for Awfully sophisticated proof for simple facts

2010-10-17T20:12:37Z

If two elements in a poset have the same lower bounds then they are equal by Yoneda lemma. (I actually said this in a seminar two weeks ago, and of course I explained I killed a mosquito with a nuke.)

Answer by Maxime Bourrigan for Awfully sophisticated proof for simple facts

2010-10-17T21:48:16Z

In his 1962 article "A unique decomposition theorem for 3-manifolds", Milnor is actually interested in the unicity of a prime decomposition. For the existence, the method is very natural: if you find an irreducible sphere, you cut the manifold along it and obtain a decomposition $M = M_1 \sharp M_2$, and you do it again with each factor, and so on.

Of course, the hard part is now to prove that this process terminates after a finite number of steps. For that, Milnor refers to Kneser but remarks that "if one assumes the Poincaré hypothesis then there is a much easier proof. Define $\rho(M)$ as the smallest number of generators for the fundamental group of M. It follows from the Gruško-Neumann theorem that $\rho(M_1\sharp M_2) = \rho(M_1) + \rho(M_2)$. Hence if $M\simeq M_1 \sharp \cdots \sharp M_k$ with $k > \rho(M)$ then some $M_i$ must satisfy $\rho(M_i)=0$, and hence must be isomorphic to $S^3$."

A nice follow-up of this proof/joke is that Perel'man's proof of Poincaré's conjecture doesn't even use Kneser-Milnor decomposition and this argument is therefore valid.

Answer by Steven Gubkin for Awfully sophisticated proof for simple facts

2010-10-18T15:17:36Z

The fundamental group of the circle is $\mathbb{Z}$ because:

It is a topological group, so its fundamental group is Abelian by the Eckmann-Hilton argument. Thus its fundamental group and first singular homology group coincide by the Hurewicz theorem. Since singular homology is the same as simplicial homology, I can just do the one line of computation to obtain the result.

Answer by Franz Lemmermeyer for Awfully sophisticated proof for simple facts

2010-10-18T15:45:52Z

There's hardly a book on class field theory that doesn't derive Kronecker-Weber as a corollary. Or quadratic reciprocity -)

Disclaimer: I like these proofs. Seeing quadratic reciprocity through the eyes of "Fearless symmetry: exposing the hidden patterns of numbers" by Ash and Gross is an experience you wouldn't want to miss.

Answer by Nate Eldredge for Awfully sophisticated proof for simple facts

2010-10-18T16:18:59Z

Proposition. Let $f$ be a bounded measurable function on $[0,1]$. Then there is a sequence of $C^\infty$ functions which converges to $f$ almost everywhere.

Proof (by flyswatter). Take the convolution of $f$ with a sequence of standard mollifiers.

Proof (by nuke). By Carleson's theorem the Fourier series of $f$ is such a sequence.

Answer by Peter Arndt for Awfully sophisticated proof for simple facts

2010-10-18T16:57:28Z

In a lecture course I saw a proof of Poincare duality by deducing it from Grothendieck duality. Proving Grothendieck duality for sheaves on topological spaces took a good part of the semester of course, and then deducing Poincare duality was still not a one liner as well, but filled an entire lecture in which we worked out what all the shrieks and derived functors were doing in terms of differential forms or singular cochains.

Answer by Joel David Hamkins for Awfully sophisticated proof for simple facts

2010-10-18T23:16:35Z

There is no largest natural number. The reason is that by Cantor's theorem, the power set of a finite set is a strictly larger set, and one can prove inductively that the power set of a finite set is still finite.
All numbers of the form $2^n$ for natural numbers $n\geq 1$ are even. The reason is that the power set of an $n$-element set has size $2^n$, proved by induction, and this is a Boolean algebra, which can be decomposed into complementary pairs $\{a,\neg a\}$. So it is a multiple of $2$.
Every finite set can be well-ordered. This follows by the Axiom of Choice via the Well-ordering Principle, which asserts that every set can be well-ordered.
Every non-empty set $A$ has at least one element. The reason is that if $A$ is nonempty, then $\{A\}$ is a family of nonempty sets, and so by the Axiom of Choice it admits a choice function $f$, which selects an element $f(A)\in A$.

Answer by François G. Dorais for Awfully sophisticated proof for simple facts

2010-10-19T12:02:40Z

Here is an example that I learned through MO!

The infinitude of completely split primes in a Galois extension K of Q is an easy consequence of Chebotarev's Density Theorem. A slightly simpler argument involves showing that the Dedekind Zeta Function ζ_K(s) has a simple pole at s = 1. However, there is a very simple arithmetic argument that accomplishes the desired task...

Answer by David MJC for Awfully sophisticated proof for simple facts

2010-10-22T20:28:05Z

A quiver whose unoriented graph is the affine D4 Dynkin diagram is tame. Therefore the moduli space of four points on a projective line is one dimensional.

Answer by Johan for Awfully sophisticated proof for simple facts

2010-11-03T21:59:08Z

The case of Fatou's theorem for H^2 can be proven as follows:

By Carleson's theorem the series $ \sum a_n e^{i \theta n} $ converges for almost all $\theta$ if $ \sum |a_n|^2 < \infty$. Now we can appeal to Abel's theorem to conclude that the function $ f(z)= \sum a_n z^n$ has radial limits almost everywhere on the unit circle. (I am not sure if we can get non-tangential limits this way.)

But Carleson's theorem is a much more difficult theorem than what we have proved here. (I got this example from a Hardy space course I am taking right now.)

Answer by Terry Tao for Awfully sophisticated proof for simple facts

2010-11-03T22:19:38Z

An example that came up in my measure theory class today:

The harmonic series $\sum_{n=1}^\infty \frac{1}{n}$ diverges, because otherwise the functions $f_n := \frac{1}{n} 1_{[0,n]}$ would be dominated by an absolutely integrable function. But $$\int_{\bf R} \lim_{n \to \infty} f_n(x)\ dx = 0 \neq 1 = \lim_{n \to \infty} \int_{\bf R} f_n(x)\ dx,$$ contradicting the dominated convergence theorem.

Answer by Gerry Myerson for Awfully sophisticated proof for simple facts

2010-11-03T22:55:22Z

D J Lewis, Diophantine equations: $p$-adic methods, in W J LeVeque, ed., Studies In Number Theory, 25-75, published by the MAA in 1969, stated on page 26, "The equation $x^3-117y^3=5$ is known to have at most 18 integral solutions but the exact number is unknown." No proof or reference is given.

R Finkelstein and H London, On D. J. Lewis's equation $x^3+117y^3=5$, Canad Math Bull 14 (1971) 111, prove the equation has no integral solutions, using ${\bf Q}(\root3\of{117})$.

Then Valeriu St. Udrescu, On D. J. Lewis's equation $x^3+117y^3=5$, Rev Roumaine Math Pures Appl 18 (1973) 473, pointed out that the equation reduces, modulo 9, to $x^3\equiv5\pmod9$, which has no solution.

I suspect Lewis was the victim of a typo, and some other equation was meant, but Finkelstein and London appear to have given an inadvertently sophisticated proof for a simple fact.

Answer by Gil Kalai for Awfully sophisticated proof for simple facts

2010-11-03T23:44:32Z

There is a Fourier analytic proof for Sperner's theorem which is much more complicated than the combinatorial proof (and give less in certain respects). This was part pf the polymath1 project.

A general point is that sometime trying to prove a Theorem X using method Y is valuable even if the proof is much more complicated than needed. So while simplification of complicated proofs is a noble endeavor, complicafication of simple theorems is also not without merit!

Here is another example (taken from lecture notes by Spencer): Suppose you want to prove that there is always a 1-1 function from a (finite) set |A| to a set |B| when |B|>=|A|. But you want to prove it using the probabilistic method. Write |A|=n. If |B| is larger than n^2 or so you can show that a 1-1 map exist by considering a random function and applying the union bound. If |B| is larger than 6n or so you can apply the much more sophisticated Lovasz Local Lemma to get a proof. I am not aware of probabilistic proofs of this nature which works when |B| is smaller and this is an interesting challenge.

Answer by Michael Greinecker for Awfully sophisticated proof for simple facts

2010-11-04T11:12:03Z

Baryshnikov gave a topological proof of Arrow's impossibility theorem, a result for which there are well known short and elementary proofs.

Answer by Peter Krautzberger for Awfully sophisticated proof for simple facts

2010-11-08T22:18:36Z

Every finite semigroup contains an idempotent element.

You can nuke this problem using a theorem by Ellis that every compact, semi-topological semigroup contains an idempotent (which uses Zorn's Lemma).

Answer by Jason for Awfully sophisticated proof for simple facts

2010-12-09T04:54:47Z

Theorem (ZFC + "There exists a supercompact cardinal."): There is no largest cardinal.

Proof: Let $\kappa$ be a supercompact cardinal, and suppose that there were a largest cardinal $\lambda$. Since $\kappa$ is a cardinal, $\lambda \geq \kappa$. By the $\lambda$-supercompactness of $\kappa$, let $j: V \rightarrow M$ be an elementary embedding into an inner model $M$ with critical point $\kappa$ such that $M^{\lambda} \subseteq M$ and $j(\kappa) > \lambda$. By elementarity, $M$ thinks that $j(\lambda) \geq j(\kappa) > \lambda$ is a cardinal. Then since $\lambda$ is the largest cardinal, $j(\lambda)$ must have size $\lambda$ in $V$. But then since $M$ is closed under $\lambda$ sequences, it also thinks that $j(\lambda)$ has size $\lambda$. This contradicts the fact that $M$ thinks that $j(\lambda)$, which is strictly greater than $\lambda$, is a cardinal.

Answer by Pete L. Clark for Awfully sophisticated proof for simple facts

2010-12-19T22:13:14Z

I claim that the rational canonical model of the modular curve $X(1) = \operatorname{SL}_2(\mathbb{Z}) \backslash \overline{\mathcal{H}}$ is isomorphic over $\mathbb{Q}$ to the projective line $\mathbb{P}^1$.

Indeed, by work of Igusa on integral canonical models, the corresponding moduli problem (for elliptic curves) extends to give a smooth model over $\mathbb{Z}$. By a celebrated 1985 theorem of Fontaine, this implies that $X(1)$ has genus zero. Therefore it is a Severi-Brauer conic, which by Hensel's Lemma and the Riemann Hypothesis for curves over finite fields is smooth over $\mathbb{Q}_p$ iff it has a $\mathbb{Q}_p$-rational point. By the reciprocity law in the Brauer group of $\mathbb{Q}$, this implies that $X(1)$ also has $\mathbb{R}$-rational points and then by the Hasse-Minkowski theorem it has $\mathbb{Q}$-rational points. Finally, it is an (unfortunately!) very elementary fact that a smooth genus zero curve with a rational point must be isomorphic to $\mathbb{P}^1$.

I did actually give an argument like this in a class I taught on Shimura varieties. Like many of the other answers here, it is ridiculous overkill in the situation described but begins to be less silly when looked at more generally, e.g. in the context of Shimura curves over totally real fields.

Answer by Ramsey for Awfully sophisticated proof for simple facts

2011-01-28T20:56:18Z

In the recent paper by Ono and Bruinier (it's currently on the AIM web site) "An algebraic formula for the partition function" they use their formula to determine the number of partitions of 1.

This calculation involves CM points, evaluating a certain weak Maass form at these points, the Hilbert class field of $\mathbb{Q}(\sqrt{-23})$, ... etc.

Answer by William Hale for Awfully sophisticated proof for simple facts

2011-01-29T03:00:35Z

Using character theory, any group of order 4 is abelian since the only way to write 4 as a sum of squares is 4 = 1^2 + 1^2 + 1^2 + 1^2.

Answer by Peter for Awfully sophisticated proof for simple facts

2011-01-29T05:27:43Z

No finite field $\mathbb{F}_q$ is algebraically closed:

Let $k$ be an algebraically closed field. Then every element of $GL_2(k)$ has an eigenvector, and hence is similar to an upper triangular matrix. Therefore $GL_2(k)$ is the union of the conjugates of its proper subgroup $T$ of upper triangular matrices. No finite group is the union of the conjugates of a proper subgroup, so $GL_2(k)$ is not finite. Hence $k$ is not finite either.

Answer by Michael Blackmon for Awfully sophisticated proof for simple facts

2011-01-29T11:55:08Z

Because for some reason no one has mentioned it.

Russell's proof that 1+1=2.

http://quod.lib.umich.edu/cgi/t/text/pageviewer-idx?c=umhistmath;cc=umhistmath;rgn=full%20text;idno=AAT3201.0001.001;didno=AAT3201.0001.001;view=pdf;seq=00000412

Answer by Artur for Awfully sophisticated proof for simple facts

2011-03-25T12:38:33Z

In 1993 D.Christodoulou and S.Klainerman proved Global Nonlinear Stability of the Minkowski Space. It was one of the most important results of the mathematical General Relativity. Their proof was published in an over 500-pages book. It stated, that any initial data "sufficiently close" to that corresponding to the Minkowski space will remain so forever. Most of mathematical physicists belived this frankly simple fact but the proof was one of the most sophisticated in the field.

Answer by Timothy Chow for Awfully sophisticated proof for simple facts

2011-05-05T18:52:45Z

There is a simple pigeonhole argument for the following fact, due to Erdős and Szekeres I believe:

In any sequence $a_1, a_2, \ldots, a_{mn+1}$ of $mn+1$ distinct integers, there must exist either an increasing subsequence of length $m+1$ or a decreasing subsequence of length $n+1$ (or both).

The "sophisticated" proof of this fact is that any Young tableau with $mn+1$ boxes must either have more than $m$ columns or more than $n$ rows, and so the result follows because the number of columns/rows corresponds to the length of the longest increasing/decreasing subsequence of the corresponding permutation under the Robinson--Schensted correspondence.

Answer by anonymous for Awfully sophisticated proof for simple facts

2011-05-05T19:37:11Z

Here's a topological proof that $\mathbb{Z}$ is a PID.

Let $p,q$ be relatively prime. Then the line from the origin to the point $(p,q)\in\mathbb{R}^2$ does not pass through any lattice point, and therefore defines a simple closed curve in the torus $\mathbb{T}=\mathbb{R}^2/\mathbb{Z}^2$. Cut the torus along this curve. By classification of surfaces, the resulting surface is a cylinder. Therefore, we can reglue it to get a torus, but where our simple closed curve is now a "stupid" such thing, i.e., a ring around the torus.

Which is all to say that in this case, there exists an automorphism of the torus which takes $(p,q)\in\mathbb{Z}^2=\pi_1(\mathbb{T})$ to $(1,0)$. But this gives a matrix $\begin{bmatrix} p & x \\ q & y \end{bmatrix}\in GL_2(\mathbb{Z})$ , so $py-qx\in\mathbb{Z}^{\times}$, i.e., $py-qx=\pm 1$.

The only two things this proof needs are the computation of the homology of a torus and the classification of surfaces, neither of which actually relies on $\mathbb{Z}$ being a PID!

Answer by Jan Weidner for Awfully sophisticated proof for simple facts

2011-05-05T20:22:28Z

Every finite dimensional complex representation of a finite cyclic group decomposes into a direct sum of irreducible representations. This can be deduced from the decomposition theorem for perverse sheaves as follows: It is enough to show that the group algebra is semi simple. To check this it is enough to lift the regular representation of $\mathbb Z/n$ to $\mathbb Z=\pi^1(\mathbb C^*)$ and show that it decomposes into a direct sum of irreducible representations of $\mathbb Z$.

Consider the covering $z \mapsto z^n$ of $\mathbb C^*$ by itself.

It is easy to see, that the monodromy action on the pushforward of the constant sheaf $\mathbb C[1]$ along this map coincides with the regular representation. On the other hand since the map is small, the decomposition theorem guarantees that the pushforward decomposes into a direct sum of IC complexes. Since our map is a covering and our space is smooth these are actually irreducible local systems on $\mathbb C^*$. But irreducible local systems correspond to irreducible representation of the fundamental group.

Answer by Jan Weidner for Awfully sophisticated proof for simple facts

2011-05-05T20:34:00Z

The space $C[0,1]$ is not reflexive. If it was, it also had a predual. But then it would be a von Neumann algebra. However von Neumann algebras correspond to very strange topological spaces which have the property that closures of open subsets are again open. Clearly this is not the case for $[0,1]$.

Answer by Manuel Araújo for Awfully sophisticated proof for simple facts

2011-06-05T14:45:35Z

Every finite integral domain is a field:

Let $D$ be a finite integral domain. Being finite, it is Artinian an Noetherian and therefore has Krull dimension zero. But $(0)$ is a prime ideal, because $D$ is a domain, therefore $(0)$ is a maximal ideal and $D$ is a field.

Answer by Bill Johnson for Awfully sophisticated proof for simple facts

2011-06-05T16:54:15Z

If $0\le f_n \le 1$ is a sequence of continuous functions on $[0,1]$ that converges pointwise to $0$, then $\int_0^1 f_n(t) dt $ converges to $0$. Understandable by freshman, the statement is hard to prove using only the tools of calculus but is immediate from the dominated convergence theorem.

Answer by MTS for Awfully sophisticated proof for simple facts

2011-06-15T13:32:28Z

One can use the continuous functional calculus of a C$^*$-algebra (namely $M_N(\mathbb{C})$) to prove that a normal matrix is diagonalizable.

Answer by godelian for Awfully sophisticated proof for simple facts

2011-06-19T19:17:26Z

There is an elementary problem that goes more or less like this: you have a special telephone keyboard with nine lighted buttons (one for each number from $1$ to $9$); when pushing each button other than number $5$ (the central button) then this switches the state of the lights of the button itself and of all its surrounding buttons; pushing number $5$ only switches the state of the lights of its surrounding buttons, but not of itself. Starting with all lights off, the question asks whether we can get all lights on by pushing buttons. The obvious solution to the negative answer relies on the fact that the parity of lighted buttons at every state of the keyboard is an invariant. But there is also a sophisticated solution.

Take the set $X$ of $9$ elements and think of $\mathcal{P}(X)$ as a vector space over the field $\mathbb{Z}_2$ with the sum being the symmetric difference and the product given by $0.v=\emptyset$ and $1.v=v$. Then we can identify each state of the keyboard with a corresponding vector in this space, while pushing the button $i$ corresponds to summing a special vector $v_i$ (associated to the button) to the vector representing the state of the keyboard. Thus, we are wondering if there are some scalars $\alpha_i$ such that $\sum_{i=1}^{9} \alpha_iv_i=X$. Writing each $v_i$ and $X$ in the base of the space given by the singleton elements $1, ..., 9$, we get a system of linear equations which can be seen to have no solutions by computing the $9 \times 9$ determinant and verifying it is null.

Answer by Benjamin Steinberg for Awfully sophisticated proof for simple facts

2011-07-09T00:17:00Z

Proving the Banach fixed point theorem for compact metric spaces using the structure of monothetic compact semigroups.

Thm. Let $X$ be a compact metric space and $f\colon X\to X$ a strict contraction, meaning $d(f(x),f(y))< d(x,y)$ for $x\neq y$. Then $f$ has a unique fixed point and for any $x_0\in X$, the iterates $f^n(x_0)$ converges to the fixed point. Pf. Contractions are clearly equicontinuous, so by the Arzelà–Ascoli theorem, the closed subsemigroup $S$ generated by $f$ is compact in the compact-open topology. Now, a monothetic compact semigroup has a unique minimal ideal $I$, which is a compact abelian group. Moreover, either $S$ is finite and $I$ consists of all sufficiently high powers of $f$ or $S$ is infinite and $I$ consists of all limit points of the sequence $f^n$. In either case, $I$ consists of strict contractions, being in the ideal generated by $f$. Thus the identity element $e$ of $I$ is a constant map, being an idempotent strict contraction. Thus $I={e}$, being a group. Thus $f^n$ converges to a constant map to some point $y$. Clearly $y$ is the unique fixed point of $f$.

Answer by Gil Kalai for Awfully sophisticated proof for simple facts

2011-08-29T09:36:35Z

Arrow's theorem is a basic result in social choice theory which has several simple proofs. (For three proofs see this paper: Three Brief Proofs of Arrow's Impossibility Theorem by J. Geanakoplos)

It also has a few complicated proofs: The paper by Tang, Pingzhong and Lin, Fangzhen Computer-aided proofs of Arrow's and other impossibility theorems, Artificial Intelligence 173 (2009), no. 11, 1041–1053. Gives an inductive proof based on rather complicted inductive step and a computerized check for the base case. The paper by Yuliy Baryshnikov, Unifying impossibility theorems: a topological approach. Adv. in Appl. Math. 14 (1993), 404–415, gives a proof based on algebraic topology. My paper: A Fourier-theoretic perspective on the Condorcet paradox and Arrow's theorem. Adv. in Appl. Math. 29 (2002), 412–426, gives a fairly complicated Fourier-theoretic proof but only to a special case of the theorem.

(A complicated proof to a related theorem is by Shelah, Saharon, On the Arrow property, Adv. in Appl. Math. 34 (2005), 217–251.)

Answer by Woett for Awfully sophisticated proof for simple facts

2011-10-13T03:28:07Z

An olympiad-type question I once tried to solve was: prove that all integers $>1$ can be written as a sum of two squarefree integers$^{[1]}$. The proof I came up with (which uses at least $3$ non-trivial results!) went as follows:

We can check that it holds for $n \le 10^4$. Now, let $S$ be the set of all squarefree integers, except for the primes larger than $10^4$. Then by the fact that the Schnirelmann density of the set of squarefree integers equals $\dfrac{53}{88}$ $^{[2]}$ and some decent estimate on the prime counting function$^{[3]}$, we have that the Schnirelmann density of $S$ must be larger than $\dfrac{1}{2}$. By Mann's Theorem$^{[4]}$ we now have that every positive integer can be written as sum of at most $2$ elements of $S$. In particular, every prime number can be written as sum of $2$ elements of $S$, and every integer that is not squarefree can be written as sum of $2$ elements of $S$. All there is now left, is proving the theorem for composite squarefree numbers; $n = pq = (p_1 + p_2)q = p_1q + p_2q$, where $p$ is the smallest prime dividing $n$ and $p_1, p_2$ are squarefree integers.

$^{[1]}$ http://www.artofproblemsolving.com/Forum/viewtopic.php?f=470&t=150908 $^{[2]}$ http://www.jstor.org/pss/2034736 $^{[3]}$ http://en.wikipedia.org/wiki/Prime-counting_function#Inequalities $^{[4]}$ http://mathworld.wolfram.com/MannsTheorem.html

Answer by Matthias Künzer for Awfully sophisticated proof for simple facts

2011-10-13T06:42:05Z

(1) Let $G$ be a finite group. Let $H\leqslant G$ be a subgroup of index $2$. Let us prove that $H$ is normal in $G$. Let $L|K$ be a Galois extension of fields with Galois group $G$ (easily constructed via a representation of $G$ as a permutation group, taking $L$ to be a function field in suitably many variables on which $G$ acts and $K$ to be the fixed field under $G$). Let $F$ be the fixed field in $L$ under $H$. Then $F|K$ is a quadratic extension, hence normal. By the Main Theorem of Galois Theory, it follows that $H$ is normal in $G$.

(2) Let $G$ be a finite group. Let $K$ be a finite field of characteristic not dividing $|G|$. Let us prove Maschke's Theorem in this situation: $KG$ is semisimple. Given two finite dimensional $KG$-modules $X$ and $Y$, it suffices to show that $\text{Ext}^1_{KG}(X,Y) = 0$. But $\text{Ext}^1_{KG}(X,Y) = \text{H}^1(G,\text{Hom}_K(X,Y)) = 0$, since $|G|$ and $|\text{Hom}_K(X,Y)|$ are coprime.

(Well, not sure whether any of these arguments are really awfully sophisticated. It's rather breaking a butterfly on a small wheel.)

Answer by none for Awfully sophisticated proof for simple facts

2011-10-14T07:27:32Z

Dan Bernstein, "A New Proof that 83 is prime", http://cr.yp.to/talks/2003.03.23/slides.pdf

Answer by Zsbán Ambrus for Awfully sophisticated proof for simple facts

2012-03-04T13:54:46Z

The following theorem has several essentially different proofs that need quite different levels of mathematical background, ranging from high school to graduate level. Which proof is most natural depends on who you ask, but many people (including me) will find at least some proof unnecessarily complicated.

There exists a set $ A $ that is everywhere dense on the square $ [0, 1]^2 $, but such that for any real number $ x $, the intersections $ A \cap (\{x\} \times [0, 1]) $ and $ A \cap ([0, 1] \times \{x\}) $ are both finite.

(This is a variant of a homework problem posed by Sági Gábor.)

Here's the idea of a few proofs.

$ A = \{(p/r, q/r) \mid p, q, r \in \mathbb{Z} \text{ and } \gcd(p,r) = \gcd(q,r) = 1 \} $ is dense because if you subdivide the square to $ 2^n $ times $ 2^n $ squares, $ A $ contains the center of each square; and has only as many points on each horizontal or vertical line as the denominator of $ x $.
$ A = \{(x + y\sqrt3, y - x\sqrt3) \mid x, y\in\mathbb{Q} \} $ is dense because it's a scaled rotation of $ \mathbb{Q}^2 $, but has at most one point on every horizontal or vertical line otherwise $ \sqrt3 $ would be rational.
Choose $ a_0, b_0, a_1, b_1 $ as four reals linear independent over rationals, this is possible because of cardinalities. $ A = \{(ma_0 + na_1, mb_0 + nb_1) \mid m, n \in \mathbb{Q}\} $ has no two points sharing coordinates because of rational independence, and $ A $ is dense because it's a non-singular affine image of $ \mathbb{Q}^2 $.
A is the set of a countably infinite sequence of random points independent and uniform on the square. This is almost surely dense, but almost surely has no two points that share a coordinate.
Choose a countable topological base of the square, then choose a point from each of its elements inductively such that you never choose a point that shares a coordinate with any point chosen previously.
Choose a continuum (or smaller) size topological base of the square, then choose a point from each by transfinite induction such that when you choose a point, the cardinality of points chosen previously is less than continuum, thus you can avoid sharing coordinates with those points.
Choose $ a, b $ as reals such that $ a, b, 1 $ are linear independent over rationals, possible because of cardinalities. Let $ A = \{((ma + nb) \bmod 1, (ma - nb) \bmod 1) \mid m, n \in \mathbb{Z}\} $ . No two points share coordinates because of rational independence. Looking on the torus, A is dense somewhere on the square and the difference of any two points of A is in A so it must be dense in the origin. As A is closed to addition, it must be dense on a line passing through the origin. As it's also closed to rotation by $ \pi/2 $, it's also dense on the rotation of that line, thus, because it's closed to addition, dense everywhere.
Choose $ a, b $ like above. Let $ A = \{(an \bmod 1, bn \bmod 1) \mid n \in \mathbb{Z}\} $ . Prove A is dense by ergodic theory and Fourier analysis.

Update: Edited the drafts of proofs to somewhat cleaner. Permuted proofs. Also fixed typo in last proof.

Answer by unknown (google) for Awfully sophisticated proof for simple facts

2012-08-12T10:02:54Z

The density Hales-Jewett theorem implies that there cannot exist perfect magic hypercubes of fixed side length $k$ and arbitrarily high dimension $n$ whose cells are filled with the consecutive numbers $1,2,\dots,k^n$ and for which the numbers in cells along any geometric line sum to the magic constant $\frac{k(k^n+1)}{2}$.

For, take the cells with numbers $ 1,2,\dots,\left\lfloor\frac{k^n}{2}\right\rfloor $ .

This always has density about $1/2$, and so by the density Hales-Jewett theorem, will contain a hyperline for sufficiently large $n$. But no $k$ numbers from this set of density about $1/2$ can ever sum to the magic constant.

Answer by unknown (google) for Awfully sophisticated proof for simple facts

2012-08-20T00:28:41Z

Not really sure if this should count, but: From Chebyshev's proof using the central binomial coefficient that there exists some constant $C>0$ such that

$$ \pi(x) < C\frac{x}{\log x} $$

for sufficiently large $x$, and from the infinitude of primes, we get that

$$ \log x \ll x. $$

Answer by Kjetil B Halvorsen for Awfully sophisticated proof for simple facts

2012-08-20T17:54:23Z

The Herbert Simon (Nobel Price Winner, Economics, 1978)--- Karl Egil Aubert Dispute, see

http://www.tandfonline.com/doi/abs/10.1080/00201748208601972

Aubert criticizes Simon for irrelevant use of mathematics for his "Application", but also for the fact that he uses the Brouwer fixed point theorem for a proof, when the Intermediate Value Theorem would be enough.

Answer by Pål GD for Awfully sophisticated proof for simple facts

2012-11-09T08:45:07Z

$Forest$ is in $P$. Given a finite undirected graph $G$ one can in polynomial time decide whether the input is a forest. The class of all finite forests is a minor-closed property and by the Robertson–Seymour theorem, there are finitely many forbidden minors. We can in $O(n^3)$ time test whether $G$ contains a forbidden minor and if not, output yes.

Answer by John Stalfos for Awfully sophisticated proof for simple facts

2012-11-09T14:32:52Z

This is quite late(and just a restatement of the regular proof in fancy terms), but I came around this while goofing off one day:

Theorem: Let $X$ a space, and $\mathscr{F}$ a sheaf of (not necessarily abelian) groups, and denote by $\pi$ the projection from the étalé space $Sp\acute{e}(\mathscr{F})$. Then $\Gamma(X,\mathscr{F})$ inject into $\mathrm{Aut}(\pi)$(taken in the category of spaces étalé over $X$).

Proof: Straightforward and not difficult(but there are a bunch of things to check).

Theorem: (Cayley's theorem) Let $G$ a finite group, then $G$ is a subgroup of a symmetric group.

Proof. Let $X$ a nonempty, connected topological space and take $\mathbb{G}$ the constant sheaf associated to $G$ on $X$. Apply previous theorem and notice that $Sp\acute{e}(\mathbb{G})$ is a globally trivial covering space, and homeomorphic(over $X$) to $\coprod_{|G|} X$, so that $G$ injects into the group of deck transformations of this covering space, which is just $\mathfrak{S}_{|G|}$!

Answer by Ramón Barral for Awfully sophisticated proof for simple facts

2012-11-09T19:32:14Z

Seen on http://legauss.blogspot.com.es/2012/05/para-rir-ou-para-chorar-parte-13.html

Theorem: $5!/2$ is even.

Proof: $5!/2$ is the order of the group $A_5$. It is known that $A_5$ is a non-abelian simple group. Therefore $A_5$ is not solvable. But the Feit-Thompson Theorem asserts that every finite group with odd cardinal is solvable, so $5!/2$ must be an even number.

Answer by ACL for Awfully sophisticated proof for simple facts

2012-12-21T09:02:02Z

Liouville remarked that the fundamental theorem of algebra could be derived from his theorem that elliptic functions (doubly periodic meromorphic functions of one complex variable) must have poles. The proof goes by substituting the inverse of a polynomial as the argument of, say, Weierstrass $\wp$-function with large enough periods, and observing that it has no poles.

Of course, the proof of Liouville's theorem on elliptic functions requires the same kind of arguments used for proving the famous Liouville theorem (due to Cauchy) that bounded holomorphic functions are bounded and, apparently, already used before by Cauchy for algebraic functions.

But Liouville's observation is really more complicated than the present proof. What it simplifies, however, is the compactness argument. For elliptic functions, or for algebraic functions, one has at hand a compact Riemann surface on which some holomorphic function is bounded, hence achieves its supremum, etc. This may be the reason why the general form of Liouville theorem came only after the case of algebraic or elliptic functions.

Answer by Pablo Zadunaisky for Awfully sophisticated proof for simple facts

2012-12-21T14:28:42Z

The skew-field of quaternions $\mathbb H$ is isomorphic to its opposite algebra.

Indeed, by a theorem of Frobenius, division algebras over the reals are isomorphic to either $\mathbb R, \mathbb C$ or $\mathbb H$. Since $\mathbb H^\mathsf{opp}$ is again a division algebra, it must be isomorphic to one of these. There are several ways to conclude: since it is four dimensional, or since it is not commutative, or since it has more than two square roots of $-1$, etc., we conclude that the only possibility is $\mathbb H \cong \mathbb H^\mathsf{opp}$.

If you are only interested in Morita equivalence between these two algebras, you can do better: the Brauer group of $\mathbb R$ is isomorphic to $\mathbb Z_2$, and so all elements are of order $2$. This implies that the class of $\mathbb H$ coincides with its inverse, which is the class of $\mathbb H^{\mathsf{opp}}$. Thus $\mathbb H$ and $\mathbb H^\mathsf{opp}$ are Morita equivalent.

Answer by Johannes Ebert for Awfully sophisticated proof for simple facts

2012-12-21T19:13:22Z

The fundamental theorem of algebra holds because:

For each degree $n$ normed polynomial $p$ over the complex numbers, there is an $n \times n$ matrix $A$ with characteristic polynomial $\pm p$.
We show that $A$ has an eigenvector.
We may assume that $0$ is not an eigenvalue of $A$ (otherwise $p(0)=0$), so $A \in GL_n (\mathbb{C})$.
$A$ induces a self-map $f_A$ of $CP^{n-1}$, and the eigenspaces of $A$ correspond to the fixed points of $f_A$; so we need to show that $A$ has a fixed point.
As $GL_n (\mathbb{C})$ is connected, $f_A$ is homotopic to the identity (this does not depend on the fundamental theorem of algebra; if $A \in GL_n (\mathbb{C})$, then $ z 1 + (1-z )A$ is invertible except for a finite number of values of $z$; and the complement of a finite set of points of the plane is path-connected (this follows, for example, from the transversality theorem).
The Lefschetz number of the identity on $CP^{n-1}$ equals $n\neq 0$, thus the Lefschetz number of $f_A$ is not zero.
By the Lefschetz fixed point theorem, $f_A$ has a fixed point.

Answer by Ron Maimon for Awfully sophisticated proof for simple facts

2012-12-30T02:09:50Z

Kn is non-planar for n>4: it contradicts the four-color theorem.

Answer by Benjamin Steinberg for Awfully sophisticated proof for simple facts

2012-12-30T04:58:58Z

Here is a Ramsey theory proof every finite semigroup has an idempotent. Let S be a finite semigroup with finite generating set A. Choose an infinite word $a_1a_2\cdots$ over A. Color the complete graph on 0,1,2... by coloring the edge from i to j with $i\lneq j$ by the image in S of $a_{i+1}\cdots a_j$. By Ramsey's theorem there is a monochromatic clique $i\lneq j\lneq k$. This means $$a_{i+1}\cdots a_j=a_{j+1}\cdots a_k=a_{i+1}\cdots a_k$$ is an idempotent.

This proof, generalized to larger clique sizes, actually shows any infinite word contains arbitrarily long consecutive subwords mapping to the same idempotent of S, which is used in studying automata over infinite words.

Answer by Martin Brandenburg for Awfully sophisticated proof for simple facts

2013-01-25T13:46:50Z

$5/2 = 2 \frac{1}{2}$ since both are the groupoid cardinality of the following action:

Thinking about this, it is actually quite enlightening. For more information, see the wonderful paper From Finite Sets to Feynman Diagrams by John Baez and James Dolan.

Answer by J. H. S. for Awfully sophisticated proof for simple facts

2013-02-05T08:18:33Z

I think that the following proof of the fact that every subgroup of index $2$ of a given group is normal might count too. When I first came up with it (sometime during my sophomore year), I believed that I had just found the entrance to a royal road to mathematics.

Let $H\leq G$ be such that $[G:H]=2$. We'll prove that every right coset of $H$ is equal to a left coset of $H$.

Since $[G:H]=2$, $G$ is both the union of two disjoint right cosets of $H$ and the union of two disjoint left cosets of $H$. Let us suppose that $G=He \cup Hx = eH \cup yH$ where $x,y\in G\setminus H$ and $e$ denotes the identity element of $G$. According to standard lore regarding the symmetric difference of sets,

$He \cup Hx = He \triangle Hx \triangle (He \cap Hx) = He \triangle Hx \triangle \emptyset = H \triangle (Hx\triangle \emptyset) = H\triangle Hx$

and

$eH \cup yH = eH \triangle yH \triangle (eH \cap yH) = eH \triangle yH \triangle \emptyset = H \triangle (yH \triangle \emptyset) = H \triangle yH$.

Therefore, $H\triangle Hx = H\triangle yH$. Canceling $H$ on both sides of the latter equality—which is perfectly valid given that $(2^G, \triangle)$ is a group—we conclude that $Hx=yH$. Done.

If you consider that the prior argument doesn't qualify as awfully sophisticated, there is still another fancy way to derive the result in question. As a consequence of P. Hall's famous marriage theorem, M. Hall proves in Theorem 5.1.7 of his Combinatorial Theory that if $H$ is a finite index subgroup of $G$, there exists a set of elements that are simultaneously representatives for the right cosets of $H$ and the left cosets of $H$ (once he's proven the said theorem, he adds: "Simultaneous right-and-left coset representatives exist for a subgroup in a variety of other circumstances. This problem has been investigated by Ore 1."). In the case $[G:H]=2$, this implies at once that every right coset of $H$ is equal to a left coset of $H$ and we are done...

Last but not least, $[G:H]=2 \Rightarrow H \trianglelefteq G$ in the case when $|G|<\infty$ can also be seen a consequence of the well-known fact according to which any subgroup of a finite group whose index is equal to the smallest prime that divides the order of the group is of necessity a normal subgroup of the group. B. R. Gelbaum showcases in one of his books an action-free proof of this fact. He attributes both the fact and the action-free proof to Ernst G. Straus. Does any of you know on what grounds he did so? I have a Xerox copy of the relevant page in the book here. This is exactly what Gelbaum writes therein:

At some time in the early 1940s Ernst G. Straus, sitting in a group theory class, saw the proof of the ... result [i.e., $[G:H]=2 \Rightarrow H \trianglelefteq G$] ... and immediately conjectured (and proved that night): ... IF G:H [sic] IS THE SMALLEST PRIME DIVISOR P of #(G) THEN H IS A NORMAL SUBGROUP.

P.S. The Galois-theoretic proof given by Matthias Künzer is just fabulous!

Answer by Alexander Gruber for Awfully sophisticated proof for simple facts

2013-02-23T01:41:41Z

This is kind of an elementary example, but I always thought it was funny to prove that $S_3$ is isomorphic to a subgroup of $S_6$ using Cayley's theorem.

Answer by Brendan McKay for Awfully sophisticated proof for simple facts

2013-02-24T07:35:39Z

The sum of the degrees of the vertices of a graph is even.

Proof: The number $N$ of graphs with degrees $d_1,\ldots,d_n$ is the coefficient of $x_1^{d_1}\cdots x_n^{d_n}$ in the generating function $\prod_{j\lt k}(1+x_jx_k)$. Now apply Cauchy's Theorem in $n$ complex dimensions to find that $$N = \frac{1}{(2\pi i)^n} \oint\cdots\oint \frac{\prod_{j\lt k}(1+x_jx_k)}{x_1^{d_1+1}\cdots x_n^{d_n+1}} dx_1\cdots dx_n,$$ where each integral is a simple closed contour enclosing the origin once. Choosing the circles $x_j=e^{i\theta_j}$, we get $$N = \frac{1}{(2\pi)^n} \int_{-\pi}^\pi\cdots\int_{-\pi}^\pi \frac{\prod_{j\lt k}(1+e^{\theta_j+\theta_k})}{e^{i(d_1\theta_1+\cdots +d_n\theta_n)}} d\theta_1\cdots d\theta_n.$$ Alternatively, choosing the circles $x_j=e^{i(\theta_j+\pi)}$, we get $$N = \frac{1}{(2\pi)^n} \int_{-\pi}^\pi\cdots\int_{-\pi}^\pi \frac{\prod_{j\lt k}(1+e^{\theta_j+\theta_k})}{e^{i(d_1\theta_1+\cdots +d_n\theta_n+k\pi)}} d\theta_1\cdots d\theta_n,$$ where $k=d_1+\cdots+d_n$. Since $e^{ik\pi}=-1$ when $k$ is an odd integer, we can add these two integrals to get $2N=0$.

Answer by practical for Awfully sophisticated proof for simple facts

2013-02-24T19:57:30Z

There exists transcendantal numbers because:

-- $x\mapsto \frac{1}{[{\mathbb Q}(x):{\mathbb Q}]}{\rm Tr}_{{\mathbb Q}(x)/{\mathbb Q}}x$ is a well defined, non zero, linear form from $\bar{\mathbb Q}$ to ${\mathbb Q}$.

-- The kernel of a non zero linear form form ${\mathbb R}$ to ${\mathbb Q}$ is not measurable.

-- By Solovay, every subset of ${\mathbb R}$ can be assumed to be measurable.

Conclusion: ${\mathbb R}\neq \bar{\mathbb Q}$.

Answer by Dietrich Burde for Awfully sophisticated proof for simple facts

2013-03-28T10:37:42Z

One can also show with Fermat's last theorem that $\sqrt{2}$ is irrational - the answer of mt did $2^{1/n}$ for $n\ge 3$. Suppose that $\sqrt{2}$ is rational. Then there is a right-angled triangle with rational sides $(a,b,c)=(\sqrt{2},\sqrt{2},2)$ and area 1. Hence $1$ would be a congruent number. This contradicts Fermat's last theorem with exponent $4$.

Answer by Wlodzimierz Holsztynski for Awfully sophisticated proof for simple facts

2013-05-11T04:02:26Z

Around year 1970 a popular way to compute cohomology groups of the finite cyclic groups was by applying spectral sequences (which was quite an overkill).

Answer by Dominic Michaelis for Awfully sophisticated proof for simple facts

2013-05-14T17:47:39Z

As Helfgott uploaded a proof of the weak Goldbach conjecture it is now possible (but I guess circular) to proof that there are infinity many primes in this way.

Suppose there are only finite many primes, let $p_{\max}$ be the highest prime number, then \[ 3 p_{\max}+2 \] would be an odd number which is not the sum of 3 primes in contradiction to goldbachs weak conjecture.

Answer by Toink for Awfully sophisticated proof for simple facts

2013-05-14T20:03:11Z

Claim: $\sum\limits_{k=0}^n (-1)^k {n\choose k} = 0$ for all integers $n≥1$

Proof: Take the $n-1$-dimensional simplex $\Delta_{n-1}$. We can compute it's Euler characteristic by using simplicial homology. There are exactly $n \choose k+1$ many $k$-sub-simplexes of $\Delta_{n-1}$. Thus we get a simplicial chain complex of the form $\mathbb{Z}^{n\choose n} \to \mathbb{Z}^{n\choose n-1} \to \cdots \to \mathbb{Z}^{n\choose 2}\to\mathbb{Z}^{n\choose 1}$. So the Euler characteristic is $\chi(\Delta_{n-1}) = \sum\limits_{k=0}^{n-1} (-1)^k {n\choose k+1}=-\sum\limits_{k=1}^{n} (-1)^k {n\choose k}$
On the other hand $\Delta_{n-1}$ is contractible, and $\chi$ is homotopy-equivalence-invariant, so $\chi(\Delta_{n-1})=\chi(pt) =1$.
Putting those toghether we obtain: $0=\chi(\Delta_{n-1})-\chi(\Delta_{n-1})=1+\sum\limits_{k=1}^{n} (-1)^k {n\choose k}=\sum\limits_{k=0}^n (-1)^k {n\choose k}$