\binom{5}{2} Proofs that \binom{n}{k} ≤ \binom{n}{k+1} if k < n/2.

Doron ZEILBERGER^{1}

There is no trivial mathematics, there are only trivial mathematicians! A mathematician is trivial if he or she believes that there exists trivial mathematics. But this is not the only way for a mathematician to be trivial. Another sufficient condition for a mathematician to be trivial is not to show up to a colloquium talk with such an intriguing title and abstract! Conversely, if you do show up, you are definitely non-trivial, so, congratulations, dear audience, you are the (only!) 19 non-trivial mathematicians in Columbia University.

There are (at least) two ways to define \binom{n}{k}. One way is

\binom{n}{k} := \frac{n!}{k!(n − k)!} .

By this definition, the statement of the title is indeed trivial:

\begin{matrix}( & {n \atop k} & ) \\ ( & {n \atop k+1} & ) \end{matrix} = \frac{\frac{n!}{k!(n−k)!}}{\frac{n!}{(k+1)!(n−k−1)!}} = \frac{(k + 1)!(n − k − 1)!}{k!(n − k)!} = \frac{k + 1}{n − k} ≤ 1 ,

if k < n/2.

The other definition is a combinatorial one. \binom{n}{k} is the number of ways of choosing a set of k members out of an n-element set. It is also the number of n-letter words in the alphabet {Street, Avenue} with exactly k occurrences of the “letter” “Street”, as I was reminded when I walked, earlier today, in the (real!, not proverbial) Manhattan lattice from Pennsylvania Station to the Columbia campus (except I confess that I cheated, and walked most of the way on Broadway). If you adapt the latter definition, then \binom{n}{k} = \frac{n!}{k!(n−k)!} becomes a theorem, that can be proved, e.g., by proving that both sides satisfy the recurrence (and initial condition)

f(n, k) = f(n − 1, k − 1) + f(n − 1, k) , f(0, k) = δ_{0,k} .

(If after you walked n blocks, you are currently at the corner of k-th Street and (n − k)-th Avenue, then one block earlier, you were either at the corner of (k − 1)-th Street and (n − k)-th Avenue or the corner of k-th Street and (n − k − 1)-th Avenue).

1 Department of Mathematics, Rutgers University (New Brunswick), Hill Center-Busch Campus, 110 Frelinghuysen

Rd., Piscataway, NJ 08854-8019, USA. zeilberg at math dot rutgers dot edu , http://www.math.rutgers.edu/~zeilberg/ . March 4, 2010. An almost verbatim transcript of a mathematics Colloquium talk delivered at Columbia University (Mathematics 520), Feb. 17, 2010, 5:00-6:00pm EST. I would like to thank Mikhail Khovanov for inviting me, and the nineteen brave souls who attended (for example Walter Neumann). I also would like to express my great disappointment at the 71 (faculty)+58 (grad students)-19=110 people (for example David Bayer) who did not attend. Supported in part by the NSF. Exclusively published in the Personal Journal of Ekhad and Zeilberger (http://www.math.rutgers.edu/~zeilberg/pj.html), and arxiv.org.

In particular, it follows that \frac{n!}{k!(n−k)!} is always an integer!, which is not so obvious, (since this is a ratio of two integers, that morally should be a fraction, unless some miracle occurs), and that it is less than 2^{n}. The special case that

\frac{(2n)!}{n!^{2}} = \frac{(2n)(2n − 1) ⋯ (n + 1)}{(1) ⋯ (n)}

is an integer, and that it is less than 2^{2n}, has an enormous number-theoretical significance. It was used by Chebychev, in 1851, to “almost” prove the Prime Number Theorem. Even though this breakthrough was “superseded” by the full Prime Number Theorem, first proved at the end of the 19th-century, all the proofs of the latter, as well as the later elementary proofs of Erdős and Selberg, use Chebychev’s result as a stepping-stone for the stronger statement. More recently, it turned out to be crucial in the amazing Agrawal-Kayal-Saxena[AKS] P RIMES ∈ P proof.

Let’s recall Chebychev’s argument. Since people today are so specialized, I am willing to bet that many of you have never seen it before. Only this gem is worth the admission fee of this talk (which is an hour of your precious time, that at least % 80 of the Columbia faculty and graduate students found too exorbitant.)

Let’s look at all the prime numbers between n and 2n. They must all divide the integer \frac{(2n)!}{n!^{2}}, so

∏_{n≤p≤2n} p ≤ 2^{2n} .

Now take log of both sides, define θ(x) := ∑_{p≤x} log p, and you would get that

θ(2n) − θ(n) ≤ (2n) log 2 ,

that implies

θ(n) − θ(n/2) ≤ (n) log 2
θ(n/2) − θ(n/4) ≤ (n/2) log 2
… .

Adding these up, you get that θ(n) ≤ (2 ln 2)n, which is equivalent to π(x) ≤ C \frac{x}{log x} for C = 2 ln 2 = 1.386 . . .. Later Chebychev made C even smaller, and Sylvester got very close to 1, and analogously for lower bounds, but the full Prime Number Theorem had to wait for Hadamard and de la Vallée Poussin, in 1896.

Going back to proving (and reproving) that \binom{n}{k} ≤ \binom{n}{k+1} if k < n/2, here is an inductive proof.

\binom{n}{k} = \binom{n − 1}{k − 1} + \binom{n − 1}{k} ≤ \binom{n − 1}{k} + \binom{n − 1}{k + 1} = \binom{n}{k + 1} ,

by the induction hypothesis, (provided that the hypothesis is fulfilled!). This is always true for \binom{n−1}{k−1} ≤ \binom{n−1}{k}, since k < n/2 implies k − 1 < (n − 1)/2, but for \binom{n−1}{k} ≤ \binom{n−1}{k+1} it may happen that k < n/2 but k ≥ (n − 1)/2. This happens exactly when n = 2k + 1, and for this special case we have to separately prove:

\binom{2k + 1}{k} ≤ \binom{2k + 1}{k + 1} ,

but this follows from the even stronger fact that:

\binom{2k + 1}{k} = \binom{2k + 1}{k + 1} ,

by the symmetry of the binomial coefficients.

I admit that this is an ugly duckling of a proof (manipulatorics, induction), but by carefully tracing it, we can get a beautiful swan of a proof, by defining an explicit injection that maps, in a canonical way, an n-letter word in the alphabet {S, A} with k S’s to one with k + 1 S’s. Simply look at the last time the number of Avenues exceeded the number of Streets by exactly one, and swap Avenues and Streets, until then, and leave the rest intact.

The proof that I just gave is an example of a combinatorial proof, and the process of finding a combinatorial interpretation to an algebraic identity or inequality, using a bijection and injection respectively, is called Combinatorization. Often algebraic/inductive proofs can be “traced” and converted to beautiful bijective or injective proofs, like in the above case.

Speaking of combinatorization, this is the grand-daddy of a more recent trend, called categorification, made popular by master-blogger John Baez. Categorification became a household name when my host, Mikhail Khovanov[Kh], in 2000, astounded the mathematical world by categorifying the famous Jones polynomials, by replacing a boring polynomial by an exciting cell-complex. I strongly recommend Dror Bar-Natan’s ([B]) very lucid exposition of Khovanov’s seminal ideas, that you can easily find in arxiv.org.

Going back to combinatorics, we will meet other, even better, combinatorial proofs, later on, but let me now present to you yet another algebraic proof. Using the Zeilberger algorithm[Z2] (or otherwise^{2}), we can find the generating function

P_{n}(x) := ∑_{k=0}^{n} \binom{n}{k}x^{k} = (1 + x)^{n} ,

and another way of stating that \binom{n}{k} ≤ \binom{n}{k+1} if k < n/2 is to say that the coefficients of P_{n}(x) = (1 + x)^{n} first go up and then go down. Such a polynomial is called unimodal. In our case it is also symmetric. Let’s call a symmetric and unimodal polynomial, with non-negative integer coefficients, a Z-polynomial, and let’s call the darga of P its low-degree plus its (high)-degree. For example, the darga of x^{4} + x^{5} is 9 while the darga of x^{3} is 6.

2 Algebra is really combinatorics in disguise, when you expand (1+x)^{n} you make n independent decisions, whether

to pick the 1 or the x. The coefficient of x^{k} is the the number of ways of choosing which k of the n terms will donate its x to the common cause.

The following two simple facts (taken from my de-combinatorization ([Z1]) of Kathy O’Hara’s ([O]) seminal combinatorial proof of the unimodality of the Gaussian polynomials) are easily proved.

Fact 1: The sum of two Z-polynomials of the same darga is another Z-polynomial of that darga.

Fact 2: The product of two Z-polynomials is yet-another-one, and its darga is the sum of their dargas.

To prove Fact 2 note that the additive “atoms” of Z-polynomials are polynomials of the form

x^{i} + x^{i+1} + ... + x^{j}

and multiplying out two such atoms would yield

(x^{a}+x^{a+1}+...+x^{b})(x^{c}+x^{c+1}+...+x^{d}) = x^{a+c}+2x^{a+c+1}+3x^{a+c+2}+…+3x^{b+d−2}+2x^{b+d−1}+x^{b+d} ,

which is indeed a Z-polynomial of darga (a + c) + (b + d) = (a + b) + (c + d).

It follows immediately, by induction, that (1 + x)^{n} is a Z-polynomial, since 1 + x is. But we get, for the same price, that many other polynomials are Z-polynomials, and hence automatically unimodal. For example

(x + x^{2} + x^{3} + x^{4} + x^{5} + x^{6})^{n} ,

which has the following probabilistic interpretation. You roll a fair die n times and at each roll you win as many dollars as the number of dots that show up. Then you are more likely to win k + 1 dollars than k dollars as long as k is less than your expected gain 7n/2.

More generally:

(1 + x)^{m}(x + x^{2} + x^{3} + x^{4} + x^{5} + x^{6})^{n}(x + x^{2} + x^{3} + x^{4})^{k} ,

that also has a gambling interpretation, and many more complicated gambling scenarios, that you are welcome to make up.

Let’s take a closer look at the above combinatorial proof that \binom{n}{k} ≤ \binom{n}{k+1}, that consisted in defining an explicit injection between k-sets to (k + 1)-sets. It inputs a set S with k elements (k < n/2) and outputs a set with one more element by looking at the smallest integer r such that |S ∩ {1, 2, …, r}| = (r − 1)/2 and mapping it to the set ({1, …, r}\S) ∪ (S ∩ {r + 1, …, n}).

For example, with n = 11 and k = 4 the 4-set {1, 2, 4, 11} is mapped to the 5-set {3, 5, 6, 7, 11} (in this example r = 7). Note that for this injection the output-set does not contain the input set. It would be more desirable, and natural, if we could come-up with an injection S → S′ from the collection of k-sets to the collection of k + 1-sets that has the property that S ⊂ S′, in other words, find a “rule” that adds a new member to S, as long as k < n/2, and in such a way that no two different Ss would give the same S′.

If there would be such a mapping we would get, by iterating it, a maximal chain that ends at the middle rank. By symmetry, if we reflect this to the complement, we would get a central chain decomposition of the Boolean lattice (alias n-dimensional unit cube). Conversely, any such chaindecomposition of the Boolean lattice would give such an injection, and would yield yet-another-proof of the unimodality of the binomial coefficients.

The easiest way to construct such a chain decomposition is recursively. Take any symmetric chain of B_{n−1}

C_{r} → C_{r+1} → … → C_{n−1−r} ,

and construct two new chains in B_{n}. The first is the same

C_{r} → C_{r+1} → … → C_{n−1−r} ,

but viewed as belonging to B_{n}, and the second is

C_{r} ∪ n → C_{r+1} ∪ n → … → C_{n−1−r} ∪ n .

There is only one problem! Neither chains are legitimate symmetric chains in B_{n}. The sum of the starting rank and ending rank (in the case of the Boolean lattice, the rank of a set is its number of elements) should be n, whereas the first chain has that sum too low, namely n−1, while the second chain has that sum too high, namely n + 1. To get two new chains that are just right, we cut the last member of the second chain and put it at the end of the first chain, getting the two chains:

C_{r} → C_{r+1} → … → C_{n−1−r} → C_{n−1−r} ∪ n ,

and

C_{r} ∪ n → C_{r+1} ∪ n → … → C_{n−2−r} ∪ n .

Let’s illustrate this construction for n ≤ 3. For n = 1 we only have one chain, namely:

∅ → {1} .

This gives rise to two chains for n = 2:

∅ → {1} → {1, 2} , {2} .

The first of these gives rise to two chains for n = 3:

∅ → {1} → {1, 2} → {1, 2, 3} , {3} → {1, 3} ,

while the singleton chain {2} only gives rise to one chain (the second one is empty)

{2} → {2, 3} .

Martin Aigner came up with another way of constructing a symmetric chain decomposition for the Boolean lattice B_{n}, that may be termed lexicographic greed. Start with the empty set, and at each level look at the lexicographically first set that has not yet been committed and that contains the current tail of the emerging chain. Keep doing it until you get stuck. Surprisingly, you get a symmetric chain decomposition. Why?, because it happens to be the same as the one above. So even though many people would find Aigner’s construction more elegant and appealing, the easiest way to prove its validity is to discover the recursive construction above and then it is easy to prove by induction that it is indeed the same.

The drawback that both the recursive and Aigner’s([A]) lexicographic-greed approaches share is that you have to construct all chains, and find out how the injection acts on all sets, at once, requiring exponential time and space. What if you only care about the successor of just one individual set? Curtis Greene and Daniel Kleitman[GK] came up with a very elegant description of (essentially the same!) injection.

There is a one-to-one mapping between sets and words in the alphabet {[,]}. For any set S of n elements form the “word” (w_{1}, …, w_{n}) by the rule w_{i} = [ iff i ∈ S. For example, the empty set for n = 4 corresponds to the word ]]]] and the whole set {1, 2, 3, 4} corresponds to the word [[[[.

If you have a legal bracketing then it forms its own singleton chain. Otherwise, “compile” it to the best of your ability, matching a left-bracket “[” with a right one “]”. Once you have finished “compiling” you would get a bunch of ]’s followed by a bunch of [’s which is as illegal as it gets, possibly (and usually) interspersed with clusters of legal bracketings. Leave these legal bracketings alone, and change the last ] by a [. In symbols:

L_{1} ] L_{2} ] L_{3} ] … L_{k−1} ] L_{k} ] L_{k+1} [ L_{k+2} [ … [ L_{r}

goes to

L_{1} ] L_{2} ] L_{3} ] … L_{k−1} ] L_{k} [ L_{k+1} [ L_{k+2} [ … [ L_{r} .

If you can’t do it (i.e. k = 0), then the chain ends.

The existence of a symmetric chain decomposition for the Boolean lattice immediately implies Sperner’s theorem that the largest possible collection of subsets of {1, 2, … n} such that none of its members properly contains another one (what is called an anti-chain, or clutter) equals \binom{n}{[n/2]}. Obviously, this is sharp, since the collection of all [n/2]-sets , that has \binom{n}{[n/2]} members, is obviously an anti-chain. Can you do better? Of course not! The number of symmetric chains in any symmetric chain decomposition of the Boolean lattice (and we know that one exists) equals \binom{n}{[n/2]}, since each chain passes once through the middle-rank, and every set belongs to exactly one chain. Given any anti-chain, there can be at most one-set-per-chain, or else it would not be an anti-chain!

While the above proof of Sperner’s theorem is my personal favorite, let me remind you of another, just-as-nice proof from the book, due to David Lubell[L].

There are n! possible chains that start at the top, the empty set, and end-up at the bottom ({1, 2, … n}) (in obvious one-one correspondence with permutations). Let C be a potential antichain. For each S ∈ C, there are exactly |S|!(n − |S|)! such top-to-bottom chains that pass through S, and of course, no two different members of C can share such a top-to-bottom chain, or else they would be related!

So we have the obvious inequality

∑_{S∈C} |S|!(n − |S|)! ≤ n! ,

that implies that

∑_{S∈C} \frac{1}{\binom{n}{|S|}} ≤ 1 .

But the maximum of \binom{n}{|S|} is \binom{n}{[n/2]} (thanks to the main theorem of the present article!), so the minimum of 1/\binom{n}{|S|} is 1/\binom{n}{[n/2]}, and we have

\frac{|C|}{\binom{n}{[n/2]}} ≤ 1 ,

as claimed.

The Last (and longest! (yet the best!)) Proof

We have already presented above several combinatorial proofs of

\binom{n}{k} ≤ \binom{n}{k + 1} , if k < n/2 ,

by finding a set-theoretical injection between the collection of k-sets and the collection of (k + 1)- sets, i.e. between two sets (of sets).

In general, a combinatorial proof of

a ≤ b

consists of constructing sets A and B such a = |A| and b = |B|, and an injection

f : A → B .

But, there is yet another way, a linear-algebra proof! Come-up with two vector spaces A and B such that dim(A) = a and dim(B) = b and construct a linear transformation

T : A → B ,

and prove that T is an injection by proving that for any f ∈ A, T f = 0 implies f = 0.

Let V_{k} be the vector space spanned by all k-subsets of {1, …, n}, in other words the vector space of all “formal sums” (as they would say in algebraic topology)

∑_{|S|=k} a_{S}S ,

where a_{S} are members of your favorite field (say the field of rational numbers, or even GF (p) for any prime p larger than n).

Our proposed mapping, M : V_{k} → V_{k+1}, soon to be proved an injection, is defined on basis elements by

M(S) = ∑_{j∉S}(S ∪ j) ,

and extended linearly. What is the “meaning” of M(S)? Suppose that you enlarge your current faculty S by another member, and you can’t decide, and you want to hire everyone who is not already in S but you are only allowed to hire one person. If you live in a classical world, you would have to make-up your mind, make one new professor happy, but disappoint all the other applicants. But in the quantum world, you can have a “superposition” of all scenarios for “hiring an extra professor”.

In order to prove that M is indeed an injection, we need a “companion operator”: L : V_{k} → V_{k−1}, defined on basis elements by:

L(S) = ∑_{i∈S}(S\i) ,

and extended linearly. L(S) has an analogous meaning in a quantum world. Because of budget cuts, you have to fire one professor, but you don’t want to get anyone upset, so you have a quantumsuperposition of all firing-one-professor scenarios.

I now claim that on V_{k},

ML − LM = μ(k)I ,

where μ(k) = 2k − n (n is fixed through this proof), and I is the identity mapping. Of course, by linearity, it is enough to prove this for basis elements S ∈ V_{k}:

ML(S) − LM(S) = μ(k)S .  (1)

ML(S) is formal sum of all scenarios of fire-and-then-hire while LM(S) is the formal sum of all scenarios of hire-and-then-fire. If the guy you hired and the guy you fired are different then “hire-Smith-then-fire-Jones” yields the same set as “fire-Jones-then-hire-Smith” and so they cancel out. The only scenarios that do not cancel out are those where the guy you fired and the guy you hired are one and the same. There are k ways to fire-and-then-hire the same person, and there are n − k ways to hire-and-then-fire the same person, at each case resulting in the original set S. This gives a net contribution of k − (n − k) = 2k − n copies of S.

Next I claim that on V_{k}, for any r ≥ 1

ML^{r} − L^{r}M = (μ(k) + … + μ(k − r + 1))L^{r−1} .  (2)

This follows easily by induction on r, by using

ML^{r+1} − L^{r+1}M = (ML^{r} − L^{r}M)L + L^{r}(ML − LM) .

So, if f ∈ V_{k}, we have:

(ML^{r+1} − L^{r+1}M)f = (ML^{r} − L^{r}M)(Lf) + L^{r}(ML − LM)f .  (3)

Since Lf ∈ V_{k−1}, we have from the induction hypothesis that the first term on the right side of (3) equals

(ML^{r} −L^{r}M)(Lf) = (μ(k −1)+…+μ(k −r))L^{r−1}(Lf) = (μ(k −1)+…+μ(k −r))L^{r}f , (3a)

and since (ML − LM)f = μ(k)f, the second term of (3) is

L^{r}(ML − LM)f = L^{r}μ(k)f = μ(k)L^{r}f .  (3b)

Incorporating (3a) and (3b) into (3), we get:

(ML^{r+1} − L^{r+1}M)f = (ML^{r} − L^{r}M)(Lf) + L^{r}(ML − LM)f =
(μ(k − 1) + … + μ(k − r))L^{r}f + μ(k)L^{r}f = (μ(k) + … + μ(k − r))L^{r}f ,

that is (2) with r replaced by r + 1.

Now suppose that there is an f ∈ V_{k} such that Mf = 0. We have to prove that f = 0. By (2) we have that

ML^{r}f = (μ(k) + … + μ(k − r + 1))L^{r−1}f .

Applying M^{r−1} to both sides gives

M^{r}L^{r}f = (μ(k) + … + μ(k − r + 1))M^{r−1}L^{r−1}f .

Iterating gives:

M^{r}L^{r}f = (μ(k) + … + μ(k − r + 1))(μ(k) + … + μ(k − r + 2)) ⋯ (μ(k))f .

So we have

M^{k+1}L^{k+1}f = (Non − Zero − Number)f .

Since f ∈ V_{k}, L^{k}f is a multiple of the empty set, and hence L^{k+1}f = 0 (L∅ = 0, since in that case we get the empty sum in the definition of L∅). So we get that f = 0, as promised.

So indeed, if k < n/2, the mapping M : V_{k} → V_{k+1} is an injection, and we get dim(V_{k}) ≤ dim(V_{k+1}), and so, once again, we know that \binom{n}{k} ≤ \binom{n}{k+1} if n < k/2.

But why work so hard, if we had the former far easier proofs? One reason, is why not? Who said that an elegant proof has to be short? Another reason is that this proof extends, almost verbatim, to other lattices, for which no simple proofs of rank-unimodality and the Sperner property are known. The proof that I just presented was inspired by, and is along similar lines as-but not quite the same- as Robert Proctor’s ([P]) beautiful simplification of Richard Stanley’s([St]) seminal proof of the Sperner property for lattices of integer partitions. The main part in the proof of Spernerity, proving that M is injective (as we just did), can be traced, in an almost equivalent form (but using differential operators operating on so-called semi-invariants) to James Joseph Sylvester[Sy] way back in 1878.

References

[A] Martin Aigner, Lexicographic matching in Boolean algebras, J. Combinatorial Theory Ser. B 14 (1973), 187–194.

[AKS] Manindra Agrawal, Neeraj Kayal and Nitin Saxena, PRIMES is in P, Annals of Mathematics, Second Series 160 (2004), 781-793 .

[B] Dror Bar-Natan, On Khovanov’s categorification of the Jones polynomial, Algebr. Geom. Topol. (electronic) 2 (2002), 337–370 .

[GK] Curtis Greene and Daniel J. Kleitman, Proof techniques in the theory of finite sets, Studies in combinatorics, MAA Stud. Math. 17, (G. -C. Rota, ed.), Math. Assoc. America, Washington, D.C., 1978

[K] Mikhail Khovanov, A categorification of the Jones polynomial, Duke Math. J. 101 (2000), 359–426 .

[L] David Lubell, A short proof of Sperner’s lemma, J. Combinatorial Theory 1 (1966), 299.

[O] Kathleen M. O’Hara, Unimodality of Gaussian coefficients: a constructive proof, J. Combin. Theory Ser. A 53 (1990), 29–52.

[P] Robert A. Proctor, Solution of two difficult combinatorial problems with linear algebra, Amer. Math. Monthly 89(1982), 721–734

[St] Richard P. Stanley, Weyl groups, the hard Lefschetz theorem, and the Sperner property, SIAM J. Algebraic Discrete Methods 1(1980), 168–184.

[Sy] James Joseph Sylvester, Proof of the hitherto undemonstrated fundamental theorem of invariants, Philosophical Magazine 5 (1878), 178-188. (Also in Collected Works v. 3, 117-126, Chelsea, NY, 1973).

[Z1] Doron Zeilberger, A one-line high school algebra proof of the unimodality of the Gaussian polynomials [_{k}^{n}] for k < 20. In: q-series and partitions (Minneapolis, MN, 1988), D. Stanton, ed., 67–72, IMA Vol. Math. Appl., 18 (1989), Springer, New York.

[Z2] Doron Zeilberger, The method of creative telescoping, J. Symbolic Comput. 11 (1991), 195–204.