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Proofs of Integrality of Binomial Coefficients

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Proofs of Integrality of Binomial Coefficients PROOFS OF INTEGRALITY OF BINOMIAL COEFFICIENTS KEITH CONRAD 1. Introduction The binomial coefficients are the numbers n n! n(n − 1) ··· (n − k + 1) (1.1) := = k k!(n − k)! k! for integers n and k with 0 ≤ k ≤ n. Their name comes from their appearance as coefficients in the binomial theorem n X n (1.2) (x + y)n = xkyn−k: k k=0 n but we will use (1.1), not (1.2), as their definition. While k is, by (1.1), obviously a n positive rational number, it is not at all obvious from (1.1) that k is an integer. n Theorem 1.1. For all integers n and k with 0 ≤ k ≤ n, 2 Z. k We will give six proofs of Theorem 1.1 and then discuss a generalization of binomial coefficients called q-binomial coefficients, which have an analogue of Theorem 1.1. 2. Proof by Combinatorics Our first proof will be a proof of the binomial theorem that, at the same time, provides a combinatorial meaning of binomial coefficients. By the distributive law we can write n n X k (1 + x) = (1 + x)(1 + x) ··· (1 + x) = cn;kx ; | {z } n terms k=0 where cn;k is the number of ways of picking k items out of n items (in k of the factors 1 + x n of (1 + x) pick x and in the other factors pick 1). The numbers cn;k, by the description we just gave of them (\the number of ways :::"), are positive integers. Clearly cn;0 = cn;n = 1. To get a formula for cn;k, we have (1 + x)n+1 = (1 + x)(1 + x)n n X k = (1 + x) cn;kx k=0 n n X k X k+1 = cn;kx + cn;kx k=0 k=0 n X k n+1 = 1 + (cn;k + cn;k−1)x + x ; k=1 1 2 KEITH CONRAD and differentiation of both sides gives us n n X k−1 n (n + 1)(1 + x) = k(cn;k + cn;k−1)x + (n + 1)x : k=1 Equating coefficients of xk−1 on both sides for k = 1; : : : ; n − 1, we get n − (k − 1) (2.1) (n + 1)c = k(c + c ) =) c = c : n;k−1 n;k n;k−1 n;k k n;k−1 Iterating this a second time, (n − (k − 1))(n − (k − 2)) c = c : n;k k(k − 1) n;k−2 Continuing, we eventually get (n − k + 1)(n − k + 2) ··· (n − k + k) (n − k + 1)(n − k + 2) ··· n n c = c = = ; n;k k(k − 1) ··· (1) n;k−k k! k n so k is a positive integer. (There are other ways to show cn;k = n!=k!(n−k)! by counting.) 3. Proof by Recursion Binomial coefficients are determined by the Pascal's triangle recursion, illustrated below. 0 1 0 1 1 1 1 0 1 2 2 2 1 2 1 0 1 2 1 3 3 1 3 3 3 3 0 1 2 3 1 4 6 4 1 4 4 4 4 4 0 1 2 3 4 n n Algebraically, the recursion is 0 = n = 1 for n ≥ 0, and n n − 1 n − 1 (3.1) = + : k k − 1 k for 1 ≤ k ≤ n. The reader can check directly that (1.1) satisfies this recursion. For k ≤ n − 1 we can apply the recursion to the second term on the right in (3.1), getting n n − 1 n − 2 n − 2 = + + : k k − 1 k − 1 k Continuing, n n − 1 n − 2 n − 3 k − 1 = + + + ··· + k k − 1 k − 1 k − 1 k − 1 n−k X n − 1 − j = k − 1 j=0 n−1 X j (3.2) = : k − 1 j=k−1 This says each entry in Pascal's triangle is not just the sum of the two entries directly above it, but is the sum of the entries in all previous rows shifted over to the left by 1 position. 4 a The diagram below illustrates how 2 (where k = 2) is the sum of 1 for a = 1; 2; 3. PROOFS OF INTEGRALITY OF BINOMIAL COEFFICIENTS 3 1 1 1 1 1 1 1 11 1 2 1 121 121 1331 1331 1331 1 4641 1 4641 1 4641 n Since k is a sum of binomial coefficients with denominator k − 1, if all binomial coeffi- cients with denominator k−1 are in Z then so are all binomial coefficients with denominator n k, by (3.2). Thus the integrality of all k is proved by induction since it is clear when k = 0. 4. Proof by Calculus For jxj < 1 we have the geometric series expansion 1 X = 1 + x + x2 + x3 + ··· = xk: 1 − x k≥0 There is no obvious connection between this and binomial coefficients, but we will discover one by looking at the series expansion of powers of 1=(1 − x). For m ≥ 1, m 1 1 X = = (1 + x + x2 + x3 + ··· )m = a xk; (1 − x)m 1 − x m;k k≥0 + where am;k 2 Z by the way power series (like polynomials) multiply. Working out the first few powers of 1=(1 − x), we get the expansions shown below. 1=(1 − x) = 1 + x + x2 + x3 + x4 + x5 + ··· 1=(1 − x)2 = 1 +2 x +3 x2 +4 x3 +5 x4 + ··· 1=(1 − x)3 =1+3 x +6 x2 + 10x3 + 15x4 + ··· 1=(1 − x)4 =1+4 x + 10x2 + 20x3 + 35x4 + ··· 1=(1 − x)5 =1+5 x + 15x2 + 35x3 + 70x4 + ··· There are binomial coefficients along diagonals. For instance, the coefficients of degree 0, 1, 2, 3, 4 in 1=(1 − x)5; 1=(1 − x)4;:::; 1=(1 − x) are (tilt your head left) 1, 4, 6, 4, 1. This suggests trying to describe am;k as an explicit binomial coefficient. Then, since we already know each am;k is an integer, we will get a proof that binomial coefficients are integers. (k) m −m Taylor's formula says am;k = f (0)=k! where f(x) = 1=(1−x) = (1−x) . Successive differentiation of f(x) = (1 − x)−m with the power rule and chain rule gives f 0(x) = (−m)(1 − x)−m−1(−1) = m(1 − x)−m−1; f 00(x) = (−m − 1)m(1 − x)−m−2(−1) = (m + 1)m(1 − x)−m−2; f 000(x) = (−m − 2)(m + 1)m(1 − x)−m−3(−1) = (m + 2)(m + 1)m(1 − x)−m−3; and in general for k ≥ 0 f (k)(x) = (m + k − 1) ··· (m + 2)(m + 1)m(1 − x)−m−k; so f (k)(0) = (m + k − 1) ··· (m + 2)(m + 1)m. Thus (m + k − 1) ··· (m + 1)m m + k − 1 a = =! : m;k k! k 4 KEITH CONRAD n m+k−1 Now if we start with integers n ≥ k ≥ 0, let's make k be k by using m = n−k+1. n + This is a positive integer since n ≥ k. Therefore k = an−k+1;k 2 Z . 5. Proof by Number Theory n We observed in the introduction that k is a positive rational number. We will prove it is an integer by showing its denominator is 1 using prime factorizations. The multiplicity of a prime p in a nonzero rational number r is the power of p in the reduced form of r. For example, 40 23 · 5 = = 23 · 51 · 7−1 7 7 so we say 40=7 has 2-multiplicity 3, 5-multiplicity 1, 7-multiplicity −1, and p-multiplicity 0 for primes p other than 2, 5, and 7. Denote the p-multiplicity of r as mp(r), so the above calculations say m2(40=7) = 3, m5(40=7) = 1, m7(40=7) = −1, and mp(40=7) = 0 for primes p other than 2, 5, and 7. Since the p-multiplicity is an exponent, it behaves like a logarithm: (5.1) mp(ab) = mp(a) + mp(b); mp(a=b) = mp(a) − mp(b): For example, m2(80) = m2(16 · 5) = 4 and m2(8) + m2(10) = 3 + 1 = 4, while m2(80=14) = m2(40=7) = 3 and m2(80) − m2(14) = 4 − 1 = 3. The rules (5.1) essentially come from unique prime factorization in Z. A nonzero rational number is an integer exactly when its p-multiplicity is nonnegative for all primes p (a rational number that is not an integer has negative p-multiplicity for n primes p appearing in its reduced form denominator). We will prove k 2 Z. by showing n mp k ≥ 0 for every prime p. According to [2, p. 263], this argument is due to Andr´e[1, pp. 188-189]. For prime p and integer N ≥ 0, Legendre [3, p. 10] proved a formula for mp(N!): XjN k jN k jN k jN k m (N!) = = + + + ··· ; p pi p p2 p3 i≥1 where the sum is finite since the ith term is 0 once pi > N.1 Thus n n! m = m p k p k!(n − k)! = mp(n!) − mp(k!) − mp((n − k)!) Xj n k Xj k k Xjn − k k = − − pi pi pi i≥1 i≥1 i≥1 X j n k j k k jn − k k (5.2) = − − : pi pi pi i≥1 Each term in this last sum has the form bx + yc − bxc − byc, where x = k and y = n − k.
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