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Examples of Proofs by Induction

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Examples of Proofs by Induction EXAMPLES OF PROOFS BY INDUCTION KEITH CONRAD 1. Introduction Mathematical induction is a method that allows us to prove infinitely many similar assertions in a systematic way, by organizing the results in a definite order and showing • the first assertion is correct (\base case") • whenever an assertion in the list is correct (\inductive hypothesis"), prove the next assertion in the list is correct (\inductive step"). This tells us every assertion in the list is correct, which is analogous to falling dominos. If dominos are close enough and each domino falling makes the next domino fall, then after the first domino falls all the dominos will fall. The most basic results that are proved by induction are summation identities, such as n(n + 1) (1.1) 1 + 2 + 3 + ··· + n = 2 for all positive integers n. Think about this identity as a separate statement for each n: S(1);S(2);S(3); and so on, where S(n) is the (as yet unproved) assertion in (1.1). Definitely S(1) is true, since the left side of (1.1) when n = 1 is 1 and the right side of (1.1) when n = 1 is 1(1 + 1)=2 = 1. Next, if S(n) is true for some n, then we can show S(n + 1) is true by writing 1 + 2 + ··· + n + (n + 1) in terms of 1 + 2 + ··· + n and using the truth of S(n): 1 + 2 + ··· + n + (n + 1) = (1 + 2 + ··· + n) + (n + 1) n(n + 1) = + (n + 1) since S(n) is assumed to be true 2 n(n + 1) + 2(n + 1) = 2 (n + 1)(n + 2) = : 2 The equality of the first and last expressions here is precisely what it means for S(n + 1) to be true. We have proved S(1) is true and proved that if S(n) is true then S(n + 1) is true, so all of S(1);S(2);S(3);::: are true, which proves (1.1) for all n. Another way to prove the inductive step (if S(n) is true then S(n + 1) is true) is to add n + 1 to both sides of (1.1): n(n + 1) n(n + 1) 1 + 2 + ··· + n = =) 1 + 2 + ··· + n + (n + 1) = + (n + 1) 2 2 n(n + 1) + 2(n + 1) =) 1 + 2 + ··· + n + (n + 1) = 2 (n + 1)(n + 2) =) 1 + 2 + ··· + n + (n + 1) = : 2 1 2 KEITH CONRAD Proofs of summation identities are not how proofs by induction usually look: we can't always convert a previously settled case into the next case. The inductive hypothesis (that is, the assumed truth of previously settled cases) should help us derive the next case by some method, and that method may or may not start with the inductive hypothesis itself. Becoming comfortable with induction proofs is mostly a matter of having lots of experi- ence. We will give proofs by induction from several parts of mathematics: linear algebra, polynomial algebra, and calculus. Pay attention to the point in the inductive step where the inductive hypothesis is used. 2. Linear Algebra Theorem 2.1. Suppose B = MAM −1, where A and B are n × n matrices and M is an invertible n × n matrix. Then Bk = MAkM −1 for all integers k ≥ 0. If A and B are invertible, this equation is true for all integers k. Proof. We argue by induction on k, the exponent. (Not on n, the size of the matrix!) The equation Bk = MAkM −1 is obvious for k = 0, when both sides equal the n × n identity matrix I. When k = 1, the equation Bk = MAkM −1 says MA1M −1 = B1, which is just the original condition MAM −1 = B. Let's now prove B2 = MA2M −1: B2 = B · B = (MAM −1) · (MAM −1) = MA(M −1M)AM −1 = MAIAM −1 = MAAM −1 = MA2M −1: Now assume Bk = MAkM −1 for some exponent k ≥ 0. Then Bk+1 = Bk · B = (MAkM −1) · (MAM −1) by inductive hypothesis = MAk(M −1M)AM −1 = MAkIAM −1 = MAk · AM −1 = MAk+1M −1: Thus, the desired result is true for exponent k + 1 if it is true for exponent k. Since the base case k = 1 is true, and if the k-th case is true then we showed the (k+1)-th case is true, the theorem is true for all integers k ≥ 0. If A and B are invertible, the result holds for negative integer exponents too: for k > 0 (MAkM −1) · (MA−kM −1) = MAk(M −1M)A−kM −1 = MAkIA−kM −1 = MAkA−kM −1 = MIM −1 = MM −1 = I; EXAMPLES OF PROOFS BY INDUCTION 3 and thus MA−kM −1 = inverse of MAkM −1 = inverse of Bk by the result for general k > 0 = B−k: Theorem 2.2. Let A be a square matrix. Eigenvectors for A having distinct eigenvalues are linearly independent: if v1;:::; vr are eigenvectors for A, with Avi = λivi for distinct scalars λ1; : : : ; λr, then v1;:::; vr are linearly independent. Proof. We induct on r, the number of eigenvectors. The result is true if r = 1: eigenvectors are not 0 and a single nonzero vector is linearly independent. Suppose r > 1 and the result has been verified for all sets of fewer than r eigenvectors of A (with distinct eigenvalues). Given r eigenvectors vi of A with distinct eigenvalues λi, suppose (2.1) c1v1 + c2v2 + ··· + crvr = 0 for some scalars ci. We want to prove each ci is 0. Applying A to both sides of (2.1), we get (2.2) c1λ1v1 + c2λ2v2 + ··· + crλrvr = 0: Now multiply through (2.1) by λ1: (2.3) c1λ1v1 + c2λ1v2 + ··· + crλ1vr = 0: Subtracting (2.3) from (2.2) eliminates the first term: (2.4) c2(λ2 − λ1)v2 + ··· + cr(λr − λ1)vr = 0: By the inductive hypothesis, the r − 1 eigenvectors v2;:::; vr are linearly independent. Therefore (2.4) tells us that ci(λi − λ1) = 0 for i = 2; 3; : : : ; r. Since the eigenvalues are distinct, λi − λ1 6= 0, so ci = 0 for i = 2; 3; : : : ; r. Feeding this into (2.1) gives us c1v1 = 0, so c1 = 0 as well since v1 6= 0. Thus every ci is 0. 3. Polynomial algebra Theorem 3.1. Let f(x) be a nonconstant polynomial with real coefficients and degree d. Then f(x) has at most d real roots. We can't replace \at most d real roots" with \exactly d real roots" since there are nonconstant real polynomials like x2 + 1 that have no real roots. Proof. We induct on the degree d of f(x). Each step in the induction is about infinitely many polynomials: the theorem in degree 1, then in degree 2, then in degree 3, and so on. The base case has d = 1. A polynomial of degree 1 with real coefficients is of the form f(x) = ax + b, where a and b are real and a 6= 0. This has exactly one root, namely −b=a, and thus at most one real root. That settles the theorem for d = 1. Now assume the theorem is true for all polynomials of degree d with real coefficients. We will prove the theorem is true for all polynomials of degree d + 1 with real coefficients. A typical polynomial of degree d + 1 with real coefficients is d+1 d (3.1) f(x) = cd+1x + cdx + ··· + c1x + c0; 4 KEITH CONRAD where cj 2 R and cd+1 6= 0. If f(x) has no real roots, then we're done, since 0 ≤ d + 1. If f(x) has a real root, say r, then d+1 d (3.2) 0 = cd+1r + cdr + ··· + c1r + c0: Subtracting (3.2) from (3.1), the constant terms c0 cancel and we get d+1 d+1 d d (3.3) f(x) = cd+1(x − r ) + cd(x − r ) + ··· + c1(x − r): Each difference xj − rj for j = 1; 2; : : : ; d + 1 has x − r as a factor: xj − rj = (x − r)(xj−1 + rxj−2 + ··· + rixj−1−i + ··· + rj−2x + rj−1): Write the more complicated second factor, a polynomial of degree j − 1, as Qj;r(x). So j j (3.4) x − r = (x − r)Qj;r(x); and substituting (3.4) into (3.3) gives d+1 X f(x) = cj(x − r)Qj;r(x) j=1 d+1 X = (x − r) cjQj;r(x) j=1 degree d z }| { (3.5) = (x − r)(cd+1 Qd+1;r(x) + ··· + c1Q1;r(x)): |{z} | {z } 6=0 lower degree Since cd+1Qd+1;r(x) is a polynomial of degree d, and each lower degree polynomial does not decrease the degree when added to cd+1Qd+1;r(x), the second factor in (3.5) has degree d. Each root of f(x) is either r or a root of the second factor in (3.5). Each Qj;r(x) has real coefficients and all cj are real, so the second factor in (3.5) has real coefficients. We can therefore apply the inductive hypothesis to the second factor and conclude that the second factor in (3.5) has at most d real roots, so f(x) has at most d + 1 real roots. As f(x) was an arbitrary polynomial of degree d+1 with real coefficients, we have shown that the d-th case of the theorem being true implies the (d + 1)-th case is true. By induction on the degree, the theorem is true for all nonconstant polynomials. Our next two theorems use the truth of some earlier case to prove the next case, but not necessarily the truth of the immediately previous case to prove the next case.
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