Lecture 6 Mathematical Induction The Mathematical Induction is a very powerful tool for proving infinitely many propositions by using only a few steps. In particular, it can be used often when the propositions are involving the nonnegative integer n. Theorem I. (The First Induction) Let fp n g; n m be a family of infinitely many propositions (or statements) involving the integer n, where m is a nonnegative integer. If (i) P m can be proved to be true, and (ii) P kc must be true provided P k is true for any k m, then P n is true for all n m. Theorem II. (The Second Induction) Let fp n g; n m be a family of infinitely many propositions (or statements) involving the integer n, where m is a nonnegative integer. If (i) P m can be proved to be true, and (ii) P kc must be true provided P m ; P mc ; : : : ; P k are all true for any k m, then P n is true for all n m. Note that it seems to be more difficult for using the second induction, since it require more conditions. However it is actually not so. We can use either one freely if necessary. Variations of Mathematical Induction (i) (ii) (iii) (iv) For proving each P n, n ; ; : : :, we can prove P nk ; k ; ; : : : first by induction, and then prove P n for each n. For proving fp n ; n Ng by induction, sometimes it is convenient to prove a stronger proposition Pn 0. For proving fp n ; n n 0 g by induction, sometimes it is convenient to prove P n by induction starting from n n > n 0. The induction can be conducted backwards: (i) Prove each proposition in the subsequence fp nk ; k Ng first, and then (ii) prove that P k is true provided P kc is true.
Lecture 6 Mathematical Induction (v) (vi) Cross-Active (or Spiral) Induction: For proving propositions P n ; n 0; ; ; : : : and Q n ; n 0; ; ; : : :, it suffices to show that (i) P 0 and Q 0 are true; (ii) If P k ; Q k are true, then P k ) Q kc ) P kc ) P kc. Multiple Induction: Let fp n;m ; n; m Ng be a sequence of propositions with two independent parameters n and m. If (i) P ;m is true for all m N and P n; are true for all n N; (ii) P nc;mc is true provided P n;mc and P nc;m are true, then P n;m is true for any n; m N. Examples Example. Prove by induction that 3 C 3 C C n 3 n.n C / for each n N. Hence prove that 3 C 3 3 C 5 3 C C.n / 3 n.n / for each n N. Solution Let P n be the statement 3 C 3 C C n 3 n.n C / ; n ; ; : : :: For n, left hand side. C / right hand side, P is proven. Assume that P k is true (k ), i.e. 3 C 3 C C k 3 k.k C /, then for n k C, 3 C 3 C C.k C / 3 3 C 3 C C k 3 C.k C / 3 k.k C / C.k C / 3.k C / Œk C.k C /.k C /.k C / : Thus, P kc is also true. By induction, P n is true for each n N. Now for any positive integer n, 3 C 3 3 C C.n / 3 3 C 3 C C.n/ 3. 3 C 3 C C.n/ 3 /.n/.n C / 3. 3 C 3 C C n 3 / n.n C / 8 n.n C / n.n C / n.n C / n Œ.n C /.n C / n.n /; as desired. Example. (CWMO/008) fa n g is a sequence of real numbers, satisfying a 0 0;, a a 0 ; a nc a n. a n / for all n N. Prove that for any positive
Lecture Notes on Mathematical Olympiad 3 integer n, a 0 a a n a 0 C a C C a n : namely Solution The given condition yields a nc a n. a n /; n N 0, hence a nc a n a n. a n / a n a. a / a n a a 0 ; a nc a 0 a a n ; n ; ; : : :: We now prove the original conclusion by induction. For n, a 0 a a 0 C a a 0 C a, the conclusion is proven. Assume that the proposition is true for n k, then for n k C, a 0 a a kc C C C a 0 a a kc a 0 a a k C C C a kc C a 0 a a k a 0 a a k a kc C a 0 a a k : Thus, the proposition is true for n k C also. By induction, the conclusion is true for all n N. Example 3. (IMO/Shortlist/006) The sequence of real numbers a 0 ; a ; a ; : : : is defined recursively by a n k a 0 ; 0 for n : k C Show that a n > 0 for n. k0 k0 Solution The proof goes by induction on n. For n the formula yields a =. Take n and assume a ; : : : ; a n > 0. Write the recurrence formula for n and n C respectively as nc a k n k C 0 and X a k n k C 0: Subtraction yields nc X a k 0.n C / n k C k0.n C /a nc C k0 k0.n C / n C n k C k0 a k n k C n C a k : n k C
Lecture 6 Mathematical Induction The coefficient of a 0 vanishes, so a nc n C k n C k n C n k C n C a k n k C k.n k C /.n k C / a k: The coefficients of a ; : : : ; a n are all positive. Therefore, a ; : : : ; a n > 0 implies a nc > 0. Example. (CHNMO/00) (edited) If 0 < x x : : : x s ; 0 < y y : : : y s satisfy the condition that x r C xr C C xr s y C y r C C yr s ; for all r N; prove that x i y i ; i ; ; : : :; s. Solution We prove by induction on s. Let P s ; s ; ; : : : be the statement to be proven. For s, taking r yields x y. Assume that P s is true for s k (k ) i.e., x y ; x y ; : : : ; x k y k : then for s k C, if x kc y kc, without loss of generality we assume that x kc < y kc. Then x y kc r C C xkc y kc r y y kc r r yk C C C : y kc But on the other hand, since 0 < x i < ; i ; ; : : :; k C, letting r! C y kc yields 0, a contradiction. Hence x kc y kc, then the inductive assumption implies that x i y i ; i ; ; : : :; k. Thus, P kc is also true. By induction, P s is true for each s N. Example 5. (AM-GM Inequality) For real numbers a ; a ; : : : ; a n > 0, prove that a Ca C Ca n n np a a a n ;./ and that equality holds if and only if a a a n. Solution We first prove () for n t ; t N by induction on t. When t,./,. p p a a / 0, so it s true.
Lecture Notes on Mathematical Olympiad 5 Assume that () is true for t k, then for t k C, by the induction assumption, X kc j a j k X j a j C k X j a j C k k q k Y k j a j C k q k Y k a j C k j k q kc Y kc j a j kc q kc Y kc j a j : Thus, the proposition is proven for t k C. By induction, () is proven for all n t ; t ; ; : : :. Next we prove () for each n N. For any n N which is not of the form t, there exist an s N such that s < n < s. Since a Ca C Ca s s s p a a a s;./ By letting a nc a nc a s np a a a n in (), () yields a C a C C a n C. s n/ np a a a n s sq.a a a n /.a a a n / s n n s n p a a a n ; ) a C a C C a n n n p a a a n ; hence./ is proven for general n N. Example 6. (CMC/007) For each positive integer n >, prove that Solution n 3n C < n C C n C C C n C n < 5 : To prove the left inequality, define f.n/ n C C n C C C n C n Then f.n C / n C C n C 3 C C f.n C / f.n/ n 3n C ; n N: n C n C 3n C, so n C n C n C 3n C.n C /.n C /.3n C /.3n C / n.n C 3/.n C /.n C /.3n C /.3n C / > 0; n C 3n C
6 Lecture 6 Mathematical Induction i.e., f.n/ is an increasing function of n. Thus, for n, f.n/ f./ 3 C 7 8 > 0; the left inequality is proven. Below we prove the right inequality by proving a stronger one by induction. Let the proposition P n ; n be is 5 n C C n C C C n C n < 5 n C : When n, the left hand side of P is 3 C 7, and the right hand side 9 7, so P is true. Assume P n is true for n k (k ), i.e., k C C k C C C k C k < 5 Then for n k C, the left hand side of P kc is Since k C 5 0, therefore k C C C k C k C C C k C 5 5 k C C k C C k C k C 5 C k C 5 k C k C C C k C < 5 k C k C C k C k C : k C 3 k C k C :.k C /.k C /.k C /.k C 5/ < k C 5. Thus, P kc is true also. The inductive proof is completed. Example 7. Suppose that a ; a ; ; a n ; are real numbers with 0 a i for i ; ;. Prove that for every n, i C a i n C np a a a n :./ Solution As the first step, we prove the conclusion for all n with the form m by induction on m. For m, the inequality () becomes C C a C a C p a a
Lecture Notes on Mathematical Olympiad 7 or equivalently:. C a /. C a /. C a C a /. p a a C /; a C a C a a. C a C a / p a a ; p.a C a /. a a / p p a a. a a / 0;. p p a a / p. a a / 0: Therefore the conclusion is true. Assume that for m k (k ) the conclusion is true, i.e. Then, for m k C X kc i C a i k X i k X i C a i k p : C k a a a k C C a i C a i k X i q pa p C k a a3 a p a kc a kc kc k p : C kc a a a kc C p a i a i Therefore the conclusion is true for all n m ; m N. Next, to show that () is true for all n N, it suffices to show that () is true p for n k.k / if () is true for n k. In fact, by letting a k k a a a k, we have kx i C C a i C k p a a a k k C kp a a a k p k a a a k therefore we have kx i C a i k C k p a a a k : By backward induction, () is true for each n N. k C k p a a a k ; Example 8. Let ff n g, n N be the Fibonacci sequence. Prove that F nc CF n F nc for each n N.
8 Lecture 6 Mathematical Induction Solution Let P n W FnC C F n F nc; Q n W F nc F n C FnC F nc for n ; ; : : :. When n, then P ; Q are obvious since F F. When P k ; Q k are true for n k (k ), i.e., F kc C F k F kc and F kc F k C F kc F kc, then F kc C F kc.f kc C F k / C F kc.f kc C F kcf k / C.F k C F kc / F kc C F kc F kc3 F.kC/C ; i.e., P kc is true. Further, P kc and Q k are true yields F kc F kc C F kc.f kc C F k /F kc C F kc.f kc C F kc / C.F kc C F kcf k / F kc3 C F kc F kc F.kC/C ; namely Q kc is true. By the cross-active induction, P n ; Q n are true for each n N. Example 9. m boxes are arranged in a row. Prove that there are different ways to put n identical balls into these boxes..n C m /Š nš.m /Š Solution By F.n; m/ we denote the number of ways to put n identical balls into m boxes. The propositions to be proven are F.n; m/.n C m /Š nš.m /Š for n; m N. For n, we have F.; m/ m mš, hence the conclusion is true.m /Š for n and all m N. For m and any n N, we have F.n; / nš nš, therefore the conclusion is true for m and all n N..n C m/š.n C m/š Suppose that F.nC; m/ and F.n; mc/..n C /Š.m /Š nšmš To evaluate F.n C ; m C /, by considering whether the number of balls in the m C st box is 0 or greater than 0, we have F.n C ; m C /.n C m/š F.n C ; m/ C F.n; m C /.n C /Š.m /Š.n C m/š.m C n C /.m C n C /Š.n C /ŠmŠ.n C /Š.mŠ/ The inductive proof is completed. C.n C m/š nšmš Œ.n C / C.m C / Š :.n C /Š.m C /Š
Lecture Notes on Mathematical Olympiad 9 Testing Questions (A) : (USAMO/00) Let S be a set with 00 elements, and let N be an integer with 0 N 00. Prove that it is possible to color every subset of S either blue or red so that the following conditions hold: (a) (b) (c) the union of any two red subsets is red; the union of any two blue subsets is blue; there are exactly N red subsets. : (CHNMO/TST/005) Let k be a positive integer. Prove that it is always possible to partition the set f0; ; ; 3; : : :; kc g into two disjoint subsets fx ; x ; : : : ; x kg and fy ; y ; : : : ; y kg such that for each m f; ; : : :; kg. X k xi m i X k yi m i holds 3: (CMC/006)Let f.x/ x Ca. efine f.x/ f.x/; f n.x/ f.f n.x//, n ; 3; : : : and M fa Rj jf n.0/j for all n Ng. Prove that M ; : : (BELARUS/007) For positive real numbers x ; x ; : : : ; x nc, prove that C x C x x C C x x x n. x x x nc /: x x x 3 x nc 5: (CMC/00) Given the sequence fa n g, a and a nc a a a n C for n ; ; : : :. Prove that a C a C C a n C C 8 C C n n N: 6: (AUSTRIA/009) Prove that 3 n >.nš/ for all positive integers n. 7: (CHNMO/TST/008) Let a and b be positive integers with.a; b/, and a; b having different parities. Let the set S have the following properties: () a; b S; (ii) If x; y; z S, then x C y C z S. Prove that all the integers greater than ab are in S. 8: (USAMO/007) Let S be a set containing n C n elements, for some positive integer n. Suppose that the n-element subsets of S are partitioned into two classes. Prove that there are at least n pairwise disjoint sets in the same class.
0 Lecture 6 Mathematical Induction Testing Questions (B) : (CHNMO/TST/00) It is given that the real numbers a i ; b i (i 0; ; : : :; n) satisfy the following three conditions: (i) a i C a ic 0 for i 0; ; : : :; n ; (ii) a j C 0 for j 0; ; : : :; n ; (iii) For any integers p; q with 0 p q n, Prove that. / i a i b i 0, and find the condition under which the equality holds. i0 Xq kp b k > 0. : (USAMO/007) Prove that for every nonnegative integer n, the number 7 7n C is the product of at least n C 3 (not necessarily distinct) primes. 3: (BULGARIA/007) Let a > ; a nc p.n C /a n C ; n ; ; : : :. Prove that (i) a n > n is convergent. n ; n N; (ii) The sequence b n n a n n ; n N : (USAMO/00) There are n students standing in a circle, one behind the other. The students have heights h < h < : : : < h n. If a student with height h k is standing directly behind a student with height h k or less, the two students are permitted to switch places. Prove that it is not possible to make more than n 3 such switches before reaching a position in which no further switches are possible. 5: (IMO/00) In each of six boxes B ; B ; B 3 ; B ; B 5 ; B 6 there is initially one coin. There are two types of operation allowed: Type : Choose a nonempty box B j with j 5. Remove one coin from B j and add two coins to B j C. Type : Choose a nonempty box B k with k. Remove one coin from B k and exchange the contents of (possibly empty) boxes B kc and B kc. etermine whether there is a finite sequence of such operations that results in boxes B ; B ; B 3 ; B ; B 5 being empty and box B 6 containing exactly 00 0000 coins. (Note that a bc a.bc/.)