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1 MATHEMATICAL INDUCTION Mathematical Induction is a powerful and elegant technique for proving certain types of mathematical statements: general propositions which assert that something is true for all positive integers or for all positive integers from some point on. Why is this important? Statement: If p is any prime number, 2p − 1 is also a prime. True? Not true? p 2 3 5 7 2p − 1 3 7 31 127 Since 3, 7, 31, 127 are all primes, we may be satisο¬ed the result is always true. But if we try the next prime, 11, we ο¬nd that 211 − 1 = 2047 = 23 × 89. So it is not a prime, and our general assertion is therefore FALSE. General proof? Mathematical induction is one way of doing this. Principle of Mathematical Induction (takes three steps) TASK: Prove that the statement is true for all n∈ π΅ 1. Check that the statement Pn is true for n = 1. (Or, if the assertion is that the statement is true for n ≥ a, prove it for n = a.) 2. Assume that the statement is true for n = k (inductive hypothesis) 3. Prove that if the statement is true for n = k, then it must also be true for n = k + 1 Automatic “statement proving” machine: We checked the proposition for n = 1. By the inductive step, since it is true for n = 1, it is also true for n = 2. Again, by the inductive step, since it is true for n = 2, it is also true for n = 3. And since it is true for n = 3, it is also true for n = 4, and so on. No recipe for how to do this. It varies from problem to problem, depending on the mathematical content. You have to use your ingenuity, common sense and knowledge of mathematics here. π EXAMPLE: Prove that for π ≥ 1 ∑ ππ! = (π + 1)! − 1 π=1 1. π = 1 1 ∑ ππ! = 1 β 1! = (1 + 1)! − 1 = 2 − 1 = 1 π=1 π+1 Statement is true for π = 1 ππππ: ππππ£π ∑ ππ! = (π + 1 + 1)! − 1 2. Assume the statement is true for π = π π=1 π π ∑ ππ! = (π + 1)! − 1 π π‘πππ‘ ππππ πΏπ, πππ π’π π ∑ ππ! = (π + 1)! − 1 π=1 π=1 π+1 3. ππ πππππ π‘π ππππ£π π‘βππ‘ πΏπ = π π π ∑ ππ! = ∑ ππ! + (π + 1)(π + 1)! = (π + 1)! (1 + π + 1) − 1 = (π + 2)! − 1 π=1 π=1 (π + 1)! − 1 The statement is true for π = 1, assuming that is true for π = π, statement is true for π = π + 1 ∴ ππ‘ππ‘πππππ‘ ππ π‘ππ’π πππ πππ π ≥ 1, ππ¦ πππ‘β ππππ’ππ‘πππ π. πΈ. π· 2 1 EXAMPLE: P r o v e b y m a t h i n d u c t i o n t h a t the sum of the first positive integers is n(n + 1). 2 1. π = 1 1 1 = 1(1 + 1) = 1 2 Statement is true for π = 1 2. Assume the statement is true for π = π 1 1 + 2 + β― + π = π(π + 1) 2 3. 1 1 1 1 + 2 + β― + π + (π + 1) = π(π + 1) + (π + 1) = (π + 1) ( π + 1) = (π + 1)(π + 2) 2 2 2 1 π(π + 1) 2 The statement is true for π = 1, assuming that is true for π = π, statement is true for π = π + 1 ∴ ππ‘ππ‘πππππ‘ ππ π‘ππ’π πππ πππ π ≥ 1, ππ¦ πππ‘β ππππ’ππ‘πππ EXAMPLE: For every positive integer n ≥ 2, n3 − n is a multiple of 6. Statement: n3 – n = 6m π ∈ π 1. n = 2, n3 − n = 23 − 2 = 6 which is a multiple of 6 (6 × 1). Statement is true for π = 2 2. Assume the statement is true for π = π k3 – k= 6m for some integer m 3. 3 3 2 (k + 1) − (k + 1) = k + 3k + 3k + 1 − k − 1 3 2 = (k − k) + (3k + 3k) = 6m + 3k(k + 1) = 6m + 6β = 6(m + β) k(k + 1) is even number, k(k + 1)= 2β π, β ∈ π The statement is true for π = 2, assuming that is true for π = π, statement is true for π = π + 1 ∴ ππ‘ππ‘πππππ‘ ππ π‘ππ’π πππ πππ π ≥ 2, ππ¦ πππ‘β ππππ’ππ‘πππ π. πΈ. π· π. πΈ. π· 3 EXAMPLE: P r o v e t h a t 2π > π2 for all natural numbers n ≥ 5. 1. n = 5, 25 = 32 > π2 = 25 Statement is true for π = 5 2. Assume true for π = π 2π > π 2 3. 2π+1 = 2(2π ) > 2π 2 o u r g o a l i s t o p r o v e 2π+1 > (π + 1)2 s o w e h a v e t o s h o w t h a t 2π 2 > (π + 1)2 o r 2π 2 − π 2 + 2π + 1 > 0 π(π − 2) − 1 > 0 πππ π ≥ 5 π. πΈ. π·. 2π+1 > 2π 2 > (π + 1)2 π. πΈ. π· The statement is true for π = 5, assuming that it is true for π = π, statement is true for π = π + 1 ∴ ππ‘ππ‘πππππ‘ ππ π‘ππ’π πππ πππ π ≥ 5, ππ¦ πππ‘β ππππ’ππ‘πππ 4