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MATH 226 رياضيات متقطعة لعلوم الحاسب Chapter 5 What’s is Induction About? Many statements assert that a property is an universal true – i.e., all the elements of the universe exhibit that property; Examples: 1. For every positive integer n: n! ≤ nn 2. For every set with n elements, the cardinality of its power set is 2n. Induction is one of the most important techniques for proving statements about universal properties. 3 We know that: 1. We can reach the first rung of this ladder; 2. If we can reach a particular rung of the ladder, then we can reach the next rung of the ladder. Can we reach every step of this infinite ladder? Yes, using Mathematical Induction which is a rule of inference that tells us: P(1) k (P(k) P(k+1)) ------------------------- n (P(n) EXAMPLE 1 Show that if n is a positive integer, then 1 + 2+· · ·+n = n(n + 1(/2 . Solution: Let P(n), 1 + 2+· · · n = n(n+1)/2 BASIS STEP: P(1) is true, because 1 = 1(1 + 1)/2 INDUCTIVE STEP: we assume that P(k) holds for an arbitrary positive integer k. 1 + 2+· · ·+k = k(k + 1)/2 Under this assumption, it must be shown that P(k + 1) is true, namely, that 1 + 2+· · ·+k + (k + 1) = (k + 1)[(k + 1) + 1]/2 = (k + 1)(k + 2)/2 add k + 1 to both sides of the equation in P(k), we obtain 1 + 2+· · ·+k + (k + 1) = k(k + 1)/2+ (k + 1) =( k(k + 1) + 2(k + 1))/2 = (k + 1)(k + 2)/2 . Example 2 :Use induction to prove that the 1 + 2 + 22 + … + 2n = 2n+1 - 1 for all non-negative integers n. 1 – Hypothesis? P(n) = 1 + 2 + 22 + … + 2n = 2 n+1 – 1 for all non-negative integers n. 2 - Base case? n = 0 10 = 21-1. 3 – Inductive Hypothesis Assume P(k) = 1 + 2 + 22 + … + 2k = 2 k+1 – 1 7 not n=1! The base case can be negative, zero, or positive Example 3. Use MI to prove that the sum of the first n odd positive integers is n2 n Pf : Let P(n) denote the proposition that 2 ( 2 i 1 ) n i 1 Basis step : P(1) is true , since 1=12 Inductive step : Suppose that P(k) is true for a positive integer k, i.e., 1+3+5+…+(2k-1)=k2 Note that 1+3+5+…+(2k-1)+(2k+1) = k2+2k+1= (k+1)2 ∴ P(k+1) is true By induction, P(n) is true for all nZ+ Example 4Use MI to prove the inequality n < 2n for all nZ+ pf : Let P(n) be the proposition “ n < 2n ”. Basis step : P(1) is true since 1 < 21 . Inductive step : Assume that P(k) is true for a positive integer k, i.e., k < 2k. Consider P(k+1) : k + 1 < 2k + 1 2k + 2k =2k + 1 ∴ P(k+1) is true. By MI, P(n) is true for all nZ+. Ch4-9