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Section 2.2 Induction Proof by induction: to prove p(n) true,  n

Section 2.2 Induction Proof by induction: to prove p(n) true,  n Idea: like playing the domino game. Suppose dominos are placed correctly, then hitting the 1 st domino, when it falls, we know the rest of them will also fall.

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Section 2.2 Induction Proof by induction: to prove p(n) true,  n

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  1. Section 2.2 Induction Proof by induction:to prove p(n) true, n Idea: like playing the domino game. Suppose dominos are placed correctly, then hitting the 1st domino, when it falls, we know the rest of them will also fall. ( the 2nd will fall, in general, when the kth one falls, it implies the (k+1)th falls. Since k is any arbitrary number, actually every domino falls.)

  2. Induction: Step1: (inductive base) or IB is to show that p(1) true Step2: (inductive step) or IS is to show that p(k)  p(k + 1), where p(k) is called inductive hypothesis or IH.

  3. e.g. show 1+3+5+…(2n-1) = n2, n  1 IB: when n = 1, LHS = 1 = 12 = RHS IS: Assume 1+3+5+….(2k-1)=k2 and want to Show 1+3+5+…(2k-1) + [ 2(k+1)-1] =(k+1)2 1+3+5+…(2k –1) + (2k + 1) = k2 + 2k+ 1 = (k +1)2 k2 (by IH)

  4. n(n+1) 2 e.g. Show 1 + 2 + 3 + … n = ,  n  1 IB: when n = 1, LHS = 1 = = RHS IS: Assume 1 + 2 + 3 + …k = and want to show 1 + 2 + 3 + …+ k + (k+1) = 1(1+1) 2 k(k+1) 2 (k+1)(k+2) 2

  5. k(k+1) 2 1 + 2 + 3 + …+ k + (k+1) = + (k+1) = = by direct proof: Let x = 1 + 2 + 3 + …+ k +) x = k + k-1 + k-2 + … + 1 2x = (k+1)+(k+1)+ … +(k+1)  x = k(k+1) 2 (by IH) k(k+1) + 2(k+1) 2 (k+1) (k+2) 2 k of them k(k+1) 2

  6. eg: Show 1+ 21+ 22 + … + 2n = 2n+1 – 1,  n  IB: when n = 1, LHS = 1+ 2 = 22 - 1 = RHS IS: Assume 1 + 2 + 22 + … 2k = 2k+1 – 1, and Show 1 + 2 + 22 + … 2k + 2k+1 = 2k+2 – 1  1 + 2 + 22 + … 2k + 2k+1 = 2k+1 – 1 + 2k+1 = 2*2k+1 – 1 = 2k+2 – 1 2k+1– 1 (by IH)

  7. eg: Show 22n – 1 is divisible by 3,  n  1 IB: when n = 1, LHS = 22 – 1 = 3 divisible by 3 IS: Assume 22k – 1 is divisible by 3 (22k – 1 = 3m for some integer m), and want to show 22(k+1) – 1 is divisible by 3.  22(k+1) – 1 = 22k+2 – 1 = 22 * 22k – 1 = 22(3m+1) – 1 ( 22k – 1 = 3m 22k = 3m + 1) =12m + 3 = 3(4m + 1) where 4m + 1 is an integer  is divisible by 3

  8. k__ k+1 (by IH)

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