k 1 k 1 1 2 ie to show 1 2 3 k k 1 k 1 k 2 2 ie by using the induction

K 1 k 1 1 2 ie to show 1 2 3 k k 1 k 1 k 2 2 ie by

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= ( k + 1)[( k + 1) + 1] 2 , i.e., to show 1 + 2 + 3 + ... + k + k + 1 ? = ( k + 1)( k + 2) 2 , i.e., by using the induction assumption (1), to show k ( k + 1) 2 + k + 1 ? = ( k + 1)( k + 2) 2 , or k ( k + 1) + 2( k + 1) 2 ? = ( k + 1)( k + 2) 2 , or k 2 + 3 k + 2 2 ? = ( k + 1)( k + 2) 2 , or k 2 + 3 k + 2 2 ? = ( k + 1)( k + 2) 2 , or, to show ( k + 1)( k + 2) 2 ? = ( k + 1)( k + 2) 2 , Y es ! = ( k + 1) S. S = N . [Example 2] Prove that 1 + 2 + 2 2 + 2 3 + ... + 2 n - 1 = 2 n - 1 , n N . Proof: Let S be the set of positive integers for which the equation holds. Step 1. Show that 1 S 1 ? = 2 1 - 1 Y es ! = 1 S. 4
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Step 2. Assume that k S Namely, we assume that 1 + 2 + 2 2 + 2 3 + ... + 2 k - 1 = 2 k - 1 . (2) Step 3. Prove that k + 1 S for the k in Step 2 To show: 1 + 2 + 2 2 + 2 3 + ... + 2 k - 1 + 2 k ? = 2 k +1 - 1 , i.e., to show 1 + 2 + 2 2 + 2 3 + ... + 2 k - 1 + 2 k ? = 2 k +1 - 1 i.e., by using the induction assumption (2), to show (2 k - 1) + 2 k ? = 2 k +1 - 1 , or 2 · 2 k - 1 ? = 2 k +1 - 1 , or, to show 2 k +1 - 1 ? = 2 k +1 - 1 Y es ! = ( k + 1) S. S = N . Some generalization Theorem 1.2 Let S be a subset of the set N = { 1 , 2 , 3 , ... } . If 1. m S 2. k S with k m implies k + 1 S . Then S = { n N | n m } . [Example 3] Prove that n 2 2 n , for any integer n 4 . Proof: Let S := { n N | n 2 2 n } . When n = 3, 3 2 = 9 2 3 = 8 does not hold. Then the smallest number in S must be greater than 3. Step 1. Show that 4 S 4 2 ? 2 4 Y es ! = 4 S. 5
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Step 2. Assume that k S with k 4 Namely, we assume that for any integer k 4, k 2 2 k . (3) Step 3. Prove that k + 1 S for the k in Step 2 To show: ( k + 1) 2 ? 2 k +1 , i.e., to show k 2 + 2 k + 1 ? 2 k +1 , i.e., by using the induction assumption (3), it suffices to show 2 2 k + 2 k + 1 ? 2 · 2 k , or, by subtracting 2 k in both sides, to show 2 k + 1 ? 2 k . (4) We use mathematical induction again to prove (4). Let e S = { n N | 2 n +1 2 n , n 4 } . Step 3a. Show 4 e S 2 · 4 + 1 = 9 2 4 = 16 Y es ! = 4 e S. Step 3b. Assume that k e S with k 4 2 k + 1 2 k . (5) Step 3c. Show k + 1 e S for the k in Step 3b To show: 2( k + 1) + 1 ? 2 k +1 , i.e., 2 k + 3 ? 2 · 2 k , 2 Here we have used a logical fact: if A B and B C , then A C . 6
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i.e., by the induction assumption (5), it suffices to show 2 k + 2 ? 2 · 2 k , or, by subtracting 2 k in both sides, to show 2 ? 2 k , Y es ! = k + 1 e S. Then e S = { n N | n 4 } , and hence (4) is proved. We conclude S = { n N | n 4 } Appendix Some problems did in the classroom 3.1.1. True or false: (a) If S is a non-empty subset of N , then there exists an element m S such that m k for all k S .
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  • Natural number, Prime number

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