 # 1 Intro to Induction Supplementary Notes Prepared by Raymond Wong Presented by Raymond Wong.

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1 Intro to Induction Supplementary Notes Prepared by Raymond Wong Presented by Raymond Wong

2 e.g.1 (Page 4) Illustration of “Proof by Contradiction” We are going to prove that a claim C is correct Proof by Contradiction: Suppose “ NOT C ” ….…. Derive some results, which may contradict to 1. “ NOT C ”, OR 2. some facts e.g., we derived that C is true finally e.g., we derived that “1 = 4”

3 e.g.1 Illustration of “Proof by smallest counter example” We are going to prove the following claim C: statement P(m) is true for each non-negative integer m, namely 0, 1, 2, … P(0)true P(1)true P(2)true P(3)true P(4)true If we can prove that statement P(m) is true for each non-negative integer separately, then we can prove the above claim C is correct. Suppose that I want to prove that the above claim is correct by “Proof by Contradiction”. …true

4 e.g.1 Illustration of “Proof by smallest counter example” We are going to prove the following claim C: statement P(m) is true for each non-negative integer m, namely 0, 1, 2, … P(0)true P(1)true P(2)true P(3)true P(4)true We can assume that there exists a non- negative integer k’ such that P(k’) is false Suppose that I want to prove that the above claim is correct by “Proof by Contradiction”. …true false Suppose “NOT C”. There may exist another non-negative integer k such that P(k) is false false

5 e.g.1 Illustration of “Proof by smallest counter example” We are going to prove the following claim C: statement P(m) is true for each non-negative integer m, namely 0, 1, 2, … P(0)true P(1)true P(2)true P(3)true P(4)true Suppose that I want to prove that the above claim is correct by “Proof by Contradiction”. …true false Suppose “NOT C”. false We can assume that there exists a smallest non-negative integer k such that P(k) is false Why? This is called by “Proof by smallest counter example”.

6 e.g.2 (Page 5) Steps for “ Proof by smallest counter example ” Suppose “NOT C”. We are going to prove the following claim C: statement P(n) is true for each non-negative integer n, namely 0, 1, 2, … P(0)true P(1)true P(2)true P(3)true P(4)true … false Step 1: Suppose that claim C is not true. Step 2: there exists a smallest non-negative integer m such that P(m) is false. Step 3: We want to show that this value m must be greater than the smallest value (i.e., 0) Step 4: We derive that P(i) is true for 0  i < m Step 5: We consider a special case that P(m-1) is true. Step 6: Consider the LHS (or some components) of P(m) Prove that P(m) is true (by using P(m-1)) Step 7: We have a contradiction that P(m) is false Step 8: Thus, by the principle of proof by contradiction, claim C is correct. e.g., P(n) is “0+1+2+…+n = ” n(n+1) 2

7 e.g.3 (Page 14) Illustration of “Proof by mathematical induction” We are going to prove the following claim C: statement P(n) is true for each non-negative integer n, namely 0, 1, 2, … P(0)true P(1)true P(2)true P(3)true P(4)true If we can prove that statement P(n) is true for each non-negative integer separately, then we can prove the above claim C is correct. …true

8 e.g.3 Illustration of “Proof by mathematical induction” We are going to prove the following claim C: statement P(n) is true for each non-negative integer n, namely 0, 1, 2, … P(0)true P(1) P(2) P(3) P(4) … Step 1: Prove that P(0) (i.e., the base case) is true. Verify that P(0) is true

9 e.g.3 Illustration of “Proof by mathematical induction” We are going to prove the following claim C: statement P(n) is true for each non-negative integer n, namely 0, 1, 2, … P(0)true P(1) P(2) P(3) P(4) … Step 1: Prove that P(0) (i.e., the base case) is true. Step 2: Prove that “ P(n-1)  P(n) ” is true for all n > 0. true Step 2(a): Assume that P(n-1) is true for n > 0. Verify that P(0) is true Step 2(b): According to P(n-1), we deduce that P(n) is true. Inductive Hypothesis Inductive Step

10 e.g.4 (Page 15) Prove that  n  0, 2 n+1  n 2 +2 Step 1: Prove that P(0) (i.e., the base case) is true. Let P(n) be “ 2 n+1  n 2 +2 “ Consider 2 0+1 = 2 = 0+2 We want to show that 2 0+1  0 2 +2  0 2 +2 Thus, P(0) is true.

11 e.g.4 Prove that  n  0, 2 n+1  n 2 +2 Let P(n) be “ 2 n+1  n 2 +2 “

12 e.g.4 Prove that  n  0, 2 n+1  n 2 +2 Let P(n) be “ 2 n+1  n 2 +2 “ Step 2: Prove that “ P(n-1)  P(n) ” is true for all n > 0. Step 2(a): Assume that P(n-1) is true for n > 0. Step 2(b): According to P(n-1), we deduce that P(n) is true. That is, 2 (n-1)+1  (n-1) 2 +2 for n > 0. 2 n  (n-1) 2 +2 for n > 0. Consider 2 n+1 We want to show that 2 n+1  n 2 +2 = 2 n. 2  [(n-1) 2 +2]. 2 = 2(n-1) 2 +2. 2 = 2(n 2 -2n+1)+4 = 2n 2 -4n+2+4 = 2n 2 -4n+6 = (n 2 +n 2 ) – 4n + (2+4) What should I do next? = n 2 +2 + n 2 – 4n + 4 = (n 2 +2) + (n 2 – 4n + 4) = (n 2 +2) + (n – 2) 2  n 2 +2 (Since (n-2) 2  0) That is, P(n) is true. We prove that “ P(n-1)  P(n) ” is true for all n > 0 By Mathematical Induction,  n  0, 2 n+1  n 2 +2 Thus, 2 n+1  n 2 +2

13 e.g.5 (Page 16) Prove that  n  2, 2 n+1 > n 2 +3 Step 1: Prove that P(2) (i.e., the base case) is true. Let P(n) be “ 2 n+1 > n 2 +3 “ Consider 2 2+1 = 2 3 = 8 We want to show that 2 2+1 > 2 2 +3 > 7 Thus, P(2) is true. = 2 2 +3

14 e.g.5 Prove that  n  2, 2 n+1 > n 2 +3 Let P(n) be “ 2 n+1 > n 2 +3 “

15 e.g.5 Prove that  n  2, 2 n+1 > n 2 +3 Let P(n) be “ 2 n+1 > n 2 +3 “ Step 2: Prove that “ P(n-1)  P(n) ” is true for all n > 2. Step 2(a): Assume that P(n-1) is true for n > 2. Step 2(b): According to P(n-1), we deduce that P(n) is true. That is, 2 (n-1)+1 > (n-1) 2 +3 for n > 2. 2 n > (n-1) 2 +3 for n > 2. Consider 2 n+1 We want to show that 2 n+1 > n 2 +3 = 2 n. 2 >[(n-1) 2 +3]. 2 = 2(n-1) 2 +3. 2 = 2(n 2 -2n+1)+6 = 2n 2 -4n+2+6 = 2n 2 -4n+8 = (n 2 +n 2 ) – 4n + (3+5) What should I do next? = n 2 +3 + n 2 – 4n + 5 = (n 2 +3) + (n 2 – 4n + 5) = (n 2 +3) + (n 2 – 4n + 4 + 1) > n 2 +3 (Since (n-2) 2 + 1  0) Thus, P(n) is true. We prove that “ P(n-1)  P(n) ” is true for all n > 2 By Mathematical Induction,  n  2, 2 n+1 > n 2 +3 = (n 2 +3) + (n – 2) 2 + 1

16 e.g.6 (Page 18) Prove that  k  Z +, 1+3+5+ … +(2k-1) = k 2 Step 1: Prove that P(1) (i.e., the base case) is true. Let P(k) be “ 1+3+5+ … +(2k-1) = k 2 ” Consider 1 = 1 2 We want to show that 1 = 1 2 Thus, P(1) is true.

17 e.g.6 Prove that  k  Z +, 1+3+5+ … +(2k-1) = k 2 Let P(k) be “ 1+3+5+ … +(2k-1) = k 2 ”

18 e.g.6 Prove that  k  Z +, 1+3+5+ … +(2k-1) = k 2 Let P(k) be “ 1+3+5+ … +(2k-1) = k 2 ” Step 2: Prove that “ P(n-1)  P(n) ” is true for all n > 1. Step 2(a): Assume that P(n-1) is true for n > 1. Step 2(b): According to P(n-1), we deduce that P(n) is true. That is, 1+3+5+ … +(2(n-1)-1) = (n-1) 2 for n > 1. 1+3+5+ … +(2n-3) = (n-1) 2 for n > 1. Consider 1+3+5+ … +(2n-1) We want to show that 1+3+5+ … +(2n-1) = n 2 = 1+3+5+ … +(2n-3)+(2n-1) = (n-1) 2 + (2n-1) = (n 2 – 2n + 1) + (2n – 1) = n 2 Thus, P(n) is true. We prove that “ P(n-1)  P(n) ” is true for all n > 1 By Mathematical Induction,  k  Z +, 1+3+5+ … +(2k-1) = k 2

19 e.g.7 (Page 19) For what positive integer values of n is “ 2 n > n 2 ” ? Step 1: Prove that P( ? ) (i.e., the base case) is true. Let P(n) be “ 2 n > n 2 ” Consider P(1) We want to see whether P(1) is true. (i.e., whether “ 2 1 > 1 2 ” is true.) Consider 2 1 We don ’ t know the base case. Thus, we need to test the “ smallest ” value of n for the base case. = 2 > 1 2 Thus, P(1) is true. Consider P(2) We want to see whether P(2) is true. (i.e., whether “ 2 2 > 2 2 ” is true.) Consider 2 2 = 4 Thus, P(2) is false. > 2 2 P(1) is true. P(2) is false.

20 e.g.7 For what positive integer values of n is “ 2 n > n 2 ” ? Step 1: Prove that P( ? ) (i.e., the base case) is true. Let P(n) be “ 2 n > n 2 ” Consider P(3) We want to see whether P(3) is true. (i.e., whether “ 2 3 > 3 2 ” is true.) Consider 2 3 We don ’ t know the base case. Thus, we need to test the “ smallest ” value of n for the base case. = 8 Thus, P(3) is false. Consider P(4) We want to see whether P(4) is true. (i.e., whether “ 2 4 > 4 2 ” is true.) Consider 2 4 = 16 Thus, P(4) is false. > 4 2 > 3 2 P(1) is true. P(2) is false. P(3) is false. P(4) is false.

21 e.g.7 For what positive integer values of n is “ 2 n > n 2 ” ? Step 1: Prove that P( ? ) (i.e., the base case) is true. Let P(n) be “ 2 n > n 2 ” Consider P(5) We want to see whether P(5) is true. (i.e., whether “ 2 5 > 5 2 ” is true.) Consider 2 5 We don ’ t know the base case. Thus, we need to test the “ smallest ” value of n for the base case. = 32 Thus, P(5) is true. Consider P(6) We want to see whether P(6) is true. (i.e., whether “ 2 6 > 6 2 ” is true.) Consider 2 6 = 64 Thus, P(6) is true. > 6 2 > 5 2 Thus, we guess that P(7), P(8), … are also true. Thus, we think that the base case is P(5). P(5) P(1) is true. P(2) is false. P(3) is false. P(4) is false. P(5) is true. P(6) is true.

22 e.g.7 For what positive integer values of n is “ 2 n > n 2 ” ? Step 1: Prove that P( ? ) (i.e., the base case) is true. Let P(n) be “ 2 n > n 2 ” P(5) Prove that  n  5, 2 n > n 2 Thus, we think that the base case is P(5).

23 e.g.7 Step 1: Prove that P( ? ) (i.e., the base case) is true. Let P(n) be “ 2 n > n 2 ” P(5) Prove that  n  5, 2 n > n 2

24 e.g.7 Step 1: Prove that P( ? ) (i.e., the base case) is true. Let P(n) be “ 2 n > n 2 ” P(5) Prove that  n  5, 2 n > n 2 We want to show that “ 2 5 > 5 2 ” Consider 2 5 = 32 Thus, P(5) is true. > 5 2

25 e.g.7 Let P(n) be “ 2 n > n 2 ” Prove that  n  5, 2 n > n 2

26 e.g.7 Prove that  n  5, 2 n > n 2 Let P(n) be “ 2 n > n 2 ” Step 2: Prove that “ P(n-1)  P(n) ” is true for all n > 5. Step 2(a): Assume that P(n-1) is true for n > 5. Step 2(b): According to P(n-1), we deduce that P(n) is true. That is, 2 n-1 > (n-1) 2 for n > 5. Consider 2 n We want to show that 2 n > n 2 = 2 n-1. 2 > (n-1) 2. 2 = (n 2 – 2n + 1). 2 = 2n 2 – 4n + 2 Thus, P(n) is true. We prove that “ P(n-1)  P(n) ” is true for all n > 5 By Mathematical Induction,  n  5, 2 n > n 2 = n 2 + n 2 – 4n + 2 What should I do next? > n 2 + n 2 – 4n > n 2 + n 2 – 5. n > n 2 + n 2 – n. n (Since n > 5) = n 2 Thus, 2 n > n 2

27 e.g.7 Prove that  n  5, 2 n > n 2 Let P(n) be “ 2 n > n 2 ” Step 2: Prove that “ P(n-1)  P(n) ” is true for all n > 5. Step 2(a): Assume that P(n-1) is true for n > 5. Step 2(b): According to P(n-1), we deduce that P(n) is true. That is, 2 n-1 > (n-1) 2 for n > 5. Consider 2 n We want to show that 2 n > n 2 = 2 n-1. 2 > (n-1) 2. 2 = (n 2 – 2n + 1). 2 = 2n 2 – 4n + 2 Thus, P(n) is true. We prove that “ P(n-1)  P(n) ” is true for all n > 5 By Mathematical Induction,  n  5, 2 n > n 2 = n 2 + n 2 – 4n + 2 = n 2 + n 2 – 4n + 4 – 4 +2 = n 2 + (n 2 – 4n + 4) – 2 = n 2 + (n – 2) 2 – 2 Thus, 2 n > n 2 Alternative Derivation Since (n-2) 2 is increasing when n> 5, we have (n-2) 2 > (5-2) 2 = 3 2 = 9 > n 2 + 9 – 2 = n 2 + 7 > n 2

28 e.g.8 (Page 21) Illustration of “Proof by mathematical induction” (Weak Induction) We are going to prove the following claim C: statement P(n) is true for each non-negative integer n, namely 0, 1, 2, … P(0)true P(1) P(2) P(3) P(4) … Step 1: Prove that P(0) (i.e., the base case) is true. Verify that P(0) is true

29 e.g.8 Illustration of “Proof by mathematical induction” (Weak Induction) We are going to prove the following claim C: statement P(n) is true for each non-negative integer n, namely 0, 1, 2, … P(0)true P(1) P(2) P(3) P(4) … Step 1: Prove that P(0) (i.e., the base case) is true. Step 2: Prove that “ P(n-1)  P(n) ” is true for all n > 0. true Step 2(a): Assume that P(n-1) is true for n > 0. Verify that P(0) is true Step 2(b): According to P(n-1), we deduce that P(n) is true. Inductive Hypothesis Inductive Step

30 e.g.8 Illustration of “Proof by mathematical induction” (Strong Induction) We are going to prove the following claim C: statement P(n) is true for each non-negative integer n, namely 0, 1, 2, … P(0)true P(1) P(2) P(3) P(4) … Step 1: Prove that P(0) (i.e., the base case) is true. Verify that P(0) is true

31 e.g.8 Illustration of “Proof by mathematical induction” (Strong Induction) We are going to prove the following claim C: statement P(n) is true for each non-negative integer n, namely 0, 1, 2, … P(0)true P(1) P(2) P(3) P(4) … Step 1: Prove that P(0) (i.e., the base case) is true. Step 2: Prove that “ P(0)  P(1)  …  P(n-1)  P(n) ” is true for all n > 0. true Step 2(a): Assume that P(0)  P(1)  …  P(n-1) is true for n > 0. Verify that P(0) is true Step 2(b): According to P(0)  P(1)  …  P(n-1), we deduce that P(n) is true. Inductive Hypothesis Inductive Step

32 e.g.8 Illustration of “Proof by mathematical induction” (Strong Induction) We are going to prove the following claim C: statement P(n) is true for each non-negative integer n, namely 0, 1, 2, … P(0)true P(1) P(2) P(3) P(4) … Step 1: Prove that P(0) (i.e., the base case) is true. Step 2: Prove that “ P(0)  P(1)  …  P(n-1)  P(n) ” is true for all n > 0. true Step 2(a): Assume that P(0)  P(1)  …  P(n-1) is true for n > 0. Verify that P(0) is true Step 2(b): According to P(0)  P(1)  …  P(n-1), we deduce that P(n) is true. Inductive Hypothesis Inductive Step

33 e.g.8 Illustration of “Proof by mathematical induction” (Strong Induction) We are going to prove the following claim C: statement P(n) is true for each non-negative integer n, namely 0, 1, 2, … P(0)true P(1) P(2) P(3) P(4) … Step 1: Prove that P(0) (i.e., the base case) is true. Step 2: Prove that “ P(0)  P(1)  …  P(n-1)  P(n) ” is true for all n > 0. true Step 2(a): Assume that P(0)  P(1)  …  P(n-1) is true for n > 0. Verify that P(0) is true Step 2(b): According to P(0)  P(1)  …  P(n-1), we deduce that P(n) is true. Inductive Hypothesis Inductive Step true

34 e.g.8 Illustration of “Proof by mathematical induction” (Strong Induction) We are going to prove the following claim C: statement P(n) is true for each non-negative integer n, namely 0, 1, 2, … P(0)true P(1) P(2) P(3) P(4) … Step 1: Prove that P(0) (i.e., the base case) is true. Step 2: Prove that “ P(0)  P(1)  …  P(n-1)  P(n) ” is true for all n > 0. true Step 2(a): Assume that P(0)  P(1)  …  P(n-1) is true for n > 0. Verify that P(0) is true Step 2(b): According to P(0)  P(1)  …  P(n-1), we deduce that P(n) is true. Inductive Hypothesis Inductive Step true

35 e.g.9 (Page 26) Prove that every positive integer n is a power of a prime number OR a product of powers of prime numbers Step 1: Prove that P(1) (i.e., the base case) is true. Consider 1 = 2 0 which is a power of a prime number (i.e., 2). Thus, P(1) is true. Let P(n) be “ n is a power of a prime number OR a product of powers of prime numbers. ” We want to show that 1 is a power of a prime number OR a product of powers of prime numbers e.g., 4 = 2 2 12 = 2 2. 3

36 e.g.9 Prove that every positive integer n is a power of a prime number OR a product of powers of prime numbers Let P(n) be “ n is a power of a prime number OR a product of powers of prime numbers. ”

37 e.g.9 Prove that every positive integer n is a power of a prime number OR a product of powers of prime numbers Let P(n) be “ n is a power of a prime number OR a product of powers of prime numbers. ” Step 2: Prove that “ P(1)  P(2)  …  P(n-1)  P(n) ” is true for all n > 1. Step 2(a): Assume that P(1)  P(2)  …  P(n-1) is true for n > 1. Step 2(b): According to P(0)  P(1)  …  P(n-1), we deduce that P(n) is true. That is, 1 is a power of a prime number OR a product of powers of prime numbers. 2 is a power of a prime number OR a product of powers of prime numbers. … n-1 is a power of a prime number OR a product of powers of prime numbers. We want to show that n is a power of a prime number OR a product of powers of prime numbers. Consider two cases. Case 1: n is a prime number. Case 2: n is not a prime number. n is a power of a prime number n is a product of two smaller numbers, namely a and b (where a < n and b < n) Thus, n = a. b Note that a is a power of a prime number or a product of powers of prime numbers. Note that b is a power of a prime number or a product of powers of prime numbers Thus, n is a power of a prime number or a product of powers of prime numbers

38 e.g.9 Prove that every positive integer n is a power of a prime number OR a product of powers of prime numbers Let P(n) be “ n is a power of a prime number OR a product of powers of prime numbers. ” Step 2: Prove that “ P(1)  P(2)  …  P(n-1)  P(n) ” is true for all n > 1. Step 2(a): Assume that P(1)  P(2)  …  P(n-1) is true for n > 1. Step 2(b): According to P(0)  P(1)  …  P(n-1), we deduce that P(n) is true. That is, 1 is a power of a prime number OR a product of powers of prime numbers. 2 is a power of a prime number OR a product of powers of prime numbers. … n-1 is a power of a prime number OR a product of powers of prime numbers. Thus, P(n) is true. We prove that “ P(1)  P(2)  …  P(n-1)  P(n) ” is true for all n > 1 By Strong Mathematical Induction, every positive integer n is a power of a prime number OR a product of powers of prime numbers

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