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Supplementary Notes on Proof Techniques (Induction)

These supplementary notes prepared by Raymond Wong provide an introduction to various proof techniques, including proof by contradiction, proof by smallest counter example, and proof by mathematical induction. The notes include step-by-step examples and explanations.

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Supplementary Notes on Proof Techniques (Induction)

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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 Suppose that I want to prove that the above claim is correct by “Proof by Contradiction”. • 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 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. P(1) true P(2) true P(3) true P(4) true … true

  4. e.g.1 Suppose that I want to prove that the above claim is correct by “Proof by Contradiction”. • 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 Suppose “NOT C”. We can assume that there exists a non-negative integer k’ such that P(k’) is false P(1) true P(2) true false P(3) true There may exist another non-negative integer k such that P(k) is false P(4) true false … true

  5. e.g.1 Suppose that I want to prove that the above claim is correct by “Proof by Contradiction”. • 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 Suppose “NOT C”. P(1) true We can assume that there exists a smallest non-negative integer k such that P(k) is false P(2) true false P(3) true Why? P(4) true false This is called by “Proof by smallest counter example”. … true

  6. e.g., P(n) is “0+1+2+…+n = ” n(n+1) 2 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 false P(3) true P(4) true false … true e.g.2 (Page 5) Steps for “Proof by smallest counter example” Suppose “NOT C”. 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.

  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 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. P(1) true P(2) true P(3) true P(4) true … 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 Step 1: Prove that P(0) (i.e., the base case) is true. P(1) Verify that P(0) is true P(2) P(3) P(4) …

  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 Step 1: Prove that P(0) (i.e., the base case) is true. P(1) true Verify that P(0) is true P(2) true 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. true P(3) Inductive Hypothesis true P(4) Step 2(b): According to P(n-1), we deduce that P(n) is true. … true Inductive Step

  10. e.g.4 (Page 15) Prove that n  0, 2n+1  n2+2 We want to show that 20+1 02+2 Let P(n) be “2n+1  n2+2“ Step 1: Prove that P(0) (i.e., the base case) is true. Consider 20+1 = 2 = 0+2  02+2 Thus, P(0) is true.

  11. e.g.4 Prove that n  0, 2n+1  n2+2 Let P(n) be “2n+1  n2+2“

  12. Prove that n  0, 2n+1  n2+2 e.g.4 Let P(n) be “2n+1  n2+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. That is, 2(n-1)+1  (n-1)2+2 for n > 0. 2n  (n-1)2+2 for n > 0. We want to show that 2n+1  n2+2 Step 2(b): According to P(n-1), we deduce that P(n) is true. = (n2+2) + (n – 2)2 Consider 2n+1 = 2n . 2  n2+2  [(n-1)2+2] . 2 (Since (n-2)2  0) = 2(n-1)2+2. 2 Thus, 2n+1  n2+2 = 2(n2-2n+1)+4 That is, P(n) is true. = 2n2-4n+2+4 = 2n2-4n+6 What should I do next? We prove that “P(n-1) P(n)” is true for all n > 0 = (n2+n2) – 4n + (2+4) = n2+2 + n2– 4n + 4 = (n2+2) + (n2– 4n + 4) By Mathematical Induction, n  0, 2n+1  n2+2

  13. e.g.5 (Page 16) Prove that n  2, 2n+1 > n2+3 We want to show that 22+1> 22+3 Let P(n) be “2n+1 > n2+3“ Step 1: Prove that P(2) (i.e., the base case) is true. Consider 22+1 = 23 = 8 > 7 = 22+3 Thus, P(2) is true.

  14. e.g.5 Prove that n  2, 2n+1 > n2+3 Let P(n) be “2n+1 > n2+3“

  15. Prove that n  2, 2n+1 > n2+3 e.g.5 Let P(n) be “2n+1 > n2+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. That is, 2(n-1)+1 > (n-1)2+3 for n > 2. 2n > (n-1)2+3 for n > 2. We want to show that 2n+1 > n2+3 Step 2(b): According to P(n-1), we deduce that P(n) is true. = (n2+3) + (n2– 4n + 4 + 1) Consider 2n+1 = 2n . 2 = (n2+3) + (n – 2)2 + 1 > n2+3 >[(n-1)2+3] . 2 (Since (n-2)2 + 1 0) = 2(n-1)2+3. 2 Thus, P(n) is true. = 2(n2-2n+1)+6 We prove that “P(n-1) P(n)” is true for all n > 2 = 2n2-4n+2+6 = 2n2-4n+8 What should I do next? = (n2+n2) – 4n + (3+5) By Mathematical Induction, n  2, 2n+1 > n2+3 = n2+3 + n2– 4n + 5 = (n2+3) + (n2– 4n + 5)

  16. e.g.6 (Page 18) Prove that k  Z+, 1+3+5+…+(2k-1) = k2 Let P(k) be “1+3+5+…+(2k-1) = k2” Step 1: Prove that P(1) (i.e., the base case) is true. Consider 1 = 12 We want to show that 1 = 12 Thus, P(1) is true.

  17. e.g.6 Prove that k  Z+, 1+3+5+…+(2k-1) = k2 Let P(k) be “1+3+5+…+(2k-1) = k2”

  18. Prove that k  Z+, 1+3+5+…+(2k-1) = k2 e.g.6 Let P(k) be “1+3+5+…+(2k-1) = k2” 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. 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. We want to show that 1+3+5+…+(2n-1) = n2 Step 2(b): According to P(n-1), we deduce that P(n) is true. Consider 1+3+5+…+(2n-1) = 1+3+5+…+(2n-3)+(2n-1) = (n-1)2 + (2n-1) = (n2– 2n + 1) + (2n – 1) = n2 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) = k2

  19. > 22 P(1) is true. e.g.7 (Page 19) P(2) is false. For what positive integer values of n is “2n > n2”? We don’t know the base case. Thus, we need to test the “smallest” value of n for the base case. Let P(n) be “2n > n2” Step 1: Prove that P( ? ) (i.e., the base case) is true. We want to see whether P(1) is true. (i.e., whether “21 > 12” is true.) Consider P(1) Consider 21 = 2 > 12 Thus, P(1) is true. We want to see whether P(2) is true. (i.e., whether “22 > 22” is true.) Consider P(2) Consider 22 = 4 Thus, P(2) is false.

  20. > 32 > 42 P(1) is true. e.g.7 P(2) is false. P(3) is false. P(4) is false. For what positive integer values of n is “2n > n2”? We don’t know the base case. Thus, we need to test the “smallest” value of n for the base case. Let P(n) be “2n > n2” Step 1: Prove that P( ? ) (i.e., the base case) is true. We want to see whether P(3) is true. (i.e., whether “23 > 32” is true.) Consider P(3) Consider 23 = 8 Thus, P(3) is false. We want to see whether P(4) is true. (i.e., whether “24 > 42” is true.) Consider P(4) Consider 24 = 16 Thus, P(4) is false.

  21. P(1) is true. P(5) is true. e.g.7 P(2) is false. P(6) is true. P(3) is false. P(4) is false. For what positive integer values of n is “2n > n2”? We don’t know the base case. Thus, we need to test the “smallest” value of n for the base case. Let P(n) be “2n > n2” Step 1: Prove that P( ? ) (i.e., the base case) is true. P(5) We want to see whether P(5) is true. (i.e., whether “25 > 52” is true.) Consider P(5) Consider 25 = 32 > 52 Thus, P(5) is true. We want to see whether P(6) is true. (i.e., whether “26 > 62” is true.) Consider P(6) Consider 26 = 64 > 62 Thus, we guess that P(7), P(8), …are also true. Thus, we think that the base case is P(5). Thus, P(6) is true.

  22. e.g.7 For what positive integer values of n is “2n > n2”? Let P(n) be “2n > n2” Prove that n  5, 2n > n2 Step 1: Prove that P( ? ) (i.e., the base case) is true. P(5) Thus, we think that the base case is P(5).

  23. e.g.7 Prove that n  5, 2n > n2 Let P(n) be “2n > n2” Step 1: Prove that P( ? ) (i.e., the base case) is true. P(5)

  24. e.g.7 Prove that n  5, 2n > n2 Let P(n) be “2n > n2” Step 1: Prove that P( ? ) (i.e., the base case) is true. P(5) We want to show that “25 > 52” Consider 25 = 32 > 52 Thus, P(5) is true.

  25. e.g.7 Prove that n  5, 2n > n2 Let P(n) be “2n > n2”

  26. Prove that n  5, 2n > n2 e.g.7 Let P(n) be “2n > n2” 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. That is, 2n-1 > (n-1)2for n > 5. We want to show that 2n > n2 Step 2(b): According to P(n-1), we deduce that P(n) is true. Thus, 2n > n2 Consider 2n = 2n-1 . 2 Thus, P(n) is true. > (n-1)2. 2 = (n2– 2n + 1) . 2 What should I do next? = 2n2– 4n + 2 = n2 + n2– 4n + 2 We prove that “P(n-1) P(n)” is true for all n > 5 > n2 + n2– 4n By Mathematical Induction, n  5, 2n > n2 > n2 + n2– 5.n (Since n > 5) > n2 + n2– n.n = n2

  27. Prove that n  5, 2n > n2 e.g.7 Alternative Derivation Let P(n) be “2n > n2” 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. That is, 2n-1 > (n-1)2for n > 5. We want to show that 2n > n2 Step 2(b): According to P(n-1), we deduce that P(n) is true. = n2 + 7 Consider 2n > n2 = 2n-1 . 2 Thus, 2n > n2 > (n-1)2. 2 Thus, P(n) is true. = (n2– 2n + 1) . 2 We prove that “P(n-1) P(n)” is true for all n > 5 = 2n2– 4n + 2 By Mathematical Induction, n  5, 2n > n2 = n2 + n2– 4n + 2 = n2 + n2– 4n + 4 – 4 +2 Since (n-2)2 is increasing when n> 5, = n2 + (n2– 4n + 4) – 2 we have (n-2)2 > (5-2)2 = n2 + (n – 2)2– 2 = 32 = 9 > n2 + 9 – 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 Step 1: Prove that P(0) (i.e., the base case) is true. P(1) Verify that P(0) is true P(2) P(3) P(4) …

  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 Step 1: Prove that P(0) (i.e., the base case) is true. P(1) true Verify that P(0) is true P(2) true 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. true P(3) Inductive Hypothesis true P(4) Step 2(b): According to P(n-1), we deduce that P(n) is true. … true 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, … Step 1: Prove that P(0) (i.e., the base case) is true. P(0) true Verify that P(0) is true P(1) P(2) P(3) P(4) …

  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, … Step 1: Prove that P(0) (i.e., the base case) is true. P(0) true Verify that P(0) is true P(1) true Step 2: Prove that “P(0)P(1)…P(n-1) P(n)” is true for all n > 0. P(2) Step 2(a): Assume that P(0)P(1)…P(n-1) is true for n > 0. P(3) Inductive Hypothesis P(4) Step 2(b): According to P(0)P(1)…P(n-1), we deduce that P(n) is true. … 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, … Step 1: Prove that P(0) (i.e., the base case) is true. P(0) true Verify that P(0) is true P(1) true Step 2: Prove that “P(0)P(1)…P(n-1) P(n)” is true for all n > 0. P(2) true Step 2(a): Assume that P(0)P(1)…P(n-1) is true for n > 0. P(3) Inductive Hypothesis P(4) Step 2(b): According to P(0)P(1)…P(n-1), we deduce that P(n) is true. … 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, … Step 1: Prove that P(0) (i.e., the base case) is true. P(0) true Verify that P(0) is true P(1) true Step 2: Prove that “P(0)P(1)…P(n-1) P(n)” is true for all n > 0. P(2) true Step 2(a): Assume that P(0)P(1)…P(n-1) is true for n > 0. P(3) true Inductive Hypothesis P(4) Step 2(b): According to P(0)P(1)…P(n-1), we deduce that P(n) is true. … Inductive Step

  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, … Step 1: Prove that P(0) (i.e., the base case) is true. P(0) true Verify that P(0) is true P(1) true Step 2: Prove that “P(0)P(1)…P(n-1) P(n)” is true for all n > 0. P(2) true Step 2(a): Assume that P(0)P(1)…P(n-1) is true for n > 0. P(3) true Inductive Hypothesis P(4) true Step 2(b): According to P(0)P(1)…P(n-1), we deduce that P(n) is true. … Inductive Step

  35. e.g.9 (Page 26) e.g., 4 = 22 12 = 22 . 3 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.” We want to show that 1 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 = 20 which is a power of a prime number (i.e., 2). Thus, P(1) is true.

  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. Prove that every positive integer n is a power of a prime number OR a product of powers of prime numbers e.g.9 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. 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. Step 2(b): According to P(0)P(1)…P(n-1), we deduce that P(n) is true. We want to show that n is a power of a prime number OR a product of powers of prime numbers. Consider two cases. n is a power of a prime number Case 1: n is 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 Case 2: n is not a prime number. 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. Prove that every positive integer n is a power of a prime number OR a product of powers of prime numbers e.g.9 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. 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. Step 2(b): According to P(0)P(1)…P(n-1), we deduce that P(n) is true. 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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