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# Discrete Mathematics - Proofs

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Proof techniques, proof by contradiction, mathematical induction.

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### Discrete Mathematics - Proofs

1. 1. Discrete Mathematics ProofsH. Turgut Uyar Ay¸eg¨l Gen¸ata Yayımlı s u c Emre Harmancı 2001-2013
2. 2. License c 2001-2013 T. Uyar, A. Yayımlı, E. Harmancı You are free: to Share – to copy, distribute and transmit the work to Remix – to adapt the work Under the following conditions: Attribution – You must attribute the work in the manner speciﬁed by the author or licensor (but not in any way that suggests that they endorse you or your use of the work). Noncommercial – You may not use this work for commercial purposes. Share Alike – If you alter, transform, or build upon this work, you may distribute the resulting work only under the same or similar license to this one. Legal code (the full license): http://creativecommons.org/licenses/by-nc-sa/3.0/
3. 3. Topics 1 Basic Techniques Introduction Direct Proof Proof by Contradiction Equivalence Proofs 2 Induction Introduction Strong Induction
4. 4. Topics 1 Basic Techniques Introduction Direct Proof Proof by Contradiction Equivalence Proofs 2 Induction Introduction Strong Induction
5. 5. Brute Force Method examining all possible cases one by one Theorem Every number from the set {2, 4, 6, . . . , 26} can be written as the sum of at most 3 square numbers. Proof. 2 = 1+1 10 = 9+1 20 = 16+4 4=4 12 = 4+4+4 22 = 9+9+4 6 = 4+1+1 14 = 9+4+1 24 = 16+4+4 8 = 4+4 16 = 16 26 = 25+1 18 = 9+9
6. 6. Brute Force Method examining all possible cases one by one Theorem Every number from the set {2, 4, 6, . . . , 26} can be written as the sum of at most 3 square numbers. Proof. 2 = 1+1 10 = 9+1 20 = 16+4 4=4 12 = 4+4+4 22 = 9+9+4 6 = 4+1+1 14 = 9+4+1 24 = 16+4+4 8 = 4+4 16 = 16 26 = 25+1 18 = 9+9
7. 7. Brute Force Method examining all possible cases one by one Theorem Every number from the set {2, 4, 6, . . . , 26} can be written as the sum of at most 3 square numbers. Proof. 2 = 1+1 10 = 9+1 20 = 16+4 4=4 12 = 4+4+4 22 = 9+9+4 6 = 4+1+1 14 = 9+4+1 24 = 16+4+4 8 = 4+4 16 = 16 26 = 25+1 18 = 9+9
8. 8. Basic Rules Universal Speciﬁcation (US) ∀x p(x) ⇒ p(a) Universal Generalization (UG) p(a) for an arbitrarily chosen a ⇒ ∀x p(x)
9. 9. Basic Rules Universal Speciﬁcation (US) ∀x p(x) ⇒ p(a) Universal Generalization (UG) p(a) for an arbitrarily chosen a ⇒ ∀x p(x)
10. 10. Universal Speciﬁcation Example Example All humans are mortal. Socrates is human. Therefore, Socrates is mortal. U: all humans p(x): x is mortal ∀x p(x): All humans are mortal. a: Socrates, a ∈ U: Socrates is human. therefore, p(a): Socrates is mortal.
11. 11. Universal Speciﬁcation Example Example All humans are mortal. Socrates is human. Therefore, Socrates is mortal. U: all humans p(x): x is mortal ∀x p(x): All humans are mortal. a: Socrates, a ∈ U: Socrates is human. therefore, p(a): Socrates is mortal.
12. 12. Universal Speciﬁcation Example Example 1. ∀x [j(x) ∨ s(x) → ¬p(x)] A 2. p(m) A ∀x [j(x) ∨ s(x) → ¬p(x)] 3. j(m) ∨ s(m) → ¬p(m) US : 1 p(m) ∴ ¬s(m) 4. ¬(j(m) ∨ s(m)) MT : 3, 2 5. ¬j(m) ∧ ¬s(m) DM : 4 6. ¬s(m) AndE : 5
13. 13. Universal Speciﬁcation Example Example 1. ∀x [j(x) ∨ s(x) → ¬p(x)] A 2. p(m) A ∀x [j(x) ∨ s(x) → ¬p(x)] 3. j(m) ∨ s(m) → ¬p(m) US : 1 p(m) ∴ ¬s(m) 4. ¬(j(m) ∨ s(m)) MT : 3, 2 5. ¬j(m) ∧ ¬s(m) DM : 4 6. ¬s(m) AndE : 5
14. 14. Universal Speciﬁcation Example Example 1. ∀x [j(x) ∨ s(x) → ¬p(x)] A 2. p(m) A ∀x [j(x) ∨ s(x) → ¬p(x)] 3. j(m) ∨ s(m) → ¬p(m) US : 1 p(m) ∴ ¬s(m) 4. ¬(j(m) ∨ s(m)) MT : 3, 2 5. ¬j(m) ∧ ¬s(m) DM : 4 6. ¬s(m) AndE : 5
15. 15. Universal Speciﬁcation Example Example 1. ∀x [j(x) ∨ s(x) → ¬p(x)] A 2. p(m) A ∀x [j(x) ∨ s(x) → ¬p(x)] 3. j(m) ∨ s(m) → ¬p(m) US : 1 p(m) ∴ ¬s(m) 4. ¬(j(m) ∨ s(m)) MT : 3, 2 5. ¬j(m) ∧ ¬s(m) DM : 4 6. ¬s(m) AndE : 5
16. 16. Universal Speciﬁcation Example Example 1. ∀x [j(x) ∨ s(x) → ¬p(x)] A 2. p(m) A ∀x [j(x) ∨ s(x) → ¬p(x)] 3. j(m) ∨ s(m) → ¬p(m) US : 1 p(m) ∴ ¬s(m) 4. ¬(j(m) ∨ s(m)) MT : 3, 2 5. ¬j(m) ∧ ¬s(m) DM : 4 6. ¬s(m) AndE : 5
17. 17. Universal Speciﬁcation Example Example 1. ∀x [j(x) ∨ s(x) → ¬p(x)] A 2. p(m) A ∀x [j(x) ∨ s(x) → ¬p(x)] 3. j(m) ∨ s(m) → ¬p(m) US : 1 p(m) ∴ ¬s(m) 4. ¬(j(m) ∨ s(m)) MT : 3, 2 5. ¬j(m) ∧ ¬s(m) DM : 4 6. ¬s(m) AndE : 5
18. 18. Universal Speciﬁcation Example Example 1. ∀x [j(x) ∨ s(x) → ¬p(x)] A 2. p(m) A ∀x [j(x) ∨ s(x) → ¬p(x)] 3. j(m) ∨ s(m) → ¬p(m) US : 1 p(m) ∴ ¬s(m) 4. ¬(j(m) ∨ s(m)) MT : 3, 2 5. ¬j(m) ∧ ¬s(m) DM : 4 6. ¬s(m) AndE : 5
19. 19. Universal Generalization Example Example 1. ∀x [p(x) → q(x)] A 2. p(c) → q(c) US : 1 ∀x [p(x) → q(x)] 3. ∀x [q(x) → r (x)] A ∀x [q(x) → r (x)] ∴ ∀x [p(x) → r (x)] 4. q(c) → r (c) US : 3 5. p(c) → r (c) HS : 2, 4 6. ∀x [p(x) → r (x)] UG : 5
20. 20. Universal Generalization Example Example 1. ∀x [p(x) → q(x)] A 2. p(c) → q(c) US : 1 ∀x [p(x) → q(x)] 3. ∀x [q(x) → r (x)] A ∀x [q(x) → r (x)] ∴ ∀x [p(x) → r (x)] 4. q(c) → r (c) US : 3 5. p(c) → r (c) HS : 2, 4 6. ∀x [p(x) → r (x)] UG : 5
21. 21. Universal Generalization Example Example 1. ∀x [p(x) → q(x)] A 2. p(c) → q(c) US : 1 ∀x [p(x) → q(x)] 3. ∀x [q(x) → r (x)] A ∀x [q(x) → r (x)] ∴ ∀x [p(x) → r (x)] 4. q(c) → r (c) US : 3 5. p(c) → r (c) HS : 2, 4 6. ∀x [p(x) → r (x)] UG : 5
22. 22. Universal Generalization Example Example 1. ∀x [p(x) → q(x)] A 2. p(c) → q(c) US : 1 ∀x [p(x) → q(x)] 3. ∀x [q(x) → r (x)] A ∀x [q(x) → r (x)] ∴ ∀x [p(x) → r (x)] 4. q(c) → r (c) US : 3 5. p(c) → r (c) HS : 2, 4 6. ∀x [p(x) → r (x)] UG : 5
23. 23. Universal Generalization Example Example 1. ∀x [p(x) → q(x)] A 2. p(c) → q(c) US : 1 ∀x [p(x) → q(x)] 3. ∀x [q(x) → r (x)] A ∀x [q(x) → r (x)] ∴ ∀x [p(x) → r (x)] 4. q(c) → r (c) US : 3 5. p(c) → r (c) HS : 2, 4 6. ∀x [p(x) → r (x)] UG : 5
24. 24. Universal Generalization Example Example 1. ∀x [p(x) → q(x)] A 2. p(c) → q(c) US : 1 ∀x [p(x) → q(x)] 3. ∀x [q(x) → r (x)] A ∀x [q(x) → r (x)] ∴ ∀x [p(x) → r (x)] 4. q(c) → r (c) US : 3 5. p(c) → r (c) HS : 2, 4 6. ∀x [p(x) → r (x)] UG : 5
25. 25. Universal Generalization Example Example 1. ∀x [p(x) → q(x)] A 2. p(c) → q(c) US : 1 ∀x [p(x) → q(x)] 3. ∀x [q(x) → r (x)] A ∀x [q(x) → r (x)] ∴ ∀x [p(x) → r (x)] 4. q(c) → r (c) US : 3 5. p(c) → r (c) HS : 2, 4 6. ∀x [p(x) → r (x)] UG : 5
26. 26. Vacuous Proof vacuous proof to prove P ⇒ Q, show that P is false
27. 27. Vacuous Proof Example Theorem ∀S [∅ ⊆ S] Proof. ∅ ⊆ S ⇔ ∀x [x ∈ ∅ → x ∈ S] ∀x [x ∈ ∅] /
28. 28. Vacuous Proof Example Theorem ∀S [∅ ⊆ S] Proof. ∅ ⊆ S ⇔ ∀x [x ∈ ∅ → x ∈ S] ∀x [x ∈ ∅] /
29. 29. Vacuous Proof Example Theorem ∀S [∅ ⊆ S] Proof. ∅ ⊆ S ⇔ ∀x [x ∈ ∅ → x ∈ S] ∀x [x ∈ ∅] /
30. 30. Trivial Proof trivial proof to prove P ⇒ Q, show that Q is true
31. 31. Trivial Proof Example Theorem ∀x ∈ R [x ≥ 0 ⇒ x 2 ≥ 0] Proof. ∀x ∈ R [x 2 ≥ 0]
32. 32. Trivial Proof Example Theorem ∀x ∈ R [x ≥ 0 ⇒ x 2 ≥ 0] Proof. ∀x ∈ R [x 2 ≥ 0]
33. 33. Topics 1 Basic Techniques Introduction Direct Proof Proof by Contradiction Equivalence Proofs 2 Induction Introduction Strong Induction
34. 34. Direct Proof direct proof to prove P ⇒ Q, show that P Q
35. 35. Direct Proof Example Theorem ∀a ∈ Z [3|(a − 2) ⇒ 3|(a2 − 1)] Proof. 3|(a − 2) ⇒ ∃k ∈ N [a − 2 = 3k] ⇒ a + 1 = a − 2 + 3 = 3k + 3 = 3(k + 1) ⇒ a2 − 1 = (a + 1)(a − 1) = 3(k + 1)(a − 1)
36. 36. Direct Proof Example Theorem ∀a ∈ Z [3|(a − 2) ⇒ 3|(a2 − 1)] Proof. 3|(a − 2) ⇒ ∃k ∈ N [a − 2 = 3k] ⇒ a + 1 = a − 2 + 3 = 3k + 3 = 3(k + 1) ⇒ a2 − 1 = (a + 1)(a − 1) = 3(k + 1)(a − 1)
37. 37. Direct Proof Example Theorem ∀a ∈ Z [3|(a − 2) ⇒ 3|(a2 − 1)] Proof. 3|(a − 2) ⇒ ∃k ∈ N [a − 2 = 3k] ⇒ a + 1 = a − 2 + 3 = 3k + 3 = 3(k + 1) ⇒ a2 − 1 = (a + 1)(a − 1) = 3(k + 1)(a − 1)
38. 38. Direct Proof Example Theorem ∀a ∈ Z [3|(a − 2) ⇒ 3|(a2 − 1)] Proof. 3|(a − 2) ⇒ ∃k ∈ N [a − 2 = 3k] ⇒ a + 1 = a − 2 + 3 = 3k + 3 = 3(k + 1) ⇒ a2 − 1 = (a + 1)(a − 1) = 3(k + 1)(a − 1)
39. 39. Indirect Proof indirect proof to prove P ⇒ Q, show that ¬Q ¬P
40. 40. Indirect Proof Example Theorem ∀x, y ∈ N [x · y > 25 ⇒ (x > 5) ∨ (y > 5)] Proof. ¬Q ⇔ (0 ≤ x ≤ 5) ∧ (0 ≤ y ≤ 5) x · y ≤ 5 · 5 = 25
41. 41. Indirect Proof Example Theorem ∀x, y ∈ N [x · y > 25 ⇒ (x > 5) ∨ (y > 5)] Proof. ¬Q ⇔ (0 ≤ x ≤ 5) ∧ (0 ≤ y ≤ 5) x · y ≤ 5 · 5 = 25
42. 42. Indirect Proof Example Theorem ∀x, y ∈ N [x · y > 25 ⇒ (x > 5) ∨ (y > 5)] Proof. ¬Q ⇔ (0 ≤ x ≤ 5) ∧ (0 ≤ y ≤ 5) x · y ≤ 5 · 5 = 25
43. 43. Indirect Proof Example Theorem ∀a, b ∈ N ∃k ∈ N [ab = 2k] ⇒ (∃i ∈ N [a = 2i]) ∨ (∃j ∈ N [b = 2j]) Proof. ¬Q ⇔ (¬∃i ∈ N [a = 2i]) ∧ (¬∃j ∈ N [b = 2j]) ⇒ (∃x ∈ N [a = 2x + 1]) ∧ (∃y ∈ N [b = 2y + 1]) ⇒ ab = (2x + 1)(2y + 1) ⇒ ab = 4xy + 2x + 2y + 1 ⇒ ab = 2(2xy + x + y ) + 1 ⇒ ¬(∃k ∈ N [ab = 2k])
44. 44. Indirect Proof Example Theorem ∀a, b ∈ N ∃k ∈ N [ab = 2k] ⇒ (∃i ∈ N [a = 2i]) ∨ (∃j ∈ N [b = 2j]) Proof. ¬Q ⇔ (¬∃i ∈ N [a = 2i]) ∧ (¬∃j ∈ N [b = 2j]) ⇒ (∃x ∈ N [a = 2x + 1]) ∧ (∃y ∈ N [b = 2y + 1]) ⇒ ab = (2x + 1)(2y + 1) ⇒ ab = 4xy + 2x + 2y + 1 ⇒ ab = 2(2xy + x + y ) + 1 ⇒ ¬(∃k ∈ N [ab = 2k])
45. 45. Indirect Proof Example Theorem ∀a, b ∈ N ∃k ∈ N [ab = 2k] ⇒ (∃i ∈ N [a = 2i]) ∨ (∃j ∈ N [b = 2j]) Proof. ¬Q ⇔ (¬∃i ∈ N [a = 2i]) ∧ (¬∃j ∈ N [b = 2j]) ⇒ (∃x ∈ N [a = 2x + 1]) ∧ (∃y ∈ N [b = 2y + 1]) ⇒ ab = (2x + 1)(2y + 1) ⇒ ab = 4xy + 2x + 2y + 1 ⇒ ab = 2(2xy + x + y ) + 1 ⇒ ¬(∃k ∈ N [ab = 2k])
46. 46. Indirect Proof Example Theorem ∀a, b ∈ N ∃k ∈ N [ab = 2k] ⇒ (∃i ∈ N [a = 2i]) ∨ (∃j ∈ N [b = 2j]) Proof. ¬Q ⇔ (¬∃i ∈ N [a = 2i]) ∧ (¬∃j ∈ N [b = 2j]) ⇒ (∃x ∈ N [a = 2x + 1]) ∧ (∃y ∈ N [b = 2y + 1]) ⇒ ab = (2x + 1)(2y + 1) ⇒ ab = 4xy + 2x + 2y + 1 ⇒ ab = 2(2xy + x + y ) + 1 ⇒ ¬(∃k ∈ N [ab = 2k])
47. 47. Indirect Proof Example Theorem ∀a, b ∈ N ∃k ∈ N [ab = 2k] ⇒ (∃i ∈ N [a = 2i]) ∨ (∃j ∈ N [b = 2j]) Proof. ¬Q ⇔ (¬∃i ∈ N [a = 2i]) ∧ (¬∃j ∈ N [b = 2j]) ⇒ (∃x ∈ N [a = 2x + 1]) ∧ (∃y ∈ N [b = 2y + 1]) ⇒ ab = (2x + 1)(2y + 1) ⇒ ab = 4xy + 2x + 2y + 1 ⇒ ab = 2(2xy + x + y ) + 1 ⇒ ¬(∃k ∈ N [ab = 2k])
48. 48. Indirect Proof Example Theorem ∀a, b ∈ N ∃k ∈ N [ab = 2k] ⇒ (∃i ∈ N [a = 2i]) ∨ (∃j ∈ N [b = 2j]) Proof. ¬Q ⇔ (¬∃i ∈ N [a = 2i]) ∧ (¬∃j ∈ N [b = 2j]) ⇒ (∃x ∈ N [a = 2x + 1]) ∧ (∃y ∈ N [b = 2y + 1]) ⇒ ab = (2x + 1)(2y + 1) ⇒ ab = 4xy + 2x + 2y + 1 ⇒ ab = 2(2xy + x + y ) + 1 ⇒ ¬(∃k ∈ N [ab = 2k])
49. 49. Indirect Proof Example Theorem ∀a, b ∈ N ∃k ∈ N [ab = 2k] ⇒ (∃i ∈ N [a = 2i]) ∨ (∃j ∈ N [b = 2j]) Proof. ¬Q ⇔ (¬∃i ∈ N [a = 2i]) ∧ (¬∃j ∈ N [b = 2j]) ⇒ (∃x ∈ N [a = 2x + 1]) ∧ (∃y ∈ N [b = 2y + 1]) ⇒ ab = (2x + 1)(2y + 1) ⇒ ab = 4xy + 2x + 2y + 1 ⇒ ab = 2(2xy + x + y ) + 1 ⇒ ¬(∃k ∈ N [ab = 2k])
50. 50. Topics 1 Basic Techniques Introduction Direct Proof Proof by Contradiction Equivalence Proofs 2 Induction Introduction Strong Induction
51. 51. Proof by Contradiction proof by contradiction to prove P, show that ¬P Q ∧ ¬Q
52. 52. Proof by Contradiction Example Theorem There is no largest prime number. Proof. ¬P: There is a largest prime number. Q: The largest prime number is S. prime numbers: 2, 3, 5, 7, 11, . . . , S 2 · 3 · 5 · 7 · 11 · · · S + 1 is not divisible by a prime number in the range [2, S] 1 either it is prime itself: ¬Q 2 or it is divisible by a prime number greater than S: ¬Q
53. 53. Proof by Contradiction Example Theorem There is no largest prime number. Proof. ¬P: There is a largest prime number. Q: The largest prime number is S. prime numbers: 2, 3, 5, 7, 11, . . . , S 2 · 3 · 5 · 7 · 11 · · · S + 1 is not divisible by a prime number in the range [2, S] 1 either it is prime itself: ¬Q 2 or it is divisible by a prime number greater than S: ¬Q
54. 54. Proof by Contradiction Example Theorem There is no largest prime number. Proof. ¬P: There is a largest prime number. Q: The largest prime number is S. prime numbers: 2, 3, 5, 7, 11, . . . , S 2 · 3 · 5 · 7 · 11 · · · S + 1 is not divisible by a prime number in the range [2, S] 1 either it is prime itself: ¬Q 2 or it is divisible by a prime number greater than S: ¬Q
55. 55. Proof by Contradiction Example Theorem There is no largest prime number. Proof. ¬P: There is a largest prime number. Q: The largest prime number is S. prime numbers: 2, 3, 5, 7, 11, . . . , S 2 · 3 · 5 · 7 · 11 · · · S + 1 is not divisible by a prime number in the range [2, S] 1 either it is prime itself: ¬Q 2 or it is divisible by a prime number greater than S: ¬Q
56. 56. Proof by Contradiction Example Theorem There is no largest prime number. Proof. ¬P: There is a largest prime number. Q: The largest prime number is S. prime numbers: 2, 3, 5, 7, 11, . . . , S 2 · 3 · 5 · 7 · 11 · · · S + 1 is not divisible by a prime number in the range [2, S] 1 either it is prime itself: ¬Q 2 or it is divisible by a prime number greater than S: ¬Q
57. 57. Proof by Contradiction Example Theorem There is no largest prime number. Proof. ¬P: There is a largest prime number. Q: The largest prime number is S. prime numbers: 2, 3, 5, 7, 11, . . . , S 2 · 3 · 5 · 7 · 11 · · · S + 1 is not divisible by a prime number in the range [2, S] 1 either it is prime itself: ¬Q 2 or it is divisible by a prime number greater than S: ¬Q
58. 58. Proof by Contradiction Example Theorem There is no largest prime number. Proof. ¬P: There is a largest prime number. Q: The largest prime number is S. prime numbers: 2, 3, 5, 7, 11, . . . , S 2 · 3 · 5 · 7 · 11 · · · S + 1 is not divisible by a prime number in the range [2, S] 1 either it is prime itself: ¬Q 2 or it is divisible by a prime number greater than S: ¬Q
59. 59. Proof by Contradiction Example Theorem √ a ¬∃a, b ∈ Z+ [ 2 = b ] Proof. √ a ¬P: ∃a, b ∈ Z+ [ 2 = b ] Q: gcd(a, b) = 1 a2 ⇒ 4j 2 = 2b2 ⇒ 2= b2 ⇒ b2 = 2j 2 ⇒ a = 2b2 2 ⇒ ∃k ∈ Z+ [b2 = 2k] ⇒ ∃i ∈ Z+ [a2 = 2i] + ⇒ ∃l ∈ Z+ [b = 2l] ⇒ ∃j ∈ Z [a = 2j] ⇒ gcd(a, b) ≥ 2 : ¬Q
60. 60. Proof by Contradiction Example Theorem √ a ¬∃a, b ∈ Z+ [ 2 = b ] Proof. √ a ¬P: ∃a, b ∈ Z+ [ 2 = b ] Q: gcd(a, b) = 1 a2 ⇒ 4j 2 = 2b2 ⇒ 2= b2 ⇒ b2 = 2j 2 ⇒ a = 2b2 2 ⇒ ∃k ∈ Z+ [b2 = 2k] ⇒ ∃i ∈ Z+ [a2 = 2i] + ⇒ ∃l ∈ Z+ [b = 2l] ⇒ ∃j ∈ Z [a = 2j] ⇒ gcd(a, b) ≥ 2 : ¬Q
61. 61. Proof by Contradiction Example Theorem √ a ¬∃a, b ∈ Z+ [ 2 = b ] Proof. √ a ¬P: ∃a, b ∈ Z+ [ 2 = b ] Q: gcd(a, b) = 1 a2 ⇒ 4j 2 = 2b2 ⇒ 2= b2 ⇒ b2 = 2j 2 ⇒ a = 2b2 2 ⇒ ∃k ∈ Z+ [b2 = 2k] ⇒ ∃i ∈ Z+ [a2 = 2i] + ⇒ ∃l ∈ Z+ [b = 2l] ⇒ ∃j ∈ Z [a = 2j] ⇒ gcd(a, b) ≥ 2 : ¬Q
62. 62. Proof by Contradiction Example Theorem √ a ¬∃a, b ∈ Z+ [ 2 = b ] Proof. √ a ¬P: ∃a, b ∈ Z+ [ 2 = b ] Q: gcd(a, b) = 1 a2 ⇒ 4j 2 = 2b2 ⇒ 2= b2 ⇒ b2 = 2j 2 ⇒ a = 2b2 2 ⇒ ∃k ∈ Z+ [b2 = 2k] ⇒ ∃i ∈ Z+ [a2 = 2i] + ⇒ ∃l ∈ Z+ [b = 2l] ⇒ ∃j ∈ Z [a = 2j] ⇒ gcd(a, b) ≥ 2 : ¬Q
63. 63. Proof by Contradiction Example Theorem √ a ¬∃a, b ∈ Z+ [ 2 = b ] Proof. √ a ¬P: ∃a, b ∈ Z+ [ 2 = b ] Q: gcd(a, b) = 1 a2 ⇒ 4j 2 = 2b2 ⇒ 2= b2 ⇒ b2 = 2j 2 ⇒ a = 2b2 2 ⇒ ∃k ∈ Z+ [b2 = 2k] ⇒ ∃i ∈ Z+ [a2 = 2i] + ⇒ ∃l ∈ Z+ [b = 2l] ⇒ ∃j ∈ Z [a = 2j] ⇒ gcd(a, b) ≥ 2 : ¬Q
64. 64. Proof by Contradiction Example Theorem √ a ¬∃a, b ∈ Z+ [ 2 = b ] Proof. √ a ¬P: ∃a, b ∈ Z+ [ 2 = b ] Q: gcd(a, b) = 1 a2 ⇒ 4j 2 = 2b2 ⇒ 2= b2 ⇒ b2 = 2j 2 ⇒ a = 2b2 2 ⇒ ∃k ∈ Z+ [b2 = 2k] ⇒ ∃i ∈ Z+ [a2 = 2i] + ⇒ ∃l ∈ Z+ [b = 2l] ⇒ ∃j ∈ Z [a = 2j] ⇒ gcd(a, b) ≥ 2 : ¬Q
65. 65. Proof by Contradiction Example Theorem √ a ¬∃a, b ∈ Z+ [ 2 = b ] Proof. √ a ¬P: ∃a, b ∈ Z+ [ 2 = b ] Q: gcd(a, b) = 1 a2 ⇒ 4j 2 = 2b2 ⇒ 2= b2 ⇒ b2 = 2j 2 ⇒ a = 2b2 2 ⇒ ∃k ∈ Z+ [b2 = 2k] ⇒ ∃i ∈ Z+ [a2 = 2i] + ⇒ ∃l ∈ Z+ [b = 2l] ⇒ ∃j ∈ Z [a = 2j] ⇒ gcd(a, b) ≥ 2 : ¬Q
66. 66. Proof by Contradiction Example Theorem √ a ¬∃a, b ∈ Z+ [ 2 = b ] Proof. √ a ¬P: ∃a, b ∈ Z+ [ 2 = b ] Q: gcd(a, b) = 1 a2 ⇒ 4j 2 = 2b2 ⇒ 2= b2 ⇒ b2 = 2j 2 ⇒ a = 2b2 2 ⇒ ∃k ∈ Z+ [b2 = 2k] ⇒ ∃i ∈ Z+ [a2 = 2i] + ⇒ ∃l ∈ Z+ [b = 2l] ⇒ ∃j ∈ Z [a = 2j] ⇒ gcd(a, b) ≥ 2 : ¬Q
67. 67. Proof by Contradiction Example Theorem √ a ¬∃a, b ∈ Z+ [ 2 = b ] Proof. √ a ¬P: ∃a, b ∈ Z+ [ 2 = b ] Q: gcd(a, b) = 1 a2 ⇒ 4j 2 = 2b2 ⇒ 2= b2 ⇒ b2 = 2j 2 ⇒ a = 2b2 2 ⇒ ∃k ∈ Z+ [b2 = 2k] ⇒ ∃i ∈ Z+ [a2 = 2i] + ⇒ ∃l ∈ Z+ [b = 2l] ⇒ ∃j ∈ Z [a = 2j] ⇒ gcd(a, b) ≥ 2 : ¬Q
68. 68. Proof by Contradiction Example Theorem √ a ¬∃a, b ∈ Z+ [ 2 = b ] Proof. √ a ¬P: ∃a, b ∈ Z+ [ 2 = b ] Q: gcd(a, b) = 1 a2 ⇒ 4j 2 = 2b2 ⇒ 2= b2 ⇒ b2 = 2j 2 ⇒ a = 2b2 2 ⇒ ∃k ∈ Z+ [b2 = 2k] ⇒ ∃i ∈ Z+ [a2 = 2i] + ⇒ ∃l ∈ Z+ [b = 2l] ⇒ ∃j ∈ Z [a = 2j] ⇒ gcd(a, b) ≥ 2 : ¬Q
69. 69. Proof by Contradiction Example Theorem √ a ¬∃a, b ∈ Z+ [ 2 = b ] Proof. √ a ¬P: ∃a, b ∈ Z+ [ 2 = b ] Q: gcd(a, b) = 1 a2 ⇒ 4j 2 = 2b2 ⇒ 2= b2 ⇒ b2 = 2j 2 ⇒ a = 2b2 2 ⇒ ∃k ∈ Z+ [b2 = 2k] ⇒ ∃i ∈ Z+ [a2 = 2i] + ⇒ ∃l ∈ Z+ [b = 2l] ⇒ ∃j ∈ Z [a = 2j] ⇒ gcd(a, b) ≥ 2 : ¬Q
70. 70. Topics 1 Basic Techniques Introduction Direct Proof Proof by Contradiction Equivalence Proofs 2 Induction Introduction Strong Induction
71. 71. Equivalence Proofs to prove P ⇔ Q, both P ⇒ Q and Q ⇒ P must be proven a method to prove P1 ⇔ P2 ⇔ · · · ⇔ Pn : P1 ⇒ P2 ⇒ · · · ⇒ Pn ⇒ P1
72. 72. Equivalence Proofs to prove P ⇔ Q, both P ⇒ Q and Q ⇒ P must be proven a method to prove P1 ⇔ P2 ⇔ · · · ⇔ Pn : P1 ⇒ P2 ⇒ · · · ⇒ Pn ⇒ P1
73. 73. Equivalence Proof Example Theorem a, b, n, q1 , r1 , q2 , r2 ∈ Z+ a = q1 · n + r1 b = q2 · n + r2 r1 = r2 ⇔ n|(a − b)
74. 74. Equivalence Proof Exampler1 = r2 ⇒ n|(a − b). n|(a − b) ⇒ r1 = r2 . a−b = (q1 · n + r1 ) a−b = (q1 · n + r1 ) −(q2 · n + r2 ) −(q2 · n + r2 ) = (q1 − q2 ) · n = (q1 − q2 ) · n +(r1 − r2 ) +(r1 − r2 ) r1 = r2 ⇒ r1 − r2 = 0 n|(a − b) ⇒ r1 − r2 = 0 ⇒ a − b = (q1 − q2 ) · n ⇒ r1 = r2
75. 75. Equivalence Proof Exampler1 = r2 ⇒ n|(a − b). n|(a − b) ⇒ r1 = r2 . a−b = (q1 · n + r1 ) a−b = (q1 · n + r1 ) −(q2 · n + r2 ) −(q2 · n + r2 ) = (q1 − q2 ) · n = (q1 − q2 ) · n +(r1 − r2 ) +(r1 − r2 ) r1 = r2 ⇒ r1 − r2 = 0 n|(a − b) ⇒ r1 − r2 = 0 ⇒ a − b = (q1 − q2 ) · n ⇒ r1 = r2
76. 76. Equivalence Proof Exampler1 = r2 ⇒ n|(a − b). n|(a − b) ⇒ r1 = r2 . a−b = (q1 · n + r1 ) a−b = (q1 · n + r1 ) −(q2 · n + r2 ) −(q2 · n + r2 ) = (q1 − q2 ) · n = (q1 − q2 ) · n +(r1 − r2 ) +(r1 − r2 ) r1 = r2 ⇒ r1 − r2 = 0 n|(a − b) ⇒ r1 − r2 = 0 ⇒ a − b = (q1 − q2 ) · n ⇒ r1 = r2
77. 77. Equivalence Proof Exampler1 = r2 ⇒ n|(a − b). n|(a − b) ⇒ r1 = r2 . a−b = (q1 · n + r1 ) a−b = (q1 · n + r1 ) −(q2 · n + r2 ) −(q2 · n + r2 ) = (q1 − q2 ) · n = (q1 − q2 ) · n +(r1 − r2 ) +(r1 − r2 ) r1 = r2 ⇒ r1 − r2 = 0 n|(a − b) ⇒ r1 − r2 = 0 ⇒ a − b = (q1 − q2 ) · n ⇒ r1 = r2
78. 78. Equivalence Proof Exampler1 = r2 ⇒ n|(a − b). n|(a − b) ⇒ r1 = r2 . a−b = (q1 · n + r1 ) a−b = (q1 · n + r1 ) −(q2 · n + r2 ) −(q2 · n + r2 ) = (q1 − q2 ) · n = (q1 − q2 ) · n +(r1 − r2 ) +(r1 − r2 ) r1 = r2 ⇒ r1 − r2 = 0 n|(a − b) ⇒ r1 − r2 = 0 ⇒ a − b = (q1 − q2 ) · n ⇒ r1 = r2
79. 79. Equivalence Proof Exampler1 = r2 ⇒ n|(a − b). n|(a − b) ⇒ r1 = r2 . a−b = (q1 · n + r1 ) a−b = (q1 · n + r1 ) −(q2 · n + r2 ) −(q2 · n + r2 ) = (q1 − q2 ) · n = (q1 − q2 ) · n +(r1 − r2 ) +(r1 − r2 ) r1 = r2 ⇒ r1 − r2 = 0 n|(a − b) ⇒ r1 − r2 = 0 ⇒ a − b = (q1 − q2 ) · n ⇒ r1 = r2
80. 80. Equivalence Proof Exampler1 = r2 ⇒ n|(a − b). n|(a − b) ⇒ r1 = r2 . a−b = (q1 · n + r1 ) a−b = (q1 · n + r1 ) −(q2 · n + r2 ) −(q2 · n + r2 ) = (q1 − q2 ) · n = (q1 − q2 ) · n +(r1 − r2 ) +(r1 − r2 ) r1 = r2 ⇒ r1 − r2 = 0 n|(a − b) ⇒ r1 − r2 = 0 ⇒ a − b = (q1 − q2 ) · n ⇒ r1 = r2
81. 81. Equivalence Proof Exampler1 = r2 ⇒ n|(a − b). n|(a − b) ⇒ r1 = r2 . a−b = (q1 · n + r1 ) a−b = (q1 · n + r1 ) −(q2 · n + r2 ) −(q2 · n + r2 ) = (q1 − q2 ) · n = (q1 − q2 ) · n +(r1 − r2 ) +(r1 − r2 ) r1 = r2 ⇒ r1 − r2 = 0 n|(a − b) ⇒ r1 − r2 = 0 ⇒ a − b = (q1 − q2 ) · n ⇒ r1 = r2
82. 82. Equivalence Proof Example Theorem A⊆B ⇔ A∪B =B ⇔ A∩B =A ⇔ B⊆A
83. 83. Equivalence Proof Example A ⊆ B ⇒ A ∪ B = B. A∪B =B ⇔A∪B ⊆B ∧B ⊆A∪B B ⊆A∪B x ∈A∪B ⇒ x ∈A∨x ∈B A⊆B ⇒ x ∈B ⇒ A∪B ⊆B
84. 84. Equivalence Proof Example A ⊆ B ⇒ A ∪ B = B. A∪B =B ⇔A∪B ⊆B ∧B ⊆A∪B B ⊆A∪B x ∈A∪B ⇒ x ∈A∨x ∈B A⊆B ⇒ x ∈B ⇒ A∪B ⊆B
85. 85. Equivalence Proof Example A ⊆ B ⇒ A ∪ B = B. A∪B =B ⇔A∪B ⊆B ∧B ⊆A∪B B ⊆A∪B x ∈A∪B ⇒ x ∈A∨x ∈B A⊆B ⇒ x ∈B ⇒ A∪B ⊆B
86. 86. Equivalence Proof Example A ⊆ B ⇒ A ∪ B = B. A∪B =B ⇔A∪B ⊆B ∧B ⊆A∪B B ⊆A∪B x ∈A∪B ⇒ x ∈A∨x ∈B A⊆B ⇒ x ∈B ⇒ A∪B ⊆B
87. 87. Equivalence Proof Example A ⊆ B ⇒ A ∪ B = B. A∪B =B ⇔A∪B ⊆B ∧B ⊆A∪B B ⊆A∪B x ∈A∪B ⇒ x ∈A∨x ∈B A⊆B ⇒ x ∈B ⇒ A∪B ⊆B
88. 88. Equivalence Proof Example A ∪ B = B ⇒ A ∩ B = A. A∩B =A⇔A∩B ⊆A∧A⊆A∩B y ∈A ⇒ y ∈A∪B A∩B ⊆A A∪B =B ⇒ y ∈B ⇒ y ∈A∩B ⇒ A⊆A∩B
89. 89. Equivalence Proof Example A ∪ B = B ⇒ A ∩ B = A. A∩B =A⇔A∩B ⊆A∧A⊆A∩B y ∈A ⇒ y ∈A∪B A∩B ⊆A A∪B =B ⇒ y ∈B ⇒ y ∈A∩B ⇒ A⊆A∩B
90. 90. Equivalence Proof Example A ∪ B = B ⇒ A ∩ B = A. A∩B =A⇔A∩B ⊆A∧A⊆A∩B y ∈A ⇒ y ∈A∪B A∩B ⊆A A∪B =B ⇒ y ∈B ⇒ y ∈A∩B ⇒ A⊆A∩B
91. 91. Equivalence Proof Example A ∪ B = B ⇒ A ∩ B = A. A∩B =A⇔A∩B ⊆A∧A⊆A∩B y ∈A ⇒ y ∈A∪B A∩B ⊆A A∪B =B ⇒ y ∈B ⇒ y ∈A∩B ⇒ A⊆A∩B
92. 92. Equivalence Proof Example A ∪ B = B ⇒ A ∩ B = A. A∩B =A⇔A∩B ⊆A∧A⊆A∩B y ∈A ⇒ y ∈A∪B A∩B ⊆A A∪B =B ⇒ y ∈B ⇒ y ∈A∩B ⇒ A⊆A∩B
93. 93. Equivalence Proof Example A ∪ B = B ⇒ A ∩ B = A. A∩B =A⇔A∩B ⊆A∧A⊆A∩B y ∈A ⇒ y ∈A∪B A∩B ⊆A A∪B =B ⇒ y ∈B ⇒ y ∈A∩B ⇒ A⊆A∩B
94. 94. Equivalence Proof Example A ∩ B = A ⇒ B ⊆ A. z ∈B ⇒ z ∈B / ⇒ z ∈A∩B / A∩B =A ⇒ z ∈A / ⇒ z ∈A ⇒ B⊆A
95. 95. Equivalence Proof Example A ∩ B = A ⇒ B ⊆ A. z ∈B ⇒ z ∈B / ⇒ z ∈A∩B / A∩B =A ⇒ z ∈A / ⇒ z ∈A ⇒ B⊆A
96. 96. Equivalence Proof Example A ∩ B = A ⇒ B ⊆ A. z ∈B ⇒ z ∈B / ⇒ z ∈A∩B / A∩B =A ⇒ z ∈A / ⇒ z ∈A ⇒ B⊆A
97. 97. Equivalence Proof Example A ∩ B = A ⇒ B ⊆ A. z ∈B ⇒ z ∈B / ⇒ z ∈A∩B / A∩B =A ⇒ z ∈A / ⇒ z ∈A ⇒ B⊆A
98. 98. Equivalence Proof Example A ∩ B = A ⇒ B ⊆ A. z ∈B ⇒ z ∈B / ⇒ z ∈A∩B / A∩B =A ⇒ z ∈A / ⇒ z ∈A ⇒ B⊆A
99. 99. Equivalence Proof Example B ⊆ A ⇒ A ⊆ B. ¬(A ⊆ B) ⇒ ∃w [w ∈ A ∧ w ∈ B] / ⇒ ∃w [w ∈ A ∧ w ∈ B] / ⇒ ¬(B ⊆ A)
100. 100. Equivalence Proof Example B ⊆ A ⇒ A ⊆ B. ¬(A ⊆ B) ⇒ ∃w [w ∈ A ∧ w ∈ B] / ⇒ ∃w [w ∈ A ∧ w ∈ B] / ⇒ ¬(B ⊆ A)
101. 101. Equivalence Proof Example B ⊆ A ⇒ A ⊆ B. ¬(A ⊆ B) ⇒ ∃w [w ∈ A ∧ w ∈ B] / ⇒ ∃w [w ∈ A ∧ w ∈ B] / ⇒ ¬(B ⊆ A)
102. 102. Topics 1 Basic Techniques Introduction Direct Proof Proof by Contradiction Equivalence Proofs 2 Induction Introduction Strong Induction
103. 103. Induction Deﬁnition S(n): a predicate deﬁned on n ∈ Z+ S(n0 ) ∧ (∀k ≥ n0 [S(k) ⇒ S(k + 1)]) ⇒ ∀n ≥ n0 S(n) S(n0 ): base step ∀k ≥ n0 [S(k) ⇒ S(k + 1)]: induction step
104. 104. Induction Deﬁnition S(n): a predicate deﬁned on n ∈ Z+ S(n0 ) ∧ (∀k ≥ n0 [S(k) ⇒ S(k + 1)]) ⇒ ∀n ≥ n0 S(n) S(n0 ): base step ∀k ≥ n0 [S(k) ⇒ S(k + 1)]: induction step
105. 105. Induction Deﬁnition S(n): a predicate deﬁned on n ∈ Z+ S(n0 ) ∧ (∀k ≥ n0 [S(k) ⇒ S(k + 1)]) ⇒ ∀n ≥ n0 S(n) S(n0 ): base step ∀k ≥ n0 [S(k) ⇒ S(k + 1)]: induction step
106. 106. Induction
107. 107. Induction Example Theorem ∀n ∈ Z+ [1 + 3 + 5 + · · · + (2n − 1) = n2 ] Proof. n = 1: 1 = 12 n = k: assume 1 + 3 + 5 + · · · + (2k − 1) = k 2 n = k + 1: 1 + 3 + 5 + · · · + (2k − 1) + (2k + 1) = k 2 + 2k + 1 = (k + 1)2
108. 108. Induction Example Theorem ∀n ∈ Z+ [1 + 3 + 5 + · · · + (2n − 1) = n2 ] Proof. n = 1: 1 = 12 n = k: assume 1 + 3 + 5 + · · · + (2k − 1) = k 2 n = k + 1: 1 + 3 + 5 + · · · + (2k − 1) + (2k + 1) = k 2 + 2k + 1 = (k + 1)2
109. 109. Induction Example Theorem ∀n ∈ Z+ [1 + 3 + 5 + · · · + (2n − 1) = n2 ] Proof. n = 1: 1 = 12 n = k: assume 1 + 3 + 5 + · · · + (2k − 1) = k 2 n = k + 1: 1 + 3 + 5 + · · · + (2k − 1) + (2k + 1) = k 2 + 2k + 1 = (k + 1)2
110. 110. Induction Example Theorem ∀n ∈ Z+ [1 + 3 + 5 + · · · + (2n − 1) = n2 ] Proof. n = 1: 1 = 12 n = k: assume 1 + 3 + 5 + · · · + (2k − 1) = k 2 n = k + 1: 1 + 3 + 5 + · · · + (2k − 1) + (2k + 1) = k 2 + 2k + 1 = (k + 1)2
111. 111. Induction Example Theorem ∀n ∈ Z+ [1 + 3 + 5 + · · · + (2n − 1) = n2 ] Proof. n = 1: 1 = 12 n = k: assume 1 + 3 + 5 + · · · + (2k − 1) = k 2 n = k + 1: 1 + 3 + 5 + · · · + (2k − 1) + (2k + 1) = k 2 + 2k + 1 = (k + 1)2
112. 112. Induction Example Theorem ∀n ∈ Z+ [1 + 3 + 5 + · · · + (2n − 1) = n2 ] Proof. n = 1: 1 = 12 n = k: assume 1 + 3 + 5 + · · · + (2k − 1) = k 2 n = k + 1: 1 + 3 + 5 + · · · + (2k − 1) + (2k + 1) = k 2 + 2k + 1 = (k + 1)2
113. 113. Induction Example Theorem ∀n ∈ Z+ , n ≥ 4 [2n < n!] Proof. n = 4: 24 = 16 < 24 = 4! n = k: assume 2k < k! n = k + 1: 2k+1 = 2 · 2k < 2 · k! < (k + 1) · k! = (k + 1)!
114. 114. Induction Example Theorem ∀n ∈ Z+ , n ≥ 4 [2n < n!] Proof. n = 4: 24 = 16 < 24 = 4! n = k: assume 2k < k! n = k + 1: 2k+1 = 2 · 2k < 2 · k! < (k + 1) · k! = (k + 1)!
115. 115. Induction Example Theorem ∀n ∈ Z+ , n ≥ 4 [2n < n!] Proof. n = 4: 24 = 16 < 24 = 4! n = k: assume 2k < k! n = k + 1: 2k+1 = 2 · 2k < 2 · k! < (k + 1) · k! = (k + 1)!
116. 116. Induction Example Theorem ∀n ∈ Z+ , n ≥ 4 [2n < n!] Proof. n = 4: 24 = 16 < 24 = 4! n = k: assume 2k < k! n = k + 1: 2k+1 = 2 · 2k < 2 · k! < (k + 1) · k! = (k + 1)!
117. 117. Induction Example Theorem ∀n ∈ Z+ , n ≥ 14 ∃i, j ∈ N [n = 3i + 8j] Proof. n = 14: 14 = 3 · 2 + 8 · 1 n = k: assume k = 3i + 8j n = k + 1: k = 3i + 8j, j > 0 ⇒ k + 1 = k − 8 + 3 · 3 ⇒ k + 1 = 3(i + 3) + 8(j − 1) k = 3i + 8j, j = 0, i ≥ 5 ⇒ k + 1 = k − 5 · 3 + 2 · 8 ⇒ k + 1 = 3(i − 5) + 8(j + 2)
118. 118. Induction Example Theorem ∀n ∈ Z+ , n ≥ 14 ∃i, j ∈ N [n = 3i + 8j] Proof. n = 14: 14 = 3 · 2 + 8 · 1 n = k: assume k = 3i + 8j n = k + 1: k = 3i + 8j, j > 0 ⇒ k + 1 = k − 8 + 3 · 3 ⇒ k + 1 = 3(i + 3) + 8(j − 1) k = 3i + 8j, j = 0, i ≥ 5 ⇒ k + 1 = k − 5 · 3 + 2 · 8 ⇒ k + 1 = 3(i − 5) + 8(j + 2)
119. 119. Induction Example Theorem ∀n ∈ Z+ , n ≥ 14 ∃i, j ∈ N [n = 3i + 8j] Proof. n = 14: 14 = 3 · 2 + 8 · 1 n = k: assume k = 3i + 8j n = k + 1: k = 3i + 8j, j > 0 ⇒ k + 1 = k − 8 + 3 · 3 ⇒ k + 1 = 3(i + 3) + 8(j − 1) k = 3i + 8j, j = 0, i ≥ 5 ⇒ k + 1 = k − 5 · 3 + 2 · 8 ⇒ k + 1 = 3(i − 5) + 8(j + 2)
120. 120. Induction Example Theorem ∀n ∈ Z+ , n ≥ 14 ∃i, j ∈ N [n = 3i + 8j] Proof. n = 14: 14 = 3 · 2 + 8 · 1 n = k: assume k = 3i + 8j n = k + 1: k = 3i + 8j, j > 0 ⇒ k + 1 = k − 8 + 3 · 3 ⇒ k + 1 = 3(i + 3) + 8(j − 1) k = 3i + 8j, j = 0, i ≥ 5 ⇒ k + 1 = k − 5 · 3 + 2 · 8 ⇒ k + 1 = 3(i − 5) + 8(j + 2)
121. 121. Topics 1 Basic Techniques Introduction Direct Proof Proof by Contradiction Equivalence Proofs 2 Induction Introduction Strong Induction
122. 122. Strong Induction Deﬁnition S(n0 ) ∧ (∀k ≥ n0 [(∀i ≤ k S(i)) ⇒ S(k + 1)]) ⇒ ∀n ≥ n0 S(n)
123. 123. Strong Induction Example Theorem ∀n ∈ Z+ , n ≥ 2 n can be written as the product of prime numbers. Proof. n = 2: 2 = 2 assume that the theorem is true for ∀i ≤ k n = k + 1: 1 if prime: n = n 2 if not prime: n = u · v u < k ∧ v < k ⇒ both u and v can be written as the product of prime numbers
124. 124. Strong Induction Example Theorem ∀n ∈ Z+ , n ≥ 2 n can be written as the product of prime numbers. Proof. n = 2: 2 = 2 assume that the theorem is true for ∀i ≤ k n = k + 1: 1 if prime: n = n 2 if not prime: n = u · v u < k ∧ v < k ⇒ both u and v can be written as the product of prime numbers
125. 125. Strong Induction Example Theorem ∀n ∈ Z+ , n ≥ 2 n can be written as the product of prime numbers. Proof. n = 2: 2 = 2 assume that the theorem is true for ∀i ≤ k n = k + 1: 1 if prime: n = n 2 if not prime: n = u · v u < k ∧ v < k ⇒ both u and v can be written as the product of prime numbers
126. 126. Strong Induction Example Theorem ∀n ∈ Z+ , n ≥ 2 n can be written as the product of prime numbers. Proof. n = 2: 2 = 2 assume that the theorem is true for ∀i ≤ k n = k + 1: 1 if prime: n = n 2 if not prime: n = u · v u < k ∧ v < k ⇒ both u and v can be written as the product of prime numbers
127. 127. Strong Induction Example Theorem ∀n ∈ Z+ , n ≥ 2 n can be written as the product of prime numbers. Proof. n = 2: 2 = 2 assume that the theorem is true for ∀i ≤ k n = k + 1: 1 if prime: n = n 2 if not prime: n = u · v u < k ∧ v < k ⇒ both u and v can be written as the product of prime numbers
128. 128. Strong Induction Example Theorem ∀n ∈ Z+ , n ≥ 14 ∃i, j ∈ N [n = 3i + 8j] Proof. n = 14: 14 = 3 · 2 + 8 · 1 n = 15: 15 = 3 · 5 + 8 · 0 n = 16: 16 = 3 · 0 + 8 · 2 n ≤ k: assume k = 3i + 8j n = k + 1: k + 1 = (k − 2) + 3
129. 129. Strong Induction Example Theorem ∀n ∈ Z+ , n ≥ 14 ∃i, j ∈ N [n = 3i + 8j] Proof. n = 14: 14 = 3 · 2 + 8 · 1 n = 15: 15 = 3 · 5 + 8 · 0 n = 16: 16 = 3 · 0 + 8 · 2 n ≤ k: assume k = 3i + 8j n = k + 1: k + 1 = (k − 2) + 3
130. 130. Strong Induction Example Theorem ∀n ∈ Z+ , n ≥ 14 ∃i, j ∈ N [n = 3i + 8j] Proof. n = 14: 14 = 3 · 2 + 8 · 1 n = 15: 15 = 3 · 5 + 8 · 0 n = 16: 16 = 3 · 0 + 8 · 2 n ≤ k: assume k = 3i + 8j n = k + 1: k + 1 = (k − 2) + 3
131. 131. Strong Induction Example Theorem ∀n ∈ Z+ , n ≥ 14 ∃i, j ∈ N [n = 3i + 8j] Proof. n = 14: 14 = 3 · 2 + 8 · 1 n = 15: 15 = 3 · 5 + 8 · 0 n = 16: 16 = 3 · 0 + 8 · 2 n ≤ k: assume k = 3i + 8j n = k + 1: k + 1 = (k − 2) + 3
132. 132. Flawed Induction Example Theorem n2 +n+2 ∀n ∈ Z+ [1 + 2 + 3 + · · · + n = 2 ] invalid base step k 2 +k+2 n = k: assume 1 + 2 + 3 + · · · + k = 2 n = k + 1: 1 + 2 + 3 + · · · + k + (k + 1) k2 + k + 2 k 2 + k + 2 2k + 2 = +k +1= + 2 2 2 k 2 + 3k + 4 (k + 1)2 + (k + 1) + 2 = = 2 2 12 +1+2 n = 1: 1 = 2 =2
133. 133. Flawed Induction Example Theorem n2 +n+2 ∀n ∈ Z+ [1 + 2 + 3 + · · · + n = 2 ] invalid base step k 2 +k+2 n = k: assume 1 + 2 + 3 + · · · + k = 2 n = k + 1: 1 + 2 + 3 + · · · + k + (k + 1) k2 + k + 2 k 2 + k + 2 2k + 2 = +k +1= + 2 2 2 k 2 + 3k + 4 (k + 1)2 + (k + 1) + 2 = = 2 2 12 +1+2 n = 1: 1 = 2 =2
134. 134. Flawed Induction Example Theorem n2 +n+2 ∀n ∈ Z+ [1 + 2 + 3 + · · · + n = 2 ] invalid base step k 2 +k+2 n = k: assume 1 + 2 + 3 + · · · + k = 2 n = k + 1: 1 + 2 + 3 + · · · + k + (k + 1) k2 + k + 2 k 2 + k + 2 2k + 2 = +k +1= + 2 2 2 k 2 + 3k + 4 (k + 1)2 + (k + 1) + 2 = = 2 2 12 +1+2 n = 1: 1 = 2 =2
135. 135. Flawed Induction Example Theorem n2 +n+2 ∀n ∈ Z+ [1 + 2 + 3 + · · · + n = 2 ] invalid base step k 2 +k+2 n = k: assume 1 + 2 + 3 + · · · + k = 2 n = k + 1: 1 + 2 + 3 + · · · + k + (k + 1) k2 + k + 2 k 2 + k + 2 2k + 2 = +k +1= + 2 2 2 k 2 + 3k + 4 (k + 1)2 + (k + 1) + 2 = = 2 2 12 +1+2 n = 1: 1 = 2 =2
136. 136. Flawed Induction Example Theorem n2 +n+2 ∀n ∈ Z+ [1 + 2 + 3 + · · · + n = 2 ] invalid base step k 2 +k+2 n = k: assume 1 + 2 + 3 + · · · + k = 2 n = k + 1: 1 + 2 + 3 + · · · + k + (k + 1) k2 + k + 2 k 2 + k + 2 2k + 2 = +k +1= + 2 2 2 k 2 + 3k + 4 (k + 1)2 + (k + 1) + 2 = = 2 2 12 +1+2 n = 1: 1 = 2 =2
137. 137. Flawed Induction Example Theorem n2 +n+2 ∀n ∈ Z+ [1 + 2 + 3 + · · · + n = 2 ] invalid base step k 2 +k+2 n = k: assume 1 + 2 + 3 + · · · + k = 2 n = k + 1: 1 + 2 + 3 + · · · + k + (k + 1) k2 + k + 2 k 2 + k + 2 2k + 2 = +k +1= + 2 2 2 k 2 + 3k + 4 (k + 1)2 + (k + 1) + 2 = = 2 2 12 +1+2 n = 1: 1 = 2 =2
138. 138. Flawed Induction Example
139. 139. Flawed Induction Example Theorem All horses are of the same color. A(n): All horses in sets of n horses are of the same color. ∀n ∈ N+ A(n)
140. 140. Flawed Induction Example Theorem All horses are of the same color. A(n): All horses in sets of n horses are of the same color. ∀n ∈ N+ A(n)
141. 141. Flawed Induction Example invalid induction step n = 1: A(1) All horses in sets of 1 horse are of the same color. n = k: assume A(k) is true All horses in sets of k horses are of the same color. A(k + 1) = {a1 , a2 , . . . , ak } ∪ {a2 , a3 , . . . , ak+1 } All horses in set {a1 , a2 , . . . , ak } are of the same color (a2 ). All horses in set {a2 , a3 , . . . , ak+1 } are of the same color (a2 ).
142. 142. Flawed Induction Example invalid induction step n = 1: A(1) All horses in sets of 1 horse are of the same color. n = k: assume A(k) is true All horses in sets of k horses are of the same color. A(k + 1) = {a1 , a2 , . . . , ak } ∪ {a2 , a3 , . . . , ak+1 } All horses in set {a1 , a2 , . . . , ak } are of the same color (a2 ). All horses in set {a2 , a3 , . . . , ak+1 } are of the same color (a2 ).
143. 143. Flawed Induction Example invalid induction step n = 1: A(1) All horses in sets of 1 horse are of the same color. n = k: assume A(k) is true All horses in sets of k horses are of the same color. A(k + 1) = {a1 , a2 , . . . , ak } ∪ {a2 , a3 , . . . , ak+1 } All horses in set {a1 , a2 , . . . , ak } are of the same color (a2 ). All horses in set {a2 , a3 , . . . , ak+1 } are of the same color (a2 ).
144. 144. Flawed Induction Examples
145. 145. References Required Reading: Grimaldi Chapter 2: Fundamentals of Logic 2.5. Quantiﬁers, Deﬁnitions, and the Proofs of Theorems Chapter 4: Properties of Integers: Mathematical Induction 4.1. The Well-Ordering Principle: Mathematical Induction Supplementary Reading: O’Donnell, Hall, Page Chapter 4: Induction