Discrete Mathematics - Proofs

29,791 views

Published on

Proof techniques, proof by contradiction, mathematical induction.

Published in: Education, Technology
1 Comment
7 Likes
Statistics
Notes
No Downloads
Views
Total views
29,791
On SlideShare
0
From Embeds
0
Number of Embeds
6
Actions
Shares
0
Downloads
546
Comments
1
Likes
7
Embeds 0
No embeds

No notes for slide

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

×