Download - Discrete Mathematics - Proofs
Discrete MathematicsProofs
H. Turgut Uyar Aysegul Gencata Yayımlı Emre Harmancı
2001-2013
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c©2001-2013 T. Uyar, A. Yayımlı, E. Harmancı
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Topics
1 Basic TechniquesIntroductionDirect ProofProof by ContradictionEquivalence Proofs
2 InductionIntroductionStrong Induction
Topics
1 Basic TechniquesIntroductionDirect ProofProof by ContradictionEquivalence Proofs
2 InductionIntroductionStrong Induction
Brute Force Method
examining all possible cases one by one
Theorem
Every number from the set {2, 4, 6, . . . , 26} can be writtenas the sum of at most 3 square numbers.
Proof.2 = 1+1 10 = 9+1 20 = 16+44 = 4 12 = 4+4+4 22 = 9+9+46 = 4+1+1 14 = 9+4+1 24 = 16+4+48 = 4+4 16 = 16 26 = 25+1
18 = 9+9
Brute Force Method
examining all possible cases one by one
Theorem
Every number from the set {2, 4, 6, . . . , 26} can be writtenas the sum of at most 3 square numbers.
Proof.2 = 1+1 10 = 9+1 20 = 16+44 = 4 12 = 4+4+4 22 = 9+9+46 = 4+1+1 14 = 9+4+1 24 = 16+4+48 = 4+4 16 = 16 26 = 25+1
18 = 9+9
Brute Force Method
examining all possible cases one by one
Theorem
Every number from the set {2, 4, 6, . . . , 26} can be writtenas the sum of at most 3 square numbers.
Proof.2 = 1+1 10 = 9+1 20 = 16+44 = 4 12 = 4+4+4 22 = 9+9+46 = 4+1+1 14 = 9+4+1 24 = 16+4+48 = 4+4 16 = 16 26 = 25+1
18 = 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)
Basic Rules
Universal Specification (US)
∀x p(x) ⇒ p(a)
Universal Generalization (UG)
p(a) for an arbitrarily chosen a ⇒ ∀x p(x)
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.
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.
Universal Specification Example
Example
∀x [j(x) ∨ s(x) → ¬p(x)]p(m)
∴ ¬s(m)
1. ∀x [j(x) ∨ s(x) → ¬p(x)] A
2. p(m) A
3. j(m) ∨ s(m) → ¬p(m) US : 1
4. ¬(j(m) ∨ s(m)) MT : 3, 2
5. ¬j(m) ∧ ¬s(m) DM : 4
6. ¬s(m) AndE : 5
Universal Specification Example
Example
∀x [j(x) ∨ s(x) → ¬p(x)]p(m)
∴ ¬s(m)
1. ∀x [j(x) ∨ s(x) → ¬p(x)] A
2. p(m) A
3. j(m) ∨ s(m) → ¬p(m) US : 1
4. ¬(j(m) ∨ s(m)) MT : 3, 2
5. ¬j(m) ∧ ¬s(m) DM : 4
6. ¬s(m) AndE : 5
Universal Specification Example
Example
∀x [j(x) ∨ s(x) → ¬p(x)]p(m)
∴ ¬s(m)
1. ∀x [j(x) ∨ s(x) → ¬p(x)] A
2. p(m) A
3. j(m) ∨ s(m) → ¬p(m) US : 1
4. ¬(j(m) ∨ s(m)) MT : 3, 2
5. ¬j(m) ∧ ¬s(m) DM : 4
6. ¬s(m) AndE : 5
Universal Specification Example
Example
∀x [j(x) ∨ s(x) → ¬p(x)]p(m)
∴ ¬s(m)
1. ∀x [j(x) ∨ s(x) → ¬p(x)] A
2. p(m) A
3. j(m) ∨ s(m) → ¬p(m) US : 1
4. ¬(j(m) ∨ s(m)) MT : 3, 2
5. ¬j(m) ∧ ¬s(m) DM : 4
6. ¬s(m) AndE : 5
Universal Specification Example
Example
∀x [j(x) ∨ s(x) → ¬p(x)]p(m)
∴ ¬s(m)
1. ∀x [j(x) ∨ s(x) → ¬p(x)] A
2. p(m) A
3. j(m) ∨ s(m) → ¬p(m) US : 1
4. ¬(j(m) ∨ s(m)) MT : 3, 2
5. ¬j(m) ∧ ¬s(m) DM : 4
6. ¬s(m) AndE : 5
Universal Specification Example
Example
∀x [j(x) ∨ s(x) → ¬p(x)]p(m)
∴ ¬s(m)
1. ∀x [j(x) ∨ s(x) → ¬p(x)] A
2. p(m) A
3. j(m) ∨ s(m) → ¬p(m) US : 1
4. ¬(j(m) ∨ s(m)) MT : 3, 2
5. ¬j(m) ∧ ¬s(m) DM : 4
6. ¬s(m) AndE : 5
Universal Specification Example
Example
∀x [j(x) ∨ s(x) → ¬p(x)]p(m)
∴ ¬s(m)
1. ∀x [j(x) ∨ s(x) → ¬p(x)] A
2. p(m) A
3. j(m) ∨ s(m) → ¬p(m) US : 1
4. ¬(j(m) ∨ s(m)) MT : 3, 2
5. ¬j(m) ∧ ¬s(m) DM : 4
6. ¬s(m) AndE : 5
Universal Generalization Example
Example
∀x [p(x) → q(x)]∀x [q(x) → r(x)]
∴ ∀x [p(x) → r(x)]
1. ∀x [p(x) → q(x)] A
2. p(c) → q(c) US : 1
3. ∀x [q(x) → r(x)] A
4. q(c) → r(c) US : 3
5. p(c) → r(c) HS : 2, 4
6. ∀x [p(x) → r(x)] UG : 5
Universal Generalization Example
Example
∀x [p(x) → q(x)]∀x [q(x) → r(x)]
∴ ∀x [p(x) → r(x)]
1. ∀x [p(x) → q(x)] A
2. p(c) → q(c) US : 1
3. ∀x [q(x) → r(x)] A
4. q(c) → r(c) US : 3
5. p(c) → r(c) HS : 2, 4
6. ∀x [p(x) → r(x)] UG : 5
Universal Generalization Example
Example
∀x [p(x) → q(x)]∀x [q(x) → r(x)]
∴ ∀x [p(x) → r(x)]
1. ∀x [p(x) → q(x)] A
2. p(c) → q(c) US : 1
3. ∀x [q(x) → r(x)] A
4. q(c) → r(c) US : 3
5. p(c) → r(c) HS : 2, 4
6. ∀x [p(x) → r(x)] UG : 5
Universal Generalization Example
Example
∀x [p(x) → q(x)]∀x [q(x) → r(x)]
∴ ∀x [p(x) → r(x)]
1. ∀x [p(x) → q(x)] A
2. p(c) → q(c) US : 1
3. ∀x [q(x) → r(x)] A
4. q(c) → r(c) US : 3
5. p(c) → r(c) HS : 2, 4
6. ∀x [p(x) → r(x)] UG : 5
Universal Generalization Example
Example
∀x [p(x) → q(x)]∀x [q(x) → r(x)]
∴ ∀x [p(x) → r(x)]
1. ∀x [p(x) → q(x)] A
2. p(c) → q(c) US : 1
3. ∀x [q(x) → r(x)] A
4. q(c) → r(c) US : 3
5. p(c) → r(c) HS : 2, 4
6. ∀x [p(x) → r(x)] UG : 5
Universal Generalization Example
Example
∀x [p(x) → q(x)]∀x [q(x) → r(x)]
∴ ∀x [p(x) → r(x)]
1. ∀x [p(x) → q(x)] A
2. p(c) → q(c) US : 1
3. ∀x [q(x) → r(x)] A
4. q(c) → r(c) US : 3
5. p(c) → r(c) HS : 2, 4
6. ∀x [p(x) → r(x)] UG : 5
Universal Generalization Example
Example
∀x [p(x) → q(x)]∀x [q(x) → r(x)]
∴ ∀x [p(x) → r(x)]
1. ∀x [p(x) → q(x)] A
2. p(c) → q(c) US : 1
3. ∀x [q(x) → r(x)] A
4. q(c) → r(c) US : 3
5. p(c) → r(c) HS : 2, 4
6. ∀x [p(x) → r(x)] UG : 5
Vacuous Proof
vacuous proof
to prove P ⇒ Q, show that P is false
Vacuous Proof Example
Theorem
∀S [∅ ⊆ S ]
Proof.
∅ ⊆ S ⇔ ∀x [x ∈ ∅ → x ∈ S ]∀x [x /∈ ∅]
Vacuous Proof Example
Theorem
∀S [∅ ⊆ S ]
Proof.
∅ ⊆ S ⇔ ∀x [x ∈ ∅ → x ∈ S ]∀x [x /∈ ∅]
Vacuous Proof Example
Theorem
∀S [∅ ⊆ S ]
Proof.
∅ ⊆ S ⇔ ∀x [x ∈ ∅ → x ∈ S ]∀x [x /∈ ∅]
Trivial Proof
trivial proof
to prove P ⇒ Q, show that Q is true
Trivial Proof Example
Theorem
∀x ∈ R [x ≥ 0 ⇒ x2 ≥ 0]
Proof.
∀x ∈ R [x2 ≥ 0]
Trivial Proof Example
Theorem
∀x ∈ R [x ≥ 0 ⇒ x2 ≥ 0]
Proof.
∀x ∈ R [x2 ≥ 0]
Topics
1 Basic TechniquesIntroductionDirect ProofProof by ContradictionEquivalence Proofs
2 InductionIntroductionStrong Induction
Direct Proof
direct proof
to prove P ⇒ Q, show that P ` Q
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)
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)
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)
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)
Indirect Proof
indirect proof
to prove P ⇒ Q, show that ¬Q ` ¬P
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
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
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
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])
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])
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])
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])
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])
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])
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])
Topics
1 Basic TechniquesIntroductionDirect ProofProof by ContradictionEquivalence Proofs
2 InductionIntroductionStrong Induction
Proof by Contradiction
proof by contradiction
to prove P, show that ¬P ` Q ∧ ¬Q
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 divisibleby a prime number in the range [2,S ]
1 either it is prime itself: ¬Q2 or it is divisible by a prime number greater than S : ¬Q
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 divisibleby a prime number in the range [2,S ]
1 either it is prime itself: ¬Q2 or it is divisible by a prime number greater than S : ¬Q
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 divisibleby a prime number in the range [2,S ]
1 either it is prime itself: ¬Q2 or it is divisible by a prime number greater than S : ¬Q
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 divisibleby a prime number in the range [2,S ]
1 either it is prime itself: ¬Q2 or it is divisible by a prime number greater than S : ¬Q
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 divisibleby a prime number in the range [2,S ]
1 either it is prime itself: ¬Q2 or it is divisible by a prime number greater than S : ¬Q
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 divisibleby a prime number in the range [2,S ]
1 either it is prime itself: ¬Q2 or it is divisible by a prime number greater than S : ¬Q
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 divisibleby a prime number in the range [2,S ]
1 either it is prime itself: ¬Q2 or it is divisible by a prime number greater than S : ¬Q
Proof by Contradiction Example
Theorem
¬∃a, b ∈ Z+ [√
2 = ab ]
Proof.
¬P: ∃a, b ∈ Z+ [√
2 = ab ]
Q: gcd(a, b) = 1
⇒ 2 =a2
b2
⇒ a2 = 2b2
⇒ ∃i ∈ Z+ [a2 = 2i ]
⇒ ∃j ∈ Z+ [a = 2j ]
⇒ 4j2 = 2b2
⇒ b2 = 2j2
⇒ ∃k ∈ Z+ [b2 = 2k]
⇒ ∃l ∈ Z+ [b = 2l ]
⇒ gcd(a, b) ≥ 2 : ¬Q
Proof by Contradiction Example
Theorem
¬∃a, b ∈ Z+ [√
2 = ab ]
Proof.
¬P: ∃a, b ∈ Z+ [√
2 = ab ]
Q: gcd(a, b) = 1
⇒ 2 =a2
b2
⇒ a2 = 2b2
⇒ ∃i ∈ Z+ [a2 = 2i ]
⇒ ∃j ∈ Z+ [a = 2j ]
⇒ 4j2 = 2b2
⇒ b2 = 2j2
⇒ ∃k ∈ Z+ [b2 = 2k]
⇒ ∃l ∈ Z+ [b = 2l ]
⇒ gcd(a, b) ≥ 2 : ¬Q
Proof by Contradiction Example
Theorem
¬∃a, b ∈ Z+ [√
2 = ab ]
Proof.
¬P: ∃a, b ∈ Z+ [√
2 = ab ]
Q: gcd(a, b) = 1
⇒ 2 =a2
b2
⇒ a2 = 2b2
⇒ ∃i ∈ Z+ [a2 = 2i ]
⇒ ∃j ∈ Z+ [a = 2j ]
⇒ 4j2 = 2b2
⇒ b2 = 2j2
⇒ ∃k ∈ Z+ [b2 = 2k]
⇒ ∃l ∈ Z+ [b = 2l ]
⇒ gcd(a, b) ≥ 2 : ¬Q
Proof by Contradiction Example
Theorem
¬∃a, b ∈ Z+ [√
2 = ab ]
Proof.
¬P: ∃a, b ∈ Z+ [√
2 = ab ]
Q: gcd(a, b) = 1
⇒ 2 =a2
b2
⇒ a2 = 2b2
⇒ ∃i ∈ Z+ [a2 = 2i ]
⇒ ∃j ∈ Z+ [a = 2j ]
⇒ 4j2 = 2b2
⇒ b2 = 2j2
⇒ ∃k ∈ Z+ [b2 = 2k]
⇒ ∃l ∈ Z+ [b = 2l ]
⇒ gcd(a, b) ≥ 2 : ¬Q
Proof by Contradiction Example
Theorem
¬∃a, b ∈ Z+ [√
2 = ab ]
Proof.
¬P: ∃a, b ∈ Z+ [√
2 = ab ]
Q: gcd(a, b) = 1
⇒ 2 =a2
b2
⇒ a2 = 2b2
⇒ ∃i ∈ Z+ [a2 = 2i ]
⇒ ∃j ∈ Z+ [a = 2j ]
⇒ 4j2 = 2b2
⇒ b2 = 2j2
⇒ ∃k ∈ Z+ [b2 = 2k]
⇒ ∃l ∈ Z+ [b = 2l ]
⇒ gcd(a, b) ≥ 2 : ¬Q
Proof by Contradiction Example
Theorem
¬∃a, b ∈ Z+ [√
2 = ab ]
Proof.
¬P: ∃a, b ∈ Z+ [√
2 = ab ]
Q: gcd(a, b) = 1
⇒ 2 =a2
b2
⇒ a2 = 2b2
⇒ ∃i ∈ Z+ [a2 = 2i ]
⇒ ∃j ∈ Z+ [a = 2j ]
⇒ 4j2 = 2b2
⇒ b2 = 2j2
⇒ ∃k ∈ Z+ [b2 = 2k]
⇒ ∃l ∈ Z+ [b = 2l ]
⇒ gcd(a, b) ≥ 2 : ¬Q
Proof by Contradiction Example
Theorem
¬∃a, b ∈ Z+ [√
2 = ab ]
Proof.
¬P: ∃a, b ∈ Z+ [√
2 = ab ]
Q: gcd(a, b) = 1
⇒ 2 =a2
b2
⇒ a2 = 2b2
⇒ ∃i ∈ Z+ [a2 = 2i ]
⇒ ∃j ∈ Z+ [a = 2j ]
⇒ 4j2 = 2b2
⇒ b2 = 2j2
⇒ ∃k ∈ Z+ [b2 = 2k]
⇒ ∃l ∈ Z+ [b = 2l ]
⇒ gcd(a, b) ≥ 2 : ¬Q
Proof by Contradiction Example
Theorem
¬∃a, b ∈ Z+ [√
2 = ab ]
Proof.
¬P: ∃a, b ∈ Z+ [√
2 = ab ]
Q: gcd(a, b) = 1
⇒ 2 =a2
b2
⇒ a2 = 2b2
⇒ ∃i ∈ Z+ [a2 = 2i ]
⇒ ∃j ∈ Z+ [a = 2j ]
⇒ 4j2 = 2b2
⇒ b2 = 2j2
⇒ ∃k ∈ Z+ [b2 = 2k]
⇒ ∃l ∈ Z+ [b = 2l ]
⇒ gcd(a, b) ≥ 2 : ¬Q
Proof by Contradiction Example
Theorem
¬∃a, b ∈ Z+ [√
2 = ab ]
Proof.
¬P: ∃a, b ∈ Z+ [√
2 = ab ]
Q: gcd(a, b) = 1
⇒ 2 =a2
b2
⇒ a2 = 2b2
⇒ ∃i ∈ Z+ [a2 = 2i ]
⇒ ∃j ∈ Z+ [a = 2j ]
⇒ 4j2 = 2b2
⇒ b2 = 2j2
⇒ ∃k ∈ Z+ [b2 = 2k]
⇒ ∃l ∈ Z+ [b = 2l ]
⇒ gcd(a, b) ≥ 2 : ¬Q
Proof by Contradiction Example
Theorem
¬∃a, b ∈ Z+ [√
2 = ab ]
Proof.
¬P: ∃a, b ∈ Z+ [√
2 = ab ]
Q: gcd(a, b) = 1
⇒ 2 =a2
b2
⇒ a2 = 2b2
⇒ ∃i ∈ Z+ [a2 = 2i ]
⇒ ∃j ∈ Z+ [a = 2j ]
⇒ 4j2 = 2b2
⇒ b2 = 2j2
⇒ ∃k ∈ Z+ [b2 = 2k]
⇒ ∃l ∈ Z+ [b = 2l ]
⇒ gcd(a, b) ≥ 2 : ¬Q
Proof by Contradiction Example
Theorem
¬∃a, b ∈ Z+ [√
2 = ab ]
Proof.
¬P: ∃a, b ∈ Z+ [√
2 = ab ]
Q: gcd(a, b) = 1
⇒ 2 =a2
b2
⇒ a2 = 2b2
⇒ ∃i ∈ Z+ [a2 = 2i ]
⇒ ∃j ∈ Z+ [a = 2j ]
⇒ 4j2 = 2b2
⇒ b2 = 2j2
⇒ ∃k ∈ Z+ [b2 = 2k]
⇒ ∃l ∈ Z+ [b = 2l ]
⇒ gcd(a, b) ≥ 2 : ¬Q
Topics
1 Basic TechniquesIntroductionDirect ProofProof by ContradictionEquivalence Proofs
2 InductionIntroductionStrong Induction
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
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
Equivalence Proof Example
Theorem
a, b, n, q1, r1, q2, r2 ∈ Z+
a = q1 · n + r1b = q2 · n + r2
r1 = r2 ⇔ n|(a− b)
Equivalence Proof Example
r1 = r2 ⇒ n|(a− b).
a− b = (q1 · n + r1)
−(q2 · n + r2)
= (q1 − q2) · n+(r1 − r2)
r1 = r2 ⇒ r1 − r2 = 0
⇒ a− b = (q1 − q2) · n
n|(a− b) ⇒ r1 = r2.
a− b = (q1 · n + r1)
−(q2 · n + r2)
= (q1 − q2) · n+(r1 − r2)
n|(a− b) ⇒ r1 − r2 = 0
⇒ r1 = r2
Equivalence Proof Example
r1 = r2 ⇒ n|(a− b).
a− b = (q1 · n + r1)
−(q2 · n + r2)
= (q1 − q2) · n+(r1 − r2)
r1 = r2 ⇒ r1 − r2 = 0
⇒ a− b = (q1 − q2) · n
n|(a− b) ⇒ r1 = r2.
a− b = (q1 · n + r1)
−(q2 · n + r2)
= (q1 − q2) · n+(r1 − r2)
n|(a− b) ⇒ r1 − r2 = 0
⇒ r1 = r2
Equivalence Proof Example
r1 = r2 ⇒ n|(a− b).
a− b = (q1 · n + r1)
−(q2 · n + r2)
= (q1 − q2) · n+(r1 − r2)
r1 = r2 ⇒ r1 − r2 = 0
⇒ a− b = (q1 − q2) · n
n|(a− b) ⇒ r1 = r2.
a− b = (q1 · n + r1)
−(q2 · n + r2)
= (q1 − q2) · n+(r1 − r2)
n|(a− b) ⇒ r1 − r2 = 0
⇒ r1 = r2
Equivalence Proof Example
r1 = r2 ⇒ n|(a− b).
a− b = (q1 · n + r1)
−(q2 · n + r2)
= (q1 − q2) · n+(r1 − r2)
r1 = r2 ⇒ r1 − r2 = 0
⇒ a− b = (q1 − q2) · n
n|(a− b) ⇒ r1 = r2.
a− b = (q1 · n + r1)
−(q2 · n + r2)
= (q1 − q2) · n+(r1 − r2)
n|(a− b) ⇒ r1 − r2 = 0
⇒ r1 = r2
Equivalence Proof Example
r1 = r2 ⇒ n|(a− b).
a− b = (q1 · n + r1)
−(q2 · n + r2)
= (q1 − q2) · n+(r1 − r2)
r1 = r2 ⇒ r1 − r2 = 0
⇒ a− b = (q1 − q2) · n
n|(a− b) ⇒ r1 = r2.
a− b = (q1 · n + r1)
−(q2 · n + r2)
= (q1 − q2) · n+(r1 − r2)
n|(a− b) ⇒ r1 − r2 = 0
⇒ r1 = r2
Equivalence Proof Example
r1 = r2 ⇒ n|(a− b).
a− b = (q1 · n + r1)
−(q2 · n + r2)
= (q1 − q2) · n+(r1 − r2)
r1 = r2 ⇒ r1 − r2 = 0
⇒ a− b = (q1 − q2) · n
n|(a− b) ⇒ r1 = r2.
a− b = (q1 · n + r1)
−(q2 · n + r2)
= (q1 − q2) · n+(r1 − r2)
n|(a− b) ⇒ r1 − r2 = 0
⇒ r1 = r2
Equivalence Proof Example
r1 = r2 ⇒ n|(a− b).
a− b = (q1 · n + r1)
−(q2 · n + r2)
= (q1 − q2) · n+(r1 − r2)
r1 = r2 ⇒ r1 − r2 = 0
⇒ a− b = (q1 − q2) · n
n|(a− b) ⇒ r1 = r2.
a− b = (q1 · n + r1)
−(q2 · n + r2)
= (q1 − q2) · n+(r1 − r2)
n|(a− b) ⇒ r1 − r2 = 0
⇒ r1 = r2
Equivalence Proof Example
r1 = r2 ⇒ n|(a− b).
a− b = (q1 · n + r1)
−(q2 · n + r2)
= (q1 − q2) · n+(r1 − r2)
r1 = r2 ⇒ r1 − r2 = 0
⇒ a− b = (q1 − q2) · n
n|(a− b) ⇒ r1 = r2.
a− b = (q1 · n + r1)
−(q2 · n + r2)
= (q1 − q2) · n+(r1 − r2)
n|(a− b) ⇒ r1 − r2 = 0
⇒ r1 = r2
Equivalence Proof Example
Theorem
A ⊆ B
⇔ A ∪ B = B
⇔ A ∩ B = A
⇔ B ⊆ A
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
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
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
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
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
Equivalence Proof Example
A ∪ B = B ⇒ A ∩ B = A.
A ∩ B = A ⇔ A ∩ B ⊆ A ∧ A ⊆ A ∩ B
A ∩ B ⊆ Ay ∈ A ⇒ y ∈ A ∪ B
A ∪ B = B ⇒ y ∈ B
⇒ y ∈ A ∩ B
⇒ A ⊆ A ∩ B
Equivalence Proof Example
A ∪ B = B ⇒ A ∩ B = A.
A ∩ B = A ⇔ A ∩ B ⊆ A ∧ A ⊆ A ∩ B
A ∩ B ⊆ Ay ∈ A ⇒ y ∈ A ∪ B
A ∪ B = B ⇒ y ∈ B
⇒ y ∈ A ∩ B
⇒ A ⊆ A ∩ B
Equivalence Proof Example
A ∪ B = B ⇒ A ∩ B = A.
A ∩ B = A ⇔ A ∩ B ⊆ A ∧ A ⊆ A ∩ B
A ∩ B ⊆ Ay ∈ A ⇒ y ∈ A ∪ B
A ∪ B = B ⇒ y ∈ B
⇒ y ∈ A ∩ B
⇒ A ⊆ A ∩ B
Equivalence Proof Example
A ∪ B = B ⇒ A ∩ B = A.
A ∩ B = A ⇔ A ∩ B ⊆ A ∧ A ⊆ A ∩ B
A ∩ B ⊆ Ay ∈ A ⇒ y ∈ A ∪ B
A ∪ B = B ⇒ y ∈ B
⇒ y ∈ A ∩ B
⇒ A ⊆ A ∩ B
Equivalence Proof Example
A ∪ B = B ⇒ A ∩ B = A.
A ∩ B = A ⇔ A ∩ B ⊆ A ∧ A ⊆ A ∩ B
A ∩ B ⊆ Ay ∈ A ⇒ y ∈ A ∪ B
A ∪ B = B ⇒ y ∈ B
⇒ y ∈ A ∩ B
⇒ A ⊆ A ∩ B
Equivalence Proof Example
A ∪ B = B ⇒ A ∩ B = A.
A ∩ B = A ⇔ A ∩ B ⊆ A ∧ A ⊆ A ∩ B
A ∩ B ⊆ Ay ∈ A ⇒ y ∈ A ∪ B
A ∪ B = B ⇒ y ∈ B
⇒ y ∈ A ∩ B
⇒ A ⊆ A ∩ B
Equivalence Proof Example
A ∩ B = A ⇒ B ⊆ A.
z ∈ B ⇒ z /∈ B
⇒ z /∈ A ∩ B
A ∩ B = A ⇒ z /∈ A
⇒ z ∈ A
⇒ B ⊆ A
Equivalence Proof Example
A ∩ B = A ⇒ B ⊆ A.
z ∈ B ⇒ z /∈ B
⇒ z /∈ A ∩ B
A ∩ B = A ⇒ z /∈ A
⇒ z ∈ A
⇒ B ⊆ A
Equivalence Proof Example
A ∩ B = A ⇒ B ⊆ A.
z ∈ B ⇒ z /∈ B
⇒ z /∈ A ∩ B
A ∩ B = A ⇒ z /∈ A
⇒ z ∈ A
⇒ B ⊆ A
Equivalence Proof Example
A ∩ B = A ⇒ B ⊆ A.
z ∈ B ⇒ z /∈ B
⇒ z /∈ A ∩ B
A ∩ B = A ⇒ z /∈ A
⇒ z ∈ A
⇒ B ⊆ A
Equivalence Proof Example
A ∩ B = A ⇒ B ⊆ A.
z ∈ B ⇒ z /∈ B
⇒ z /∈ A ∩ B
A ∩ B = A ⇒ z /∈ A
⇒ z ∈ A
⇒ B ⊆ A
Equivalence Proof Example
B ⊆ A ⇒ A ⊆ B.
¬(A ⊆ B) ⇒ ∃w [w ∈ A ∧ w /∈ B]
⇒ ∃w [w /∈ A ∧ w ∈ B]
⇒ ¬(B ⊆ A)
Equivalence Proof Example
B ⊆ A ⇒ A ⊆ B.
¬(A ⊆ B) ⇒ ∃w [w ∈ A ∧ w /∈ B]
⇒ ∃w [w /∈ A ∧ w ∈ B]
⇒ ¬(B ⊆ A)
Equivalence Proof Example
B ⊆ A ⇒ A ⊆ B.
¬(A ⊆ B) ⇒ ∃w [w ∈ A ∧ w /∈ B]
⇒ ∃w [w /∈ A ∧ w ∈ B]
⇒ ¬(B ⊆ A)
Topics
1 Basic TechniquesIntroductionDirect ProofProof by ContradictionEquivalence Proofs
2 InductionIntroductionStrong Induction
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
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
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
Induction
Induction Example
Theorem
∀n ∈ Z+ [1 + 3 + 5 + · · ·+ (2n − 1) = n2]
Proof.
n = 1: 1 = 12
n = k: assume 1 + 3 + 5 + · · ·+ (2k − 1) = k2
n = k + 1:
1 + 3 + 5 + · · ·+ (2k − 1) + (2k + 1)
= k2 + 2k + 1
= (k + 1)2
Induction Example
Theorem
∀n ∈ Z+ [1 + 3 + 5 + · · ·+ (2n − 1) = n2]
Proof.
n = 1: 1 = 12
n = k: assume 1 + 3 + 5 + · · ·+ (2k − 1) = k2
n = k + 1:
1 + 3 + 5 + · · ·+ (2k − 1) + (2k + 1)
= k2 + 2k + 1
= (k + 1)2
Induction Example
Theorem
∀n ∈ Z+ [1 + 3 + 5 + · · ·+ (2n − 1) = n2]
Proof.
n = 1: 1 = 12
n = k: assume 1 + 3 + 5 + · · ·+ (2k − 1) = k2
n = k + 1:
1 + 3 + 5 + · · ·+ (2k − 1) + (2k + 1)
= k2 + 2k + 1
= (k + 1)2
Induction Example
Theorem
∀n ∈ Z+ [1 + 3 + 5 + · · ·+ (2n − 1) = n2]
Proof.
n = 1: 1 = 12
n = k: assume 1 + 3 + 5 + · · ·+ (2k − 1) = k2
n = k + 1:
1 + 3 + 5 + · · ·+ (2k − 1) + (2k + 1)
= k2 + 2k + 1
= (k + 1)2
Induction Example
Theorem
∀n ∈ Z+ [1 + 3 + 5 + · · ·+ (2n − 1) = n2]
Proof.
n = 1: 1 = 12
n = k: assume 1 + 3 + 5 + · · ·+ (2k − 1) = k2
n = k + 1:
1 + 3 + 5 + · · ·+ (2k − 1) + (2k + 1)
= k2 + 2k + 1
= (k + 1)2
Induction Example
Theorem
∀n ∈ Z+ [1 + 3 + 5 + · · ·+ (2n − 1) = n2]
Proof.
n = 1: 1 = 12
n = k: assume 1 + 3 + 5 + · · ·+ (2k − 1) = k2
n = k + 1:
1 + 3 + 5 + · · ·+ (2k − 1) + (2k + 1)
= k2 + 2k + 1
= (k + 1)2
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)!
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)!
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)!
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)!
Induction Example
Theorem
∀n ∈ Z+, n ≥ 14 ∃i , j ∈ N [n = 3i + 8j ]
Proof.
n = 14: 14 = 3 · 2 + 8 · 1n = 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)
Induction Example
Theorem
∀n ∈ Z+, n ≥ 14 ∃i , j ∈ N [n = 3i + 8j ]
Proof.
n = 14: 14 = 3 · 2 + 8 · 1n = 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)
Induction Example
Theorem
∀n ∈ Z+, n ≥ 14 ∃i , j ∈ N [n = 3i + 8j ]
Proof.
n = 14: 14 = 3 · 2 + 8 · 1n = 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)
Induction Example
Theorem
∀n ∈ Z+, n ≥ 14 ∃i , j ∈ N [n = 3i + 8j ]
Proof.
n = 14: 14 = 3 · 2 + 8 · 1n = 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)
Topics
1 Basic TechniquesIntroductionDirect ProofProof by ContradictionEquivalence Proofs
2 InductionIntroductionStrong Induction
Strong Induction
Definition
S(n0) ∧ (∀k ≥ n0 [(∀i ≤ k S(i)) ⇒ S(k + 1)]) ⇒ ∀n ≥ n0 S(n)
Strong Induction Example
Theorem
∀n ∈ Z+, n ≥ 2n 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 = n2 if not prime: n = u · v
u < k ∧ v < k ⇒ both u and v can be writtenas the product of prime numbers
Strong Induction Example
Theorem
∀n ∈ Z+, n ≥ 2n 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 = n2 if not prime: n = u · v
u < k ∧ v < k ⇒ both u and v can be writtenas the product of prime numbers
Strong Induction Example
Theorem
∀n ∈ Z+, n ≥ 2n 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 = n2 if not prime: n = u · v
u < k ∧ v < k ⇒ both u and v can be writtenas the product of prime numbers
Strong Induction Example
Theorem
∀n ∈ Z+, n ≥ 2n 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 = n2 if not prime: n = u · v
u < k ∧ v < k ⇒ both u and v can be writtenas the product of prime numbers
Strong Induction Example
Theorem
∀n ∈ Z+, n ≥ 2n 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 = n2 if not prime: n = u · v
u < k ∧ v < k ⇒ both u and v can be writtenas the product of prime numbers
Strong Induction Example
Theorem
∀n ∈ Z+, n ≥ 14 ∃i , j ∈ N [n = 3i + 8j ]
Proof.
n = 14: 14 = 3 · 2 + 8 · 1n = 15: 15 = 3 · 5 + 8 · 0n = 16: 16 = 3 · 0 + 8 · 2n ≤ k: assume k = 3i + 8j
n = k + 1: k + 1 = (k − 2) + 3
Strong Induction Example
Theorem
∀n ∈ Z+, n ≥ 14 ∃i , j ∈ N [n = 3i + 8j ]
Proof.
n = 14: 14 = 3 · 2 + 8 · 1n = 15: 15 = 3 · 5 + 8 · 0n = 16: 16 = 3 · 0 + 8 · 2n ≤ k: assume k = 3i + 8j
n = k + 1: k + 1 = (k − 2) + 3
Strong Induction Example
Theorem
∀n ∈ Z+, n ≥ 14 ∃i , j ∈ N [n = 3i + 8j ]
Proof.
n = 14: 14 = 3 · 2 + 8 · 1n = 15: 15 = 3 · 5 + 8 · 0n = 16: 16 = 3 · 0 + 8 · 2n ≤ k: assume k = 3i + 8j
n = k + 1: k + 1 = (k − 2) + 3
Strong Induction Example
Theorem
∀n ∈ Z+, n ≥ 14 ∃i , j ∈ N [n = 3i + 8j ]
Proof.
n = 14: 14 = 3 · 2 + 8 · 1n = 15: 15 = 3 · 5 + 8 · 0n = 16: 16 = 3 · 0 + 8 · 2n ≤ k: assume k = 3i + 8j
n = k + 1: k + 1 = (k − 2) + 3
Flawed Induction Example
Theorem
∀n ∈ Z+ [1 + 2 + 3 + · · ·+ n = n2+n+22 ]
invalid base step
n = k: assume 1 + 2 + 3 + · · ·+ k = k2+k+22
n = k + 1:
1 + 2 + 3 + · · ·+ k + (k + 1)
=k2 + k + 2
2+ k + 1 =
k2 + k + 2
2+
2k + 2
2
=k2 + 3k + 4
2=
(k + 1)2 + (k + 1) + 2
2
n = 1: 1 6= 12+1+22 = 2
Flawed Induction Example
Theorem
∀n ∈ Z+ [1 + 2 + 3 + · · ·+ n = n2+n+22 ]
invalid base step
n = k: assume 1 + 2 + 3 + · · ·+ k = k2+k+22
n = k + 1:
1 + 2 + 3 + · · ·+ k + (k + 1)
=k2 + k + 2
2+ k + 1 =
k2 + k + 2
2+
2k + 2
2
=k2 + 3k + 4
2=
(k + 1)2 + (k + 1) + 2
2
n = 1: 1 6= 12+1+22 = 2
Flawed Induction Example
Theorem
∀n ∈ Z+ [1 + 2 + 3 + · · ·+ n = n2+n+22 ]
invalid base step
n = k: assume 1 + 2 + 3 + · · ·+ k = k2+k+22
n = k + 1:
1 + 2 + 3 + · · ·+ k + (k + 1)
=k2 + k + 2
2+ k + 1 =
k2 + k + 2
2+
2k + 2
2
=k2 + 3k + 4
2=
(k + 1)2 + (k + 1) + 2
2
n = 1: 1 6= 12+1+22 = 2
Flawed Induction Example
Theorem
∀n ∈ Z+ [1 + 2 + 3 + · · ·+ n = n2+n+22 ]
invalid base step
n = k: assume 1 + 2 + 3 + · · ·+ k = k2+k+22
n = k + 1:
1 + 2 + 3 + · · ·+ k + (k + 1)
=k2 + k + 2
2+ k + 1 =
k2 + k + 2
2+
2k + 2
2
=k2 + 3k + 4
2=
(k + 1)2 + (k + 1) + 2
2
n = 1: 1 6= 12+1+22 = 2
Flawed Induction Example
Theorem
∀n ∈ Z+ [1 + 2 + 3 + · · ·+ n = n2+n+22 ]
invalid base step
n = k: assume 1 + 2 + 3 + · · ·+ k = k2+k+22
n = k + 1:
1 + 2 + 3 + · · ·+ k + (k + 1)
=k2 + k + 2
2+ k + 1 =
k2 + k + 2
2+
2k + 2
2
=k2 + 3k + 4
2=
(k + 1)2 + (k + 1) + 2
2
n = 1: 1 6= 12+1+22 = 2
Flawed Induction Example
Theorem
∀n ∈ Z+ [1 + 2 + 3 + · · ·+ n = n2+n+22 ]
invalid base step
n = k: assume 1 + 2 + 3 + · · ·+ k = k2+k+22
n = k + 1:
1 + 2 + 3 + · · ·+ k + (k + 1)
=k2 + k + 2
2+ k + 1 =
k2 + k + 2
2+
2k + 2
2
=k2 + 3k + 4
2=
(k + 1)2 + (k + 1) + 2
2
n = 1: 1 6= 12+1+22 = 2
Flawed Induction Example
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)
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)
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 trueAll 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).
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 trueAll 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).
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 trueAll 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).
Flawed Induction Examples
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