Discrete Mathematics - Proofs

Discrete MathematicsProofs

H. Turgut Uyar Aysegul Gencata Yayımlı Emre Harmancı

2001-2013

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Topics

1 Basic TechniquesIntroductionDirect ProofProof by ContradictionEquivalence Proofs

2 InductionIntroductionStrong Induction

Topics

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

Theorem

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

Theorem

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.

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.

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

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

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

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

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

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

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

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

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

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

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

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

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 /∈ ∅]

Theorem

∀S [∅ ⊆ S ]

Proof.

∅ ⊆ S ⇔ ∀x [x ∈ ∅ → x ∈ S ]∀x [x /∈ ∅]

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

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)

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)

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)

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

Theorem

∀x , y ∈ N [x · y > 25 ⇒ (x > 5) ∨ (y > 5)]

Proof.

¬Q ⇔ (0 ≤ x ≤ 5) ∧ (0 ≤ y ≤ 5)

x · y ≤ 5 · 5 = 25

Theorem

∀x , y ∈ N [x · y > 25 ⇒ (x > 5) ∨ (y > 5)]

Proof.

¬Q ⇔ (0 ≤ x ≤ 5) ∧ (0 ≤ y ≤ 5)

x · y ≤ 5 · 5 = 25

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])

Theorem

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])

Theorem

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])

Theorem

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])

Theorem

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])

Theorem

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])

Theorem

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

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

Theorem

Proof.

prime numbers: 2, 3, 5, 7, 11, . . . ,S

Theorem

Proof.

prime numbers: 2, 3, 5, 7, 11, . . . ,S

Theorem

Proof.

prime numbers: 2, 3, 5, 7, 11, . . . ,S

Theorem

Proof.

prime numbers: 2, 3, 5, 7, 11, . . . ,S

Theorem

Proof.

prime numbers: 2, 3, 5, 7, 11, . . . ,S

Theorem

Proof.

prime numbers: 2, 3, 5, 7, 11, . . . ,S

Theorem

¬∃a, b ∈ Z+ [√

2 = ab ]

Proof.

¬P: ∃a, b ∈ Z+ [√

2 = ab ]

Q: gcd(a, b) = 1

⇒ 2 =a2

⇒ 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

Theorem

¬∃a, b ∈ Z+ [√

2 = ab ]

Proof.

¬P: ∃a, b ∈ Z+ [√

2 = ab ]

Q: gcd(a, b) = 1

⇒ 2 =a2

⇒ 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

Theorem

¬∃a, b ∈ Z+ [√

2 = ab ]

Proof.

¬P: ∃a, b ∈ Z+ [√

2 = ab ]

Q: gcd(a, b) = 1

⇒ 2 =a2

⇒ 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

Theorem

¬∃a, b ∈ Z+ [√

2 = ab ]

Proof.

¬P: ∃a, b ∈ Z+ [√

2 = ab ]

Q: gcd(a, b) = 1

⇒ 2 =a2

⇒ 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

Theorem

¬∃a, b ∈ Z+ [√

2 = ab ]

Proof.

¬P: ∃a, b ∈ Z+ [√

2 = ab ]

Q: gcd(a, b) = 1

⇒ 2 =a2

⇒ 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

Theorem

¬∃a, b ∈ Z+ [√

2 = ab ]

Proof.

¬P: ∃a, b ∈ Z+ [√

2 = ab ]

Q: gcd(a, b) = 1

⇒ 2 =a2

⇒ 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

Theorem

¬∃a, b ∈ Z+ [√

2 = ab ]

Proof.

¬P: ∃a, b ∈ Z+ [√

2 = ab ]

Q: gcd(a, b) = 1

⇒ 2 =a2

⇒ 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

Theorem

¬∃a, b ∈ Z+ [√

2 = ab ]

Proof.

¬P: ∃a, b ∈ Z+ [√

2 = ab ]

Q: gcd(a, b) = 1

⇒ 2 =a2

⇒ 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

Theorem

¬∃a, b ∈ Z+ [√

2 = ab ]

Proof.

¬P: ∃a, b ∈ Z+ [√

2 = ab ]

Q: gcd(a, b) = 1

⇒ 2 =a2

⇒ 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

Theorem

¬∃a, b ∈ Z+ [√

2 = ab ]

Proof.

¬P: ∃a, b ∈ Z+ [√

2 = ab ]

Q: gcd(a, b) = 1

⇒ 2 =a2

⇒ 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

Theorem

¬∃a, b ∈ Z+ [√

2 = ab ]

Proof.

¬P: ∃a, b ∈ Z+ [√

2 = ab ]

Q: gcd(a, b) = 1

⇒ 2 =a2

⇒ 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

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)

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

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

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

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

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

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

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

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

Theorem

A ⊆ B

⇔ A ∪ B = B

⇔ A ∩ B = A

⇔ B ⊆ A

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

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

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

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

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

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

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

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

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

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

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

A ∩ B = A ⇒ B ⊆ A.

z ∈ B ⇒ z /∈ B

⇒ z /∈ A ∩ B

A ∩ B = A ⇒ z /∈ A

⇒ z ∈ A

⇒ B ⊆ A

A ∩ B = A ⇒ B ⊆ A.

z ∈ B ⇒ z /∈ B

⇒ z /∈ A ∩ B

A ∩ B = A ⇒ z /∈ A

⇒ z ∈ A

⇒ B ⊆ A

A ∩ B = A ⇒ B ⊆ A.

z ∈ B ⇒ z /∈ B

⇒ z /∈ A ∩ B

A ∩ B = A ⇒ z /∈ A

⇒ z ∈ A

⇒ B ⊆ A

A ∩ B = A ⇒ B ⊆ A.

z ∈ B ⇒ z /∈ B

⇒ z /∈ A ∩ B

A ∩ B = A ⇒ z /∈ A

⇒ z ∈ A

⇒ B ⊆ A

A ∩ B = A ⇒ B ⊆ A.

z ∈ B ⇒ z /∈ B

⇒ z /∈ A ∩ B

A ∩ B = A ⇒ z /∈ A

⇒ z ∈ A

⇒ B ⊆ A

B ⊆ A ⇒ A ⊆ B.

¬(A ⊆ B) ⇒ ∃w [w ∈ A ∧ w /∈ B]

⇒ ∃w [w /∈ A ∧ w ∈ B]

⇒ ¬(B ⊆ A)

B ⊆ A ⇒ A ⊆ B.

¬(A ⊆ B) ⇒ ∃w [w ∈ A ∧ w /∈ B]

⇒ ∃w [w /∈ A ∧ w ∈ B]

⇒ ¬(B ⊆ A)

B ⊆ A ⇒ A ⊆ B.

¬(A ⊆ B) ⇒ ∃w [w ∈ A ∧ w /∈ B]

⇒ ∃w [w /∈ A ∧ w ∈ B]

⇒ ¬(B ⊆ A)

Topics

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(n0) ∧ (∀k ≥ n0 [S(k) ⇒ S(k + 1)]) ⇒ ∀n ≥ n0 S(n)

S(n0): base step

Induction

Definition

S(n0) ∧ (∀k ≥ n0 [S(k) ⇒ S(k + 1)]) ⇒ ∀n ≥ n0 S(n)

S(n0): base 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 + 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 + 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 + 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:

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:

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:

Topics

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

Theorem

Proof.

n = 2: 2 = 2

n = k + 1:

Theorem

Proof.

n = 2: 2 = 2

n = k + 1:

Theorem

Proof.

n = 2: 2 = 2

n = k + 1:

Theorem

Proof.

n = 2: 2 = 2

n = k + 1:

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

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

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

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

2k + 2

=k2 + 3k + 4

(k + 1)2 + (k + 1) + 2

n = 1: 1 6= 12+1+22 = 2

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

2k + 2

=k2 + 3k + 4

(k + 1)2 + (k + 1) + 2

n = 1: 1 6= 12+1+22 = 2

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

2k + 2

=k2 + 3k + 4

(k + 1)2 + (k + 1) + 2

n = 1: 1 6= 12+1+22 = 2

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

2k + 2

=k2 + 3k + 4

(k + 1)2 + (k + 1) + 2

n = 1: 1 6= 12+1+22 = 2

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

2k + 2

=k2 + 3k + 4

(k + 1)2 + (k + 1) + 2

n = 1: 1 6= 12+1+22 = 2

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

2k + 2

=k2 + 3k + 4

(k + 1)2 + (k + 1) + 2

n = 1: 1 6= 12+1+22 = 2

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)

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)

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

Discrete Mathematics - Proofs

Education

Transcript of Discrete Mathematics - Proofs