Chapter-4: More on Direct Proof and Proof by Contrapositive

MATH 401: LOGIC & PROOF TECHNIQUES

Copyright © Nahid Sultana 2014-2015.

Dr. Nahid Sultana

Email: [email protected]

Chapter-4: More on Direct Proof and Proof by Contrapositive

10/10/2014

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Topics 2

Proofs Involving Divisibility of Integers

Proofs Involving Congruence of Integers

Proofs Involving Real Numbers

Proofs Involving sets

Fundamental Properties of Set Operations

Proofs Involving Cartesian Products of Sets 10/10/2014 Copyright © Nahid Sultana 2014-2015.

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Definition: For integers a and b with a≠0, we say that a divides b or b is divisible by a if there is an integer c such

that b=ac, and is written as a|b. If a does not divide b, then we write a b. Example: 3|6 and -4 |28 ? Yes because 6 = 3 . 2 and 28 = (-4) . (-7).

Proofs Involving Divisibility of Integers

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10/10/2014 Copyright © Nahid Sultana 2014-2015.


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Result: Let x∈ℤ. If 2|(x2-1) , then 4|(x2-1) . Proof: (Direct Proof) Assume that 2|(x2-1). Then x2-1 = 2a for some a ∈ ℤ. Thus x2 = 2a+1, i.e x2 is odd. But we have the following theorem:

“Let x∈ℤ. Then x2 is odd iff x is odd. ” Using this theorem, x is odd too. Hence x= 2b+1, for some b ∈ ℤ . Then, x2-1 = (2b+1)2-1 = 4 b2+4b+1-1 = 4 b2+4b = 4c, where c = b2+b ∈ ℤ. Hence 4|(x2-1).



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Result: Let x∈ ℤ. If 3 xy , then 3 x and 3 y. Proof: (Proof by Contrapositive) Assume that 3|x or 3|y. WLOG assume that 3|x, then x = 3a for some a ∈ ℤ. Then xy = (3a)y = 3(ay)= 3b , where b = ay ∈ ℤ. Hence 3|xy.

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Result: Let x,y∈ ℤ. If 3 (x2-1) , then 3 |x .

Proof: (Proof by Contrapositive) Assume that 3 x . Then x= 3a+1 or x=3a+2 for some a ∈ ℤ. Therefore we need to consider two cases: Case1: when x= 3a+1 for some a ∈ ℤ. Then (x2-1) = 9a2+6a+1-1=3(3a2+2a) =3b, where b= 3a2+2a ∈ ℤ. Case2: when x= 3a+2 for some a ∈ ℤ. Then (x2-1) = 9a2+12a+4-1=3(3a2+4a+1) =3c, where c= 3a2+4a+1 ∈ ℤ. Hence 3| (x2-1).

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Result: Let x,y∈ ℤ. Prove that if 3|2a , then 3|a .

Proof: (Proof by Contrapositive) Assume that 3 a . Then a= 3x+1 or a=3x+2 for some x ∈ ℤ. Therefore we need to consider twp consider. Case1: when a= 3x+1 for some x ∈ ℤ. Then 2a = 6x+2 = 3(2x)+2 = 3y+2, where y=2x ∈ ℤ. Therefore 3 a. Case2: when a= 3x+2 for some x ∈ ℤ. Then 2a = 6x+4 = 3(2x+1)+1 = 3z+1, where z =2x+1 ∈ ℤ. where c = 3a2+4a+1 ∈ ℤ. Therefore 3 a.

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Proofs Involving Congruence of Integers

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Definition: For integers a, b, and n≥2, we say that a is congruent to b modulo n, written a ≡ b (mod n), if n | (a-b).

For example: 15≡7 (mod 4) since 4 | (15-7), but 14 4 (mod 6) since 6 (14-4).

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Result: Let a, b , k, and n be integers, where n≥2. If a ≡ b (mod n), then ka ≡ kb (mod n). Proof: (Direct Proof) Assume that a ≡ b(mod n). Then a-b =nx for some x ∈ ℤ. Now, ka-kb=k(a-b)=k(nx)=n(kx)=nl, where l=kx ∈ ℤ. Therefore, ka ≡ kb (mod n).


Proofs Involving Congruence of Integers (Cont…)

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Result: Let a, b, c, d, n ∈ ℤ, where n ≥2. If a ≡ b (mod n) and c ≡ d (mod n), the a+c ≡ b+d (mod n). Proof: Assume that a ≡ b (mod n) and c ≡ d (mod n), i.e. a-b = nx and c-d = ny for some x,y ∈ ℤ. Adding these two equations, a-b+c-d = nx+ny=n(x+y) ⇒ (a+c) –(b+d)= nz, where z=x+y ∈ ℤ. Therefore, the a+c ≡ b+d (mod n).


Proofs Involving Congruence of Integers (Cont…)

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Result: Let n ∈ ℤ. If n2 n (mod 3), then n 0(mod 3) and n 1(mod 3). Proof. Assume that n ≡ 0 (mod 3) or n ≡ 1(mod 3). We consider these two cases. Case 1. n ≡ 0(mod 3). Then n=3k for some k ∈ ℤ. Hence, n2- n=(3k)2- (3k)=3(3k2- k)=3l, where l = 3k2- k ∈ ℤ. Thus n2 ≡ n (mod 3). Case 2. n ≡ 1(mod 3). Then n -1=3m for some m ∈ ℤ. Hence, n2- n=(3m+1)2- (3m+1)=9m2+3m = 3(3m2+m) = 3p, where p = 3m2+m is an integer. Hence 3| (n2-n) and so n2 ≡ n (mod 3).

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Proofs Involving Real Numbers 14

Some facts about real numbers that can be used without justification. a2≥0 for every real number a. an≥0 for every real number a if n is a positive even integer. If a<0 and n is a positive odd integer, then an<0.

The product of two real numbers is positive iff both numbers are positive or both are negative.

If the product of two real numbers is 0, then at least one of these numbers is 0.

Let a, b, c ∈R. If a ≥b and c ≥0, then ac ≥ bc; and if c>0, then a/c ≥b/c.

If a>b and c>0, then ac>bc and a/c>b/c. If a>b and c<0, then ac<bc and a/c<b/c.


Proofs Involving Real Numbers (Cont…) 15

Result: Let x ∈ ℝ. if x3-5x2+3x=15, then x =5. Proof: (Direct Proof) Assume that x3-5x2+3x=15 ⇒ x3-5x2+3x-15 = 0 ⇒ x2(x-5) +3(x-5)=0 ⇒ (x2+3)(x-5)=0 But we have the following theorem: “If x and y are real numbers such that xy=0, then x=0 or y=0.” Therefore, (x2+3)= 0 or (x-5)=0. But x2+3 >0, so (x-5)=0, i.e. x=5. Hence x=5.


Proofs Involving Real Numbers (Cont…) 16

Result: Let x ∈R. if x5-3x4+2x3-x2+4x-1 ≥0, then x ≥0.

Proof: (Proof by Contrapositive)

Assume that x <0. Then x5<0, x4 >0 ⇒ -3x4 <0 x3<0, x2>0 ⇒ -x2<0 and 4x<0.

Thus x5-3x4+2x3-x2+4x-1<0-1<0, as desired.


Result: If x,y ∈ℝ, then x2/3+(3y2)/4 ≥ xy. Proof: (Direct Proof) Assume that x,y ∈ℝ. Now x2/3+(3y2)/4 ≥ xy ⇒ (4x2+ 9y2) ≥ 12xy ⇒ (4x2+ 9y2) - 12xy ≥ 0 ⇒(2x-3y)2 ≥ 0, this is true for

any x,y ∈ℝ. Therefore x2/3+(3y2)/4 ≥ xy.

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Proofs Involving Real Numbers (Cont…)


Proofs Involving sets 18

Recall, for set A and B contained in some universal set U,

A ∪ B = {x| x ∈ A or x ∈ B}.

A ∩ B = {x| x ∈ A and x ∈ B}.

A – B = {x| x ∈ A and x ∉ B}.

Ac = {x| x ∉ A and x ∈ U} = U-A.

To show that C ⊆ D, we need to show that every element of C is also an element of D; that is, if x ∈ C then x ∈ D.

To show the equality of two sets C and D, we need to show that C ⊆ D and D ⊆ C.


Proofs Involving sets (Cont…) 19

Result. For every two sets A and B, A-B = A ∩ Bc .

Proof. First we show that A-B ⊆ A∩Bc .

Let x ∈ A-B ⇒ x ∈A and x∉B

⇒ x ∈A and x ∈ Bc , since x∉ B

Therefore, A-B ⊆A∩ Bc .

Next we show that A∩Bc ⊆ A- B.

Let y ∈ A∩Bc ⇒ y∈A and y∈Bc

⇒ y∈A and y∉B , since y ∈Bc

Thus, A∩ Bc ⊆ A-B.


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Result. Let A and B be sets. Then A∪B= A if and only if B ⊆ A. Proof. First we prove that if A∪B= A, then B ⊆ A.

(proof by contrapositive) Assume that B ⊈ A ⇒ x ∈ B and x ∉ A ⇒ x∈ A∪B, Since x ∈ B and A∪B ={x| x∈A or x∈B}. But x ∉ A i.e. A ∪ B ≠ A. Next we prove the converse, i.e. if B ⊆ A, then A∪B=A.

(direct proof) Assume that B ⊆ A. To show A∪B= A, we need to show that A ⊆ A∪B and A ∪ B ⊆ A . But by the definition of union A ⊆ A∪B. To prove A ∪ B ⊆A, assume that y∈ A ∪ B ⇒ y∈A or y∈B. Case 1: If y∈A, then A ∪ B ⊆ A as also y ∈ A ∪ B by assumption. Case2: If y∈ B then A ∪ B ⊆ B ⇒ A ∪ B ⊆ A as B⊆ A. Thus A∪B= A.

Proofs Involving sets (Cont…)


Fundamental Properties of Set Operations

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Commutative laws Associative laws

Distributive laws


Fundamental Properties of Set Operations (Cont…)

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De Morgan’s laws

Absorption laws

Complement laws


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DeMorgan’s Law: For any sets A and B, (A ∪ B)c = Ac ∩ B c

Proof. Assume that x∈(A ∪ B)c ⇒ x ∉ A ∪ B ⇒ x∉ A and x∉ B ⇒x∈Ac and x∈Bc ⇒ x∈ Ac ∩ Bc .

Therefore, (A ∪ B)c ⊆ Ac ∩ Bc Next assume that x∈ Ac ∩ Bc ⇒ x∈Ac and x∈Bc

⇒x∉ A and x∉B ⇒ x ∉ A ∪ B ⇒x∈(A ∪ B)c

Therefore, Ac ∩ Bc ⊆(A ∪ B)c. Hence (A ∪B)c = Ac ∩ Bc.

Fundamental Properties of Set Operations (Cont…)


Proofs involving Cartesian products of sets

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Definition: The Cartesian product of two sets A and B is the set of

all ordered pairs (a, b) with a∈ A and b∈ B, i.e.

A×B = {(a,b)|a∈A and b∈B}.

Theorem: If A= Φ or B= Φ, then A×B =Φ.

Theorem: If A, B, C, and D are sets such that A ⊆ C and B ⊆ D, then A × B ⊆ C × D. Proof: Assume that A ⊆ c and B ⊆ D. Now let (x,y)∈ A × B ⇒ x∈A and y∈B ⇒ x∈C and y∈D, as A⊆ C and B ⊆ D. Therefore A×B⊆ C×D.


Proofs involving Cartesian products of sets (Cont…)

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Theorem: For sets A, B and C, A×(B∪C) = (A×B) ∪ (A×C).

Proof: Assume that (x,y)∈A×(B∪C) ⇒ x∈ A and {y∈ B or y∈C} ⇒ {x∈A and y∈B} or {x∈A and y∈C} ⇒(x,y)∈A×B or (x,y)∈A×C ⇒(x,y)∈(A×B) ∪ (A×C). Therefore, A×(B∪C) ⊆ (A×B) ∪ (A×C). Assume that (x,y)∈ (A×B) ∪ (A×C) ⇒(x,y)∈A×B or (x,y)∈A×C ⇒ {x∈A and y∈B} or {x∈A and y∈C} ⇒x∈ A and {y∈ B or y∈C} ⇒ (x,y)∈A×(B∪C). Therefore, (A×B) ∪ (A×C) ⊆ A×(B∪C) . Hence A×(B∪C) = (A×B) ∪ (A×C).


Proofs involving Cartesian products of sets (Cont…)

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Result: For sets A, B and C, A×(B-C) = (A×B) - (A×C). Proof: Assume that (x,y)∈A×(B-C) ⇒ x∈A and y∈ B-C

⇒ x∈ A and {y∈ B and y∉C} ⇒ {x∈A and y∈B} and {x∈A and y∉C} ⇒(x,y)∈A×B and (x,y)∉A×C, as y∉C ⇒(x,y)∈(A×B) - (A×C).

Therefore A×(B-C) ⊆ (A×B) - (A×C). Now Assume that (x,y)∈ (A×B)-(A×C) ⇒(x,y)∈A×B and (x,y)∉A×C

⇒ {x∈A and y∈B} and {x∈A and y∉C} as x∈ A ⇒x∈ A and {y∈ B and y∉C} ⇒ x∈A and y∈ B-C ⇒ (x,y)∈A×(B-C).

Therefore (A×B) - (A×C )⊆ A×(B-C). Hence A×(B-C) = (A×B) - (A×C).


Chapter-4: More on Direct Proof and Proof by Contrapositive

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Transcript of Chapter-4: More on Direct Proof and Proof by Contrapositive