The document is a lecture on logic and proof techniques in mathematics, focusing on direct proof and proof by contrapositive. It covers proofs involving divisibility, congruence of integers, real numbers, and set operations, illustrating key results and properties. Various examples and theorems demonstrate the application of these techniques in mathematical reasoning.

1. MATH 401: LOGIC & PROOF TECHNIQUES Copyright © Nahid Sultana 2014-2015.
Dr. Nahid Sultana Email: nszakir@ud.edu.sa
Chapter-4: More on Direct Proof and Proof by Contrapositive
10/10/2014
1

2. Topics 2
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Proofs Involving Divisibility of Integers

Proofs Involving Congruence of Integers

Proofs Involving Real Numbers

Proofs Involving sets

Fundamental Properties of Set Operations
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Proofs Involving Cartesian Products of Sets 10/10/2014 Copyright © Nahid Sultana 2014-2015.

3. 3
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
|/
Copyright © Nahid Sultana 2014-2015. 10/10/2014

4. Proofs Involving Divisibility of Integers (Cont…) 4
Result: Let a, b, and c be integers with a≠0 and b ≠0.
If a|b and b|c, then a|c.
Proof: (Direct Proof) Assume that a|b and b|c.
Then b = ax and c = by, where x, y ∈ ℤ.
Therefore, c = by
= (ax)y
= a(xy)
= a z, where z=xy ∈ ℤ.
Hence a|c.
 Copyright © Nahid Sultana 2014-2015. 10/10/2014

5. Proofs Involving Divisibility of Integers (Cont…) 5
Result: Let a, b, c and d be integers with a≠0 and b ≠0.
If a|c and b|d, then ab|cd.
Proof: (Direct Proof) Assume that a|c and b|d.
Then c = ax and d = by, where x, y ∈ ℤ.
Therefore, cd = (ax)(by)
= ab(xy)
= ab z, where z = xy ∈ ℤ.
Hence ab|cd.
 Copyright © Nahid Sultana 2014-2015. 10/10/2014

6. Proofs Involving Divisibility of Integers (Cont…) 6
Result: Let a, b, c, x, y ∈ ℤ, where a≠0.
If a|b and a|c, then a|(bx+cy).
Proof: (Direct Proof) Assume that a|b and a|c,
Then b = ar and c = as, where r, s ∈ ℤ.
Therefore, bx+cy = (ar)x+(as)y
= a(rx+sy)
= at, where t = rx+sy ∈ ℤ.
Hence a|(bx+cy).
 Copyright © Nahid Sultana 2014-2015. 10/10/2014

7. Proofs Involving Divisibility of Integers (Cont…) 7
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).
 10/10/2014 Copyright © Nahid Sultana 2014-2015.

8. Proofs Involving Divisibility of Integers (Cont…) 8
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.

|/|/|/
Copyright © Nahid Sultana 2014-2015. 10/10/2014

9. Proofs Involving Divisibility of Integers (Cont…) 9
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).

|/ |/
Copyright © Nahid Sultana 2014-2015. 10/10/2014

10. Proofs Involving Divisibility of Integers (Cont…) 10
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.

|/ |/ |/
Copyright © Nahid Sultana 2014-2015. 10/10/2014

11. Proofs Involving Congruence of Integers 11
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).
≡/|/
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).

10/10/2014 Copyright © Nahid Sultana 2014-2015.

12. Proofs Involving Congruence of Integers (Cont…) 12
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).

10/10/2014 Copyright © Nahid Sultana 2014-2015.

13. Proofs Involving Congruence of Integers (Cont…) 13
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).

≡/≡/ ≡/
Copyright © Nahid Sultana 2014-2015. 10/10/2014

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

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

16. 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.

Copyright © Nahid Sultana 2014-2015. 10/10/2014

17. 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.

17
Proofs Involving Real Numbers (Cont…) Copyright © Nahid Sultana 2014-2015. 10/10/2014

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

19. 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.

10/10/2014 Copyright © Nahid Sultana 2014-2015.

20. 20
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…) Copyright © Nahid Sultana 2014-2015. 10/10/2014

21. Fundamental Properties of Set Operations 21

Commutative laws
 Associative laws
 Distributive laws
Copyright © Nahid Sultana 2014-2015. 10/10/2014

22. Fundamental Properties of Set Operations (Cont…) 22

De Morgan’s laws

Absorption laws
Complement laws
Copyright © Nahid Sultana 2014-2015. 10/10/2014

23. 23
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…) Copyright © Nahid Sultana 2014-2015. 10/10/2014

24. Proofs involving Cartesian products of sets 24
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.

10/10/2014 Copyright © Nahid Sultana 2014-2015.

25. Proofs involving Cartesian products of sets (Cont…) 25
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).

10/10/2014 Copyright © Nahid Sultana 2014-2015.

26. Proofs involving Cartesian products of sets (Cont…) 26
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).

10/10/2014 Copyright © Nahid Sultana 2014-2015.