Discrete Mathematics with Applications MATH236 · Discrete Mathematics with Applications MATH236 ... University of KwaZulu-Natal Pietermaritzburg Campus Semester 1, 2013 ... De nitions

Discrete Mathematics with ApplicationsMATH236

Dr. Hung P. Tong-Viet

School of Mathematics, Statistics and Computer ScienceUniversity of KwaZulu-Natal

Pietermaritzburg Campus

Semester 1, 2013

Tong-Viet (UKZN) MATH236 Semester 1, 2013 1 / 21

Table of contents

1 Review of setsSpecial setsDeMorgan’s Law

2 Partitions

3 RelationDefinitions and examplesProperties of Relations


Review of sets Special sets

Some special sets

A numbers of sets occur frequently enough in mathematics that theyare given special names or symbols

The positive integers or natural numbers is the set:

N := {1, 2, 3, · · · }

The integers:

Z := {· · · ,−3,−2,−1, 0, 1, 2, 3, · · · }

The real numbers: R



Some special sets



N := {1, 2, 3, · · · }

The integers:

Z := {· · · ,−3,−2,−1, 0, 1, 2, 3, · · · }

The real numbers: R



Some special sets



N := {1, 2, 3, · · · }

The integers:

Z := {· · · ,−3,−2,−1, 0, 1, 2, 3, · · · }

The real numbers: R



Some special sets



N := {1, 2, 3, · · · }

The integers:

Z := {· · · ,−3,−2,−1, 0, 1, 2, 3, · · · }

The real numbers: R



Some special sets, (cont.)

The rational numbers:

Q = {a

b: a, b ∈ Z, b 6= 0} ⊆ R

The irrational numbers: R−Q

Example

0,−6 ∈ Z− N√

2 ∈ R−Qπ ∈ R−QN ⊆ Z ⊆ Q ⊆ R





Q = {a

b: a, b ∈ Z, b 6= 0} ⊆ R


Example

0,−6 ∈ Z− N

√2 ∈ R−Q

π ∈ R−QN ⊆ Z ⊆ Q ⊆ R





Q = {a

b: a, b ∈ Z, b 6= 0} ⊆ R


Example

0,−6 ∈ Z− N√

2 ∈ R−Q

π ∈ R−QN ⊆ Z ⊆ Q ⊆ R





Q = {a

b: a, b ∈ Z, b 6= 0} ⊆ R


Example

0,−6 ∈ Z− N√

2 ∈ R−Qπ ∈ R−Q

N ⊆ Z ⊆ Q ⊆ R





Q = {a

b: a, b ∈ Z, b 6= 0} ⊆ R


Example

0,−6 ∈ Z− N√

2 ∈ R−Qπ ∈ R−QN ⊆ Z ⊆ Q ⊆ R





Q = {a

b: a, b ∈ Z, b 6= 0} ⊆ R


Example

0,−6 ∈ Z− N√

2 ∈ R−Qπ ∈ R−QN ⊆ Z ⊆ Q ⊆ R


Review of sets DeMorgan’s Law

DeMorgan’s Law

Theorem (Demorgan’s Law)

If S1,S2, · · · , Sn are sets, then

S1 ∪ S2 · · · ∪ Sn = S1 ∩ S2 ∩ · · · ∩ Sn

Example

S1 = {1, 2, 3, 4} and S2 = {3, 4, 5, 6} are subsets of universal setU = {1, 2, 3, 4, 5, 6, 7, 8}S1 = {5, 6, 7, 8} and S2 = {1, 2, 7, 8}S1 ∩ S2 = {7, 8}S1 ∪ S2 = {1, 2, 3, 4, 5, 6} and S1 ∪ S2 = {7, 8}We have S1 ∪ S2 = {7, 8} = S1 ∩ S2



DeMorgan’s Law



S1 ∪ S2 · · · ∪ Sn = S1 ∩ S2 ∩ · · · ∩ Sn

Example

S1 = {1, 2, 3, 4} and S2 = {3, 4, 5, 6} are subsets of universal setU = {1, 2, 3, 4, 5, 6, 7, 8}

S1 = {5, 6, 7, 8} and S2 = {1, 2, 7, 8}S1 ∩ S2 = {7, 8}S1 ∪ S2 = {1, 2, 3, 4, 5, 6} and S1 ∪ S2 = {7, 8}We have S1 ∪ S2 = {7, 8} = S1 ∩ S2



DeMorgan’s Law



S1 ∪ S2 · · · ∪ Sn = S1 ∩ S2 ∩ · · · ∩ Sn

Example

S1 = {1, 2, 3, 4} and S2 = {3, 4, 5, 6} are subsets of universal setU = {1, 2, 3, 4, 5, 6, 7, 8}S1 = {5, 6, 7, 8} and S2 = {1, 2, 7, 8}

S1 ∩ S2 = {7, 8}S1 ∪ S2 = {1, 2, 3, 4, 5, 6} and S1 ∪ S2 = {7, 8}We have S1 ∪ S2 = {7, 8} = S1 ∩ S2



DeMorgan’s Law



S1 ∪ S2 · · · ∪ Sn = S1 ∩ S2 ∩ · · · ∩ Sn

Example

S1 = {1, 2, 3, 4} and S2 = {3, 4, 5, 6} are subsets of universal setU = {1, 2, 3, 4, 5, 6, 7, 8}S1 = {5, 6, 7, 8} and S2 = {1, 2, 7, 8}S1 ∩ S2 = {7, 8}

S1 ∪ S2 = {1, 2, 3, 4, 5, 6} and S1 ∪ S2 = {7, 8}We have S1 ∪ S2 = {7, 8} = S1 ∩ S2



DeMorgan’s Law



S1 ∪ S2 · · · ∪ Sn = S1 ∩ S2 ∩ · · · ∩ Sn

Example

S1 = {1, 2, 3, 4} and S2 = {3, 4, 5, 6} are subsets of universal setU = {1, 2, 3, 4, 5, 6, 7, 8}S1 = {5, 6, 7, 8} and S2 = {1, 2, 7, 8}S1 ∩ S2 = {7, 8}S1 ∪ S2 = {1, 2, 3, 4, 5, 6} and S1 ∪ S2 = {7, 8}

We have S1 ∪ S2 = {7, 8} = S1 ∩ S2



DeMorgan’s Law



S1 ∪ S2 · · · ∪ Sn = S1 ∩ S2 ∩ · · · ∩ Sn

Example




DeMorgan’s Law



S1 ∪ S2 · · · ∪ Sn = S1 ∩ S2 ∩ · · · ∩ Sn

Example




Proof of DeMorgan’s Law

To prove that X = Y , we need to prove that1 X ⊆ Y and

2 Y ⊆ X

Let A = S1 ∪ S2 · · · ∪ Sn and B = S1 ∩ S2 ∩ · · · ∩ Sn

We first prove that A ⊆ B1 Suppose that x ∈ A, that is, x ∈ S1 ∪ S2 · · · ∪ Sn

2 It follows that x 6∈ S1 ∪ S2 · · · ∪ Sn. Thus x is not a member of any ofthe sets S1,S2, · · · ,Sn

3 Hence, x ∈ Si for all i = 1, 2, · · · , n4 Therefore, x ∈ S1 ∩ S2 ∩ · · · ∩ Sn and hence x ∈ B5 So A ⊆ B as required




To prove that X = Y , we need to prove that1 X ⊆ Y and2 Y ⊆ X


















We first prove that A ⊆ B

1 Suppose that x ∈ A, that is, x ∈ S1 ∪ S2 · · · ∪ Sn


























3 Hence, x ∈ Si for all i = 1, 2, · · · , n

4 Therefore, x ∈ S1 ∩ S2 ∩ · · · ∩ Sn and hence x ∈ B5 So A ⊆ B as required



























Proof of DeMorgan’s Law (cont.)

The proof that B ⊆ A is left as an exercise

We have proved that A ⊆ B and B ⊆ A, hence A = B as wanted.

Example

Let S1 = {1, 2, 3},S2 = {2, 4, 5} and S3 = {5, 6, 7} are subsets of theuniversal set U = {1, 2, 3, 4, 5, 6, 7, 8}We have S1 ∪ S2 ∪ S3 = {1, 2, 3, 4, 5, 6, 7}. So S1 ∪ S2 ∪ S3 = {8}

S1 = {4, 5, 6, 7, 8}S2 = {1, 3, 6, 7, 8}S3 = {1, 2, 3, 4, 8}

Then S1 ∩ S2 ∩ S3 = {8} = S1 ∪ S2 ∪ S3






Example

Let S1 = {1, 2, 3},S2 = {2, 4, 5} and S3 = {5, 6, 7} are subsets of theuniversal set U = {1, 2, 3, 4, 5, 6, 7, 8}

We have S1 ∪ S2 ∪ S3 = {1, 2, 3, 4, 5, 6, 7}. So S1 ∪ S2 ∪ S3 = {8}

S1 = {4, 5, 6, 7, 8}S2 = {1, 3, 6, 7, 8}S3 = {1, 2, 3, 4, 8}

Then S1 ∩ S2 ∩ S3 = {8} = S1 ∪ S2 ∪ S3






Example


S1 = {4, 5, 6, 7, 8}S2 = {1, 3, 6, 7, 8}S3 = {1, 2, 3, 4, 8}

Then S1 ∩ S2 ∩ S3 = {8} = S1 ∪ S2 ∪ S3






Example


S1 = {4, 5, 6, 7, 8}

S2 = {1, 3, 6, 7, 8}S3 = {1, 2, 3, 4, 8}

Then S1 ∩ S2 ∩ S3 = {8} = S1 ∪ S2 ∪ S3






Example


S1 = {4, 5, 6, 7, 8}S2 = {1, 3, 6, 7, 8}

S3 = {1, 2, 3, 4, 8}

Then S1 ∩ S2 ∩ S3 = {8} = S1 ∪ S2 ∪ S3






Example


S1 = {4, 5, 6, 7, 8}S2 = {1, 3, 6, 7, 8}S3 = {1, 2, 3, 4, 8}

Then S1 ∩ S2 ∩ S3 = {8} = S1 ∪ S2 ∪ S3






Example


S1 = {4, 5, 6, 7, 8}S2 = {1, 3, 6, 7, 8}S3 = {1, 2, 3, 4, 8}

Then S1 ∩ S2 ∩ S3 = {8} = S1 ∪ S2 ∪ S3






Example


S1 = {4, 5, 6, 7, 8}S2 = {1, 3, 6, 7, 8}S3 = {1, 2, 3, 4, 8}

Then S1 ∩ S2 ∩ S3 = {8} = S1 ∪ S2 ∪ S3


Partitions

Disjoint sets

Two sets S and T are disjoint if S ∩ T = ∅A collection of subsets {S1, S2, · · · ,Sk} of a set S is called pairwisedisjoint if every two distinct subsets Si and Sj are disjoint

That is, {Si}ki=1 is pairwise disjoint if i 6= j implies Si ∩ Sj = ∅

Example

S1 = {1, 2, 3}, S2 = {5, 6, 7},S3 = {4, 8, 9} and S4 = {1, 5, 9}Then {S1,S2,S3} is pairwise disjoint

But {S1,S2,S4} is not pairwise disjoint


Partitions

Disjoint sets



Example




Partitions

Disjoint sets



Example

S1 = {1, 2, 3}, S2 = {5, 6, 7},S3 = {4, 8, 9} and S4 = {1, 5, 9}

Then {S1,S2,S3} is pairwise disjoint



Partitions

Disjoint sets



Example




Partitions

Disjoint sets



Example


But {S1,S2, S4} is not pairwise disjoint


Partitions

Disjoint sets



Example


But {S1,S2, S4} is not pairwise disjoint


Partitions

Partitions

Definition (Partitions)

A partition of a set S is a pairwise disjoint collection of nonempty subsetswhose union is S .

That is, a partition of a set S is a collection S = {Si}ki=1 of subsetsof S satisfying all three of the following criteria:

1 For all i ∈ {1, 2, · · · , k}, Si 6= ∅2 For all i 6= j ∈ {1, 2, · · · , k}, we have Si ∩ Sj = ∅

3⋃k

i=1 Si = S


Partitions

Partitions




1 For all i ∈ {1, 2, · · · , k}, Si 6= ∅

2 For all i 6= j ∈ {1, 2, · · · , k}, we have Si ∩ Sj = ∅

3⋃k

i=1 Si = S


Partitions

Partitions





3⋃k

i=1 Si = S


Partitions

Partitions





3⋃k

i=1 Si = S


Partitions

Partitions





3⋃k

i=1 Si = S


Partitions

Partitions (cont.)

If S = {Si}ki=1 is a partition of S , then each Si is called a part of thepartition S

Conditions 2 and 3 show that every element x of S is in exactly onepart of the partition

Example

Let S = {1, 2, · · · , 10} and S1 = {1, 2, 3, 8},S2 = {4, 5, 6},S3 = {7}S4 = {9, 10} and S5 = {7, 9, 10}{S1,S2, S3,S4} and {S1,S2,S5} are partitions of S

{S1,S2, S3,S4,S5} and {S2,S3,S5} are NOT partitions of S


Partitions

Partitions (cont.)



Example

Let S = {1, 2, · · · , 10} and S1 = {1, 2, 3, 8},S2 = {4, 5, 6}, S3 = {7}S4 = {9, 10} and S5 = {7, 9, 10}

{S1,S2, S3,S4} and {S1,S2,S5} are partitions of S



Partitions

Partitions (cont.)



Example

Let S = {1, 2, · · · , 10} and S1 = {1, 2, 3, 8},S2 = {4, 5, 6}, S3 = {7}S4 = {9, 10} and S5 = {7, 9, 10}{S1,S2, S3,S4} and {S1,S2, S5} are partitions of S



Partitions

Partitions (cont.)



Example


{S1,S2, S3,S4,S5} and {S2, S3,S5} are NOT partitions of S


Partitions

Partitions (cont.)



Example


{S1,S2, S3,S4,S5} and {S2, S3,S5} are NOT partitions of S


Partitions

Partitions (cont.)

Let I be the set of all irrational number. Then {Q, I} is a partition ofRFor i ∈ {0, 1, 2}, let Si be the set of all integers whose remainderwhen divided by 3 is i . i.e.,

S0 = {x ∈ Z : x = 3k for some k ∈ Z} = {· · · ,−3, 0, 3, · · · }S1 = {x ∈ Z : x = 3k + 1 for some k ∈ Z} = {· · · ,−2, 1, 4, · · · }S2 = {x ∈ Z : x = 3k + 2 for some k ∈ Z} = {· · · ,−1, 2, 5, · · · }

Then {S0,S1,S2} is a partition of Z


Partitions

Partitions (cont.)


S0 = {x ∈ Z : x = 3k for some k ∈ Z} = {· · · ,−3, 0, 3, · · · }

S1 = {x ∈ Z : x = 3k + 1 for some k ∈ Z} = {· · · ,−2, 1, 4, · · · }S2 = {x ∈ Z : x = 3k + 2 for some k ∈ Z} = {· · · ,−1, 2, 5, · · · }



Partitions

Partitions (cont.)


S0 = {x ∈ Z : x = 3k for some k ∈ Z} = {· · · ,−3, 0, 3, · · · }S1 = {x ∈ Z : x = 3k + 1 for some k ∈ Z} = {· · · ,−2, 1, 4, · · · }

S2 = {x ∈ Z : x = 3k + 2 for some k ∈ Z} = {· · · ,−1, 2, 5, · · · }



Partitions

Partitions (cont.)



Then {S0,S1, S2} is a partition of Z


Partitions

Partitions (cont.)





Partitions

Partitions (cont.)





Relation Definitions and examples

Relation

Let S and T be nonempty sets

A relation R from S to T is a subset of the cartesian product S × T

That is, R is a set of ordered pairs (s, t), where s ∈ S and t ∈ T

If (s, t) ∈ R, we say that s is related to t under R and we write sRt

If (s, t) 6∈ R, then we write s��Rt



Relation








Relation








Relation








Relation








Relation (cont.)

The domain of R is domR = {s ∈ S : sRt for some t ∈ T}

The range of R is ranR = {t ∈ T : sRt for some s ∈ S}

If S = T , then we say that R is a relation on S

A relation in which each element of the domain is related to exactlyone element of the range is a function



Relation (cont.)







Relation (cont.)







Relation (cont.)







Relations: Examples

Let S = {1, 2, 3},T = {a, b, c , d , e} andR = {(1, a), (2, b), (3, c), (2, c)}

Then R ⊆ S × T and thus it is a relation from S to T

The element 2 is related to both b and c under R, i.e., 2Rb and2Rc , so R is not a function

domR = {1, 2, 3} and ranR = {a, b, c}

We see that 1 is not related to b under R, that is, 1��Rb



Relations: Examples




domR = {1, 2, 3} and ranR = {a, b, c}




Relations: Examples




domR = {1, 2, 3} and ranR = {a, b, c}




Relations: Examples




domR = {1, 2, 3} and ranR = {a, b, c}




Relations: Examples




domR = {1, 2, 3} and ranR = {a, b, c}




Relations: Examples

Example

R1 = {(x , x2) : x ∈ R} is a relation on R. We have 2R4, 1R1 but2��R5

The set R2 = {(x , y) ∈ Z2 : x < y} is a relation on Z; R2 is therelation ‘is less than’ and we say ‘less than is a relation on Z’

Similarly, ‘≥’ is a relation on R (and on Z,Q).

Let P be the set of all subsets of S = {1, 2, 3, 4}.Then ⊆ is a relation on P.We have that {1} is related to {1, 2} under the relation ⊆



Relations: Examples

Example







Relations: Examples

Example




Let P be the set of all subsets of S = {1, 2, 3, 4}.

Then ⊆ is a relation on P.We have that {1} is related to {1, 2} under the relation ⊆



Relations: Examples

Example




Let P be the set of all subsets of S = {1, 2, 3, 4}.Then ⊆ is a relation on P.

We have that {1} is related to {1, 2} under the relation ⊆



Relations: Examples

Example







Relations: Examples

Example







Relations (cont.)

Let a and b be integers

We say that a divides b and write a | b if b = at for some t ∈ Z

If a | b, then we say a is a divisor of b; and b is a multiple of a

If a | b, then we say that a is divisible by b and written a...b.

We write a - b if a is not a divisor of b

2 | 6 but 5 - 6



Relations (cont.)






2 | 6 but 5 - 6



Relations (cont.)






2 | 6 but 5 - 6



Relations (cont.)






2 | 6 but 5 - 6



Relations (cont.)






2 | 6 but 5 - 6



Relations (cont.)

Example

Let R be a relation on Z defined by xRy if and only if 2 | (x − y)

We have 2R4, 3R5 and (−1)R(−3)

But 1��R8 since 2 - −7 = 1− 8

Example

Let R be a relation on Z defined by xRy if and only if 5 | (2x + 3y)

Is 2R2?

Is 1R3?



Relations (cont.)

Example



But 1��R8 since 2 - −7 = 1− 8

Example


Is 2R2?

Is 1R3?



Relations (cont.)

Example



But 1��R8 since 2 - −7 = 1− 8

Example


Is 2R2?

Is 1R3?



Relations (cont.)

Example



But 1��R8 since 2 - −7 = 1− 8

Example


Is 2R2?

Is 1R3?



Relations (cont.)

Example



But 1��R8 since 2 - −7 = 1− 8

Example


Is 2R2?

Is 1R3?



Relations (cont.)

Example



But 1��R8 since 2 - −7 = 1− 8

Example


Is 2R2?

Is 1R3?


Relation Properties of Relations

Properties of relations

Let R be a relation on a set S

R is reflexive if xRx for all x ∈ S

R is symmetric if for all x , y ∈ S , xRy implies yRx

R is transitive if for all x , y , z ∈ S , xRy and yRz implies xRz

R is irreflexive if there is no x ∈ S with xRx

R is antisymmetric if for all x , y ∈ S , xRy and yRx implies x = y
















































Properties of Relations (cont.)

Consider the relation R on R defined by xRy iff y = x2

(2, 2) 6∈ R since 2 6= 22 = 4, so R is NOT reflexive

(1, 1) ∈ R since 12 = 1, so R is NOT irreflexive

R is NOT symmetric since (2, 4) ∈ R but (4, 2) 6∈ R

R is NOT transitive since (3, 9) ∈ R and (9, 81) ∈ R but (3, 81) 6∈ R(32 6= 81)

R is antisymmetric: Suppose that (x , y), (y , x) ∈ R where x , y ∈ R.By definition, we have x2 = y and y 2 = x . Solving these equationssimultaneously, we obtain that (0, 0) and (1, 1) are the only realsolutions.

















































Consider the relation R on R defined by xRy iff x < y

R is NOT reflexive since no number is less than itself

R is irreflexive since (x , x) ∈∈ R for all x ∈ R

R is NOT symmetric since 2 < 4 but 4 6< 2

R is transitive since if x < y and y < z , then x < z for all x , y , z ∈ R




































Consider the relation R on Z defined by xRy iff 2 | (x − y)

R is reflexive since 2 | x − x = 0 for all x ∈ Z

R is symmetric since if (x , y) ∈ R, then 2 | x − y so2 | −(x − y) = y − x hence yRx

R is transitive since if 2 | x − y and 2 | y − z , then2 | (x − y) + (y − z) = x − z so xRz























Documents

Discrete Mathematics with Applications MATH236 · Discrete Mathematics with Applications MATH236 ... University of KwaZulu-Natal Pietermaritzburg Campus Semester 1, 2013 ... De nitions