Discrete Mathematics

Discrete Mathematics

WEN-CHING LIEN Department of Mathematics National Cheng Kung University

2008

5.1: Cartesian Products and Relations

Definition (5.1)

For sets A, B, the Cartesian product (cross product), of A and B is denoted by A×B and equals{(a,b)|a∈A,b∈B}.

Example (5.3)

A= {2,3,4},B= {4,5},C = {x,y}The cartesian product A×B can be represented by a tree diagram.

5.1: Cartesian Products and Relations

Definition (5.1)

For sets A, B, the Cartesian product (cross product), of A and B is denoted by A×B and equals{(a,b)|a∈A,b∈B}.

Example (5.3)

A= {2,3,4},B= {4,5},C = {x,y}The cartesian product A×B can be represented by a tree diagram.

Definition (5.2)

For sets A, B, any subset of A×B is called a (binary) relation from A to B. Any subset of A×A is called a (binary) relation on A.

Rules:

For finite sets A, B with|A| =m and|B| =n,there are 2^mn relations from A to B, including the empty relation as well as the relation A×B itself.

There are also 2^nm(=2^mn)relations from B to A, one of which is also∅and another of which is B×A.

The reason we get the same number of relations from B to A as we have from A to B is that any relationR₁from B to A can be obtained from a unique relationR₂from A to B by simply reversing the components of each ordered pair inR₂(and vice versa).

Definition (5.2)

For sets A, B, any subset of A×B is called a (binary) relation from A to B. Any subset of A×A is called a (binary) relation on A.

Rules:

For finite sets A, B with|A| =m and|B| =n,there are 2^mn relations from A to B, including the empty relation as well as the relation A×B itself.

There are also 2^nm(=2^mn)relations from B to A, one of which is also∅and another of which is B×A.

The reason we get the same number of relations from B to A as we have from A to B is that any relationR₁from B to A can be obtained from a unique relationR₂from A to B by simply reversing the components of each ordered pair inR₂(and vice versa).

Definition (5.2)

For sets A, B, any subset of A×B is called a (binary) relation from A to B. Any subset of A×A is called a (binary) relation on A.

Rules:

For finite sets A, B with|A| =m and|B| =n,there are 2^mn relations from A to B, including the empty relation as well as the relation A×B itself.

There are also 2^nm(=2^mn)relations from B to A, one of which is also∅and another of which is B×A.

The reason we get the same number of relations from B to A as we have from A to B is that any relationR₁from B to A can be obtained from a unique relationR₂from A to B by simply reversing the components of each ordered pair inR₂(and vice versa).

Definition (5.2)

For sets A, B, any subset of A×B is called a (binary) relation from A to B. Any subset of A×A is called a (binary) relation on A.

Rules:

For finite sets A, B with|A| =m and|B| =n,there are 2^mn relations from A to B, including the empty relation as well as the relation A×B itself.

There are also 2^nm(=2^mn)relations from B to A, one of which is also∅and another of which is B×A.

The reason we get the same number of relations from B to A as we have from A to B is that any relationR₁from B to A can be obtained from a unique relationR₂from A to B by simply reversing the components of each ordered pair inR₂(and vice versa).

Example (5.6)

For B= {1,2}, let A=P(B) = {∅,{1},{2},{1,2}}.The following is an example of a relation on A:

R= {(∅,∅),(∅,{1}),(∅,{2}),(∅,{1,2}),({1},{1}),({1},{1,2}), ({2},{2}),({2},{1,2}),({1,2},{1,2})}

We can say that the relationRis the subset relation where (C,D) ∈ Rif and only if C,D⊆B and C⊆D.

Example (5.6)

For B= {1,2}, let A=P(B) = {∅,{1},{2},{1,2}}.The following is an example of a relation on A:

R= {(∅,∅),(∅,{1}),(∅,{2}),(∅,{1,2}),({1},{1}),({1},{1,2}), ({2},{2}),({2},{1,2}),({1,2},{1,2})}

We can say that the relationRis the subset relation where (C,D) ∈ Rif and only if C,D⊆B and C⊆D.

Example (5.6)

For B= {1,2}, let A=P(B) = {∅,{1},{2},{1,2}}.The following is an example of a relation on A:

R= {(∅,∅),(∅,{1}),(∅,{2}),(∅,{1,2}),({1},{1}),({1},{1,2}), ({2},{2}),({2},{1,2}),({1,2},{1,2})}

We can say that the relationRis the subset relation where (C,D) ∈ Rif and only if C,D⊆B and C⊆D.

Example (5.8)

Let R be the subset of N×N where R = {(m,n)|n=7m}.

Consequently, among the ordered pairs in R one finds (0,0),(1,7),(11,77),and(15,105).This relation R on N can also be given recursively by

1)(0,0) ∈R;and

2) If(s,t) ∈R,then(s+1,t+7) ∈R.

We use the recursive definition to show that the ordered pair (3,21)is in R.From part (1) of the recursive definition we start with(0,0) ∈R.The part (2) of the definition gives us

i)(0,0) ∈R ⇒ (0+1,0+7) = (1,7) ∈R;

ii)(1,7) ∈R⇒ (1+1,7+7) = (2,14) ∈R;and iii)(2,14) ∈R⇒ (2+1,14+7) = (3,21) ∈R;

Example (5.8)

Let R be the subset of N×N where R = {(m,n)|n=7m}.

Consequently, among the ordered pairs in R one finds (0,0),(1,7),(11,77),and(15,105).This relation R on N can also be given recursively by

1)(0,0) ∈R;and

2) If(s,t) ∈R,then(s+1,t+7) ∈R.

We use the recursive definition to show that the ordered pair (3,21)is in R. From part (1) of the recursive definition we start with(0,0) ∈R.The part (2) of the definition gives us

i)(0,0) ∈R ⇒ (0+1,0+7) = (1,7) ∈R;

ii)(1,7) ∈R⇒ (1+1,7+7) = (2,14) ∈R;and iii)(2,14) ∈R⇒ (2+1,14+7) = (3,21) ∈R;

Example (5.8)

Let R be the subset of N×N where R = {(m,n)|n=7m}.

Consequently, among the ordered pairs in R one finds (0,0),(1,7),(11,77),and(15,105).This relation R on N can also be given recursively by

1)(0,0) ∈R;and

2) If(s,t) ∈R,then(s+1,t+7) ∈R.

We use the recursive definition to show that the ordered pair (3,21)is in R.From part (1) of the recursive definition we start with(0,0) ∈R.The part (2) of the definition gives us

i)(0,0) ∈R ⇒ (0+1,0+7) = (1,7) ∈R;

ii)(1,7) ∈R⇒ (1+1,7+7) = (2,14) ∈R;and iii)(2,14) ∈R⇒ (2+1,14+7) = (3,21) ∈R;

Example (5.8)

Let R be the subset of N×N where R = {(m,n)|n=7m}.

Consequently, among the ordered pairs in R one finds (0,0),(1,7),(11,77),and(15,105).This relation R on N can also be given recursively by

1)(0,0) ∈R;and

2) If(s,t) ∈R,then(s+1,t+7) ∈R.

We use the recursive definition to show that the ordered pair (3,21)is in R. From part (1) of the recursive definition we start with(0,0) ∈R.The part (2) of the definition gives us

i)(0,0) ∈R ⇒ (0+1,0+7) = (1,7) ∈R;

ii)(1,7) ∈R⇒ (1+1,7+7) = (2,14) ∈R;and iii)(2,14) ∈R⇒ (2+1,14+7) = (3,21) ∈R;

Example (5.8)

Let R be the subset of N×N where R = {(m,n)|n=7m}.

Consequently, among the ordered pairs in R one finds (0,0),(1,7),(11,77),and(15,105).This relation R on N can also be given recursively by

1)(0,0) ∈R;and

2) If(s,t) ∈R,then(s+1,t+7) ∈R.

We use the recursive definition to show that the ordered pair (3,21)is in R. From part (1) of the recursive definition we start with(0,0) ∈R.The part (2) of the definition gives us

i)(0,0) ∈R ⇒ (0+1,0+7) = (1,7) ∈R;

ii)(1,7) ∈R⇒ (1+1,7+7) = (2,14) ∈R;and iii)(2,14) ∈R⇒ (2+1,14+7) = (3,21) ∈R;

Example (5.8)

Let R be the subset of N×N where R = {(m,n)|n=7m}.

Consequently, among the ordered pairs in R one finds (0,0),(1,7),(11,77),and(15,105).This relation R on N can also be given recursively by

1)(0,0) ∈R;and

2) If(s,t) ∈R,then(s+1,t+7) ∈R.

We use the recursive definition to show that the ordered pair (3,21)is in R. From part (1) of the recursive definition we start with(0,0) ∈R.The part (2) of the definition gives us

i)(0,0) ∈R ⇒ (0+1,0+7) = (1,7) ∈R;

ii)(1,7) ∈R⇒ (1+1,7+7) = (2,14) ∈R;and iii)(2,14) ∈R⇒ (2+1,14+7) = (3,21) ∈R;

Example (5.8)

Let R be the subset of N×N where R = {(m,n)|n=7m}.

Consequently, among the ordered pairs in R one finds (0,0),(1,7),(11,77),and(15,105).This relation R on N can also be given recursively by

1)(0,0) ∈R;and

2) If(s,t) ∈R,then(s+1,t+7) ∈R.

We use the recursive definition to show that the ordered pair (3,21)is in R. From part (1) of the recursive definition we start with(0,0) ∈R.The part (2) of the definition gives us

i)(0,0) ∈R ⇒ (0+1,0+7) = (1,7) ∈R;

ii)(1,7) ∈R⇒ (1+1,7+7) = (2,14) ∈R;and iii)(2,14) ∈R⇒ (2+1,14+7) = (3,21) ∈R;

Theorem (5.1)

For any sets A,B,C⊆U :

a) A× (B∩C) = (A×B) ∩ (A×C) b) A× (B∪C) = (A×B) ∪ (A×C) c)(A∩B) ×C = (A×C) ∩ (B×C) d)(A∪B) ×C= (A×C) ∪ (B×C)

Theorem (5.1)

For any sets A,B,C⊆U :

a) A× (B∩C) = (A×B) ∩ (A×C) b) A× (B∪C) = (A×B) ∪ (A×C) c)(A∩B) ×C = (A×C) ∩ (B×C) d)(A∪B) ×C= (A×C) ∪ (B×C)

Theorem (5.1)

For any sets A,B,C⊆U :

a) A× (B∩C) = (A×B) ∩ (A×C) b) A× (B∪C) = (A×B) ∪ (A×C) c)(A∩B) ×C = (A×C) ∩ (B×C) d)(A∪B) ×C= (A×C) ∪ (B×C)

Theorem (5.1)

For any sets A,B,C⊆U :

a) A× (B∩C) = (A×B) ∩ (A×C) b) A× (B∪C) = (A×B) ∪ (A×C) c)(A∩B) ×C = (A×C) ∩ (B×C) d)(A∪B) ×C= (A×C) ∪ (B×C)

Theorem (5.1)

For any sets A,B,C⊆U :

a) A× (B∩C) = (A×B) ∩ (A×C) b) A× (B∪C) = (A×B) ∪ (A×C) c)(A∩B) ×C = (A×C) ∩ (B×C) d)(A∪B) ×C= (A×C) ∪ (B×C)

Thank you.