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Page 1: Math 42, Discrete Mathematics

Math 42, Discrete Mathematics

Fall 2018

Richard P. Kubelka

San Jose State University

last updated 12/05/2018 at 15:47:21

For use by students in this class only; all rights reserved.Note: some prose & some tables are taken directly from Kenneth R. Rosen, Discrete

Mathematics and Its Applications, 8th Ed., the o�cial text adopted for this course.

c© R. P. Kubelka

Page 2: Math 42, Discrete Mathematics

Math 42,Discrete

Mathematics

Richard P.Kubelka

San Jose StateUniversity

Relations &Their Properties

EquivalenceRelations

Matrices,Digraphs, &RepresentingRelations

c© R. P. Kubelka

Binary Relations from A to B

De�nition

Let A and B be sets. A binary relation from A to B is a

subset of A× B

Suppose R ⊆ A× B is a relation from A to B. If (a,b) ∈ R,we write aRb and say that a is related to b by R. If

(a,b) /∈ R, we write a�Rb.

Remark

A relation as de�ned above is called binary because it

involves ordered pairs. We could talk about ternary

relations involving ordered triples, quaternary relations

involving ordered quadruples, n-ary relations involving

ordered n-tuples, etc.

Since we will restrict our investigation to the binary case, we

will henceforth simply speak of �relations," instead of �binary

relations."

Page 3: Math 42, Discrete Mathematics

Math 42,Discrete

Mathematics

Richard P.Kubelka

San Jose StateUniversity

Relations &Their Properties

EquivalenceRelations

Matrices,Digraphs, &RepresentingRelations

c© R. P. Kubelka

Relations � Examples

Examples

1. Suppose A = {students in this Math 42 class} and

B = {countries in the world other than the US}. Let

R1 = {(a,b) |a has visited b}.

2. If f : A→ B is a function, then R2 = {(a, f(a)) |a ∈ A}is a relation, called the graph of the function.

Remarks

I Conversely, if R ⊆ A× B is a relation with the property that

every element of A is the �rst entry in precisely one ordered

pair of R, then setting f(a) = b whenever (a,b) ∈ R de�nes

a function f : A→ B.

I Note that R1 illustrates the fact that a given a ∈ A may

correspond under a relation to many elements b ∈ B�or

none.

Page 4: Math 42, Discrete Mathematics

Math 42,Discrete

Mathematics

Richard P.Kubelka

San Jose StateUniversity

Relations &Their Properties

EquivalenceRelations

Matrices,Digraphs, &RepresentingRelations

c© R. P. Kubelka

Relations � Examples

3. R3 = ∅ ⊆ A× B.4. R4 = A× B ⊆ A× B.

For the most part, we will be interested in relations where

B = A.

De�nition

A relation on A is a subset of A×A, i.e., a relation from A

to A.

Examples

The relations 3 and 4 above are relations on A when B = A.

5. For any set A, R5 = {(a,a) |a ∈ A} is a relation on

A. This relation, of course, is just equality, and we

write �a1 = a2" instead of �a1R5a2."

Page 5: Math 42, Discrete Mathematics

Math 42,Discrete

Mathematics

Richard P.Kubelka

San Jose StateUniversity

Relations &Their Properties

EquivalenceRelations

Matrices,Digraphs, &RepresentingRelations

c© R. P. Kubelka

Relations � Examples

Examples

6. R6 = {(a1,a2) |a1 6= a2}. We write �a1 6= a2" instead

of �a1R6a2."

For the following examples, let A = Z+.

7. R7 = {(a1,a2) |a1 > a2}. We write �a1 > a2" instead

of �a1R7a2."

8. R8 = {(a1,a2) |a1 > a2}. We write �a1 > a2" instead

of �a1R8a2."

Remark

Relations 7 and 8 can be de�ned for A any subset of R.

9. R9 = {(a1,a2) |a1 divides a2}. We write �a1 |a2"

instead of �a1R9a2."

Page 6: Math 42, Discrete Mathematics

Math 42,Discrete

Mathematics

Richard P.Kubelka

San Jose StateUniversity

Relations &Their Properties

EquivalenceRelations

Matrices,Digraphs, &RepresentingRelations

c© R. P. Kubelka

Relations � Examples

Examples

10. If A = P(S), for some set S, we may de�ne the relation

R10 = {(X, Y) |X ⊆ Y}. We write �X ⊆ Y" instead of

�XR10Y."

For the following examples, let A = {all triangles}.

11. R11 = {(a1,a2) |a1 is congruent to a2}. We write

�a1 ∼= a2" instead of �a1R11a2."

12. R12 = {(a1,a2) |a1 is similar to a2}. We write

�a1 ∼ a2" instead of �a1R12a2."

Remark

For technical reasons, it turns out that there is no �set of all

triangles." But at this point there's no harm in pretending

that there is.

Page 7: Math 42, Discrete Mathematics

Math 42,Discrete

Mathematics

Richard P.Kubelka

San Jose StateUniversity

Relations &Their Properties

EquivalenceRelations

Matrices,Digraphs, &RepresentingRelations

c© R. P. Kubelka

Relations � Examples

For the following two examples, let A be the set of all human

beings, living and dead.

13. R13 = {(a1,a2) |a1 is Facebook friends with a2}.

14. R14 = {(a1,a2) |a1 loves a2}.

15. Let A = {(p,q) |p,q ∈ Z ∧ q 6= 0} and de�ne

(p1,q1)R15(p2,q2) if and only if p1q2 = p2q1.

Warning

The set A is already a set of ordered pairs. So the relation

R15 is a set of ordered pairs of ordered pairs:

((p1,q1), (p2,q2)).

This relation is much more complicated than the others but

it is nonetheless very important.

Page 8: Math 42, Discrete Mathematics

Math 42,Discrete

Mathematics

Richard P.Kubelka

San Jose StateUniversity

Relations &Their Properties

EquivalenceRelations

Matrices,Digraphs, &RepresentingRelations

c© R. P. Kubelka

Properties of Relations on A

De�nitions

Let R be a relation on A.

I R is called re�exive if aRa for all a ∈ A. That is, if∀a ∈ A, (a,a) ∈ R.

I R is called symmetric if a1Ra2 implies a2Ra1. That is,

if ∀(a1,a2) ∈ A×A, ((a1,a2) ∈ R)→ ((a2,a1) ∈ R).I R is called antisymmetric if a1Ra2 and a2Ra1 implies

a1 = a2.

I R is called transitive if a1Ra2 and a2Ra3 implies

a1Ra3. That is,

if ∀(a1,a2,a3) ∈ A×A×A, (((a1,a2) ∈R)∧ (a2,a3) ∈ R))→ ((a1,a3) ∈ R)

Page 9: Math 42, Discrete Mathematics

Math 42,Discrete

Mathematics

Richard P.Kubelka

San Jose StateUniversity

Relations &Their Properties

EquivalenceRelations

Matrices,Digraphs, &RepresentingRelations

c© R. P. Kubelka

Properties of Example Relations

Table 1: Properties of Example Relations

R S T A

R3 Y Y Y

R4 Y Y Y

R5 Y Y Y Y

R6 Y

R7 Y Y Y

R8 Y Y

R9 Y Y Y

R10 Y Y Y

R11 Y Y Y

R12 Y Y Y

R13 Y

R14

R15 Y Y Y

Page 10: Math 42, Discrete Mathematics

Math 42,Discrete

Mathematics

Richard P.Kubelka

San Jose StateUniversity

Relations &Their Properties

EquivalenceRelations

Matrices,Digraphs, &RepresentingRelations

c© R. P. Kubelka

Equivalence Relations

A type of relation on A that is very important in many

branches of mathematics is an equivalence relation.

De�nition

A relation R on a set A is called an equivalence relation on

A if R is

1. re�exive,

2. symmetric, &

3. transitive.

Example

Let A = Z and �x a positive integer n. De�ne a relation R16on A by

R16 = {(r, s) | r− s = nk for some k ∈ Z}= {(r, s) |n divides r− s evenly}

(1)

Page 11: Math 42, Discrete Mathematics

Math 42,Discrete

Mathematics

Richard P.Kubelka

San Jose StateUniversity

Relations &Their Properties

EquivalenceRelations

Matrices,Digraphs, &RepresentingRelations

c© R. P. Kubelka

Equivalence Relations � Examples

Claim

R16 is an equivalence relation on Z.

Proof.

Re�exive If r ∈ Z, we must show that rR16r. But

r− r = 0 = n · 0, which is what we need.

Symmetric If rR16s, we must show sR16r. rR16s gives us

that r− s = nk for some k ∈ Z. But thens− r = −nk = n(−k), and −k ∈ Z. So we

have what we need.

Transitive If rR16s and sR16t, then r− s = nk and

s− t = nl for some k, l ∈ Z. Then

r− t = (r− s)+ (s− t) = nk+nl = n(k+ l).

And since (k+ l) ∈ Z, we have rR16t, which is

what we needed.

Page 12: Math 42, Discrete Mathematics

Math 42,Discrete

Mathematics

Richard P.Kubelka

San Jose StateUniversity

Relations &Their Properties

EquivalenceRelations

Matrices,Digraphs, &RepresentingRelations

c© R. P. Kubelka

Equivalence Relations � Examples

RemarksI If (r, s) ∈ R16�i.e., rR16s�we write

r ≡ s (mod n)

and say �r is congruent to s modulo n."

I If n = 2, r ≡ s (mod 2) if and only if r and s have the

same parity, i.e., are both odd or both even.

Page 13: Math 42, Discrete Mathematics

Math 42,Discrete

Mathematics

Richard P.Kubelka

San Jose StateUniversity

Relations &Their Properties

EquivalenceRelations

Matrices,Digraphs, &RepresentingRelations

c© R. P. Kubelka

Equivalence Relations � Examples

Example

Recall that when A = {(p,q) |p,q ∈ Z ∧ q 6= 0}, we de�ned

(p1,q1)R15(p2,q2) if and only if p1q2 = p2q1.

Claim

R15 is an equivalence relation on Z× (Z− {0}).

Proof.

Re�exive If (p,q) ∈ A, then pq = pq. So(p,q)R15(p,q), as needed.

Symmetric If (p1,q1)R15(p2,q2), then p1q2 = p2q1. Butthen p2q1 = p1q2. So (p2,q2)R15(p1,q1), asdesired.

Page 14: Math 42, Discrete Mathematics

Math 42,Discrete

Mathematics

Richard P.Kubelka

San Jose StateUniversity

Relations &Their Properties

EquivalenceRelations

Matrices,Digraphs, &RepresentingRelations

c© R. P. Kubelka

Equivalence Relations � Examples

Example

(cont'd)

Proof.

(cont'd)

Transitive If (p1,q1)R15(p2,q2) and (p2,q2)R15(p3,q3),then p1q2 = p2q1, and p2q3 = p3q2. So

p1q3q2 = p2q3q1 = p2q1q3 = p3q1q2. (2)

But the extreme left and right ends of (2)

both contain a factor of q2 6= 0. Dividing by

q2 gives p1q3 = p3q1, and hence the desired

(p1,q1)R15(p3,q3).

Page 15: Math 42, Discrete Mathematics

Math 42,Discrete

Mathematics

Richard P.Kubelka

San Jose StateUniversity

Relations &Their Properties

EquivalenceRelations

Matrices,Digraphs, &RepresentingRelations

c© R. P. Kubelka

Equivalence Relations

De�nition

Two elements a1,a2 ∈ A that are related by an equivalence

relation R�i.e., that satisfy a1Ra2�are called equivalent.

The notation a1 ∼ a2 is often used to denote that a1 and a2are equivalent with respect to a particular equivalence

relation. (Although speci�c equivalence relations like

congruence mod n, and, of course, equality, have their own

symbols, ≡ and =.)

Page 16: Math 42, Discrete Mathematics

Math 42,Discrete

Mathematics

Richard P.Kubelka

San Jose StateUniversity

Relations &Their Properties

EquivalenceRelations

Matrices,Digraphs, &RepresentingRelations

c© R. P. Kubelka

Equivalence Classes

De�nition

If ∼ is an equivalence relation on A, and if a ∈ A, then we

de�ne [a], the equivalence class of a, by

[a] = {b ∈ A |b ∼ a}. (3)

Remark

In cases where more than one equivalence relation on A is

under consideration, we write [a]R = {b ∈ A |bRa} to

emphasize the particular equivalence relation R.

Page 17: Math 42, Discrete Mathematics

Math 42,Discrete

Mathematics

Richard P.Kubelka

San Jose StateUniversity

Relations &Their Properties

EquivalenceRelations

Matrices,Digraphs, &RepresentingRelations

c© R. P. Kubelka

Equivalence Classes � Examples

Example

Recall that for a �xed positive integer n, we de�ned the

relation R16 by

R16 = {(r, s) | r− s = nk for some k ∈ Z}= {(r, s) |n divides r− s evenly}

(4)

Moreover, we proved that R16 is an equivalence relation.

So what are the equivalence classes for R16?

If m ∈ Z, we have

[m] = {k ∈ Z |k ≡ m mod n}

= {k ∈ Z |k−m = nl for some l ∈ Z}= {k ∈ Z |k = m+ nl for some l ∈ Z}= {m,m± n,m± 2n, . . .}

(5)

Page 18: Math 42, Discrete Mathematics

Math 42,Discrete

Mathematics

Richard P.Kubelka

San Jose StateUniversity

Relations &Their Properties

EquivalenceRelations

Matrices,Digraphs, &RepresentingRelations

c© R. P. Kubelka

Equivalence Classes � Examples

Example

Recall that when A = {(p,q) |p,q ∈ Z ∧ q 6= 0}, we de�ned

(p1,q1)R15(p2,q2) if and only if p1q2 = p2q1.

Furthermore, we proved that R15 is an equivalence relation.

So what are the equivalence classes for R15?

If (p,q) ∈ A, we have

[(p,q)] = {(r, s) ∈ A | (r, s) ∼ (p,q)} (6)

Case 1 (r, s) ∼ (0,q) if and only if r = 0. So

[(0,q)] = {(0, s) | s 6= 0} (7)

Page 19: Math 42, Discrete Mathematics

Math 42,Discrete

Mathematics

Richard P.Kubelka

San Jose StateUniversity

Relations &Their Properties

EquivalenceRelations

Matrices,Digraphs, &RepresentingRelations

c© R. P. Kubelka

Equivalence Classes � Examples

Example

(cont'd)

Case 2 If p 6= 0,

(r, s) ∼ (p,q)⇔ rq = ps⇔ r

p=s

q= v ∈ Q−{0}

(8)

so (r, s) = (vp, vq).

Thus,

[(p,q)] = {(r, s) | (r, s) ∼ (p,q)}

= {(vp, vq) | v ∈ Q− {0}} ∩A(9)

Page 20: Math 42, Discrete Mathematics

Math 42,Discrete

Mathematics

Richard P.Kubelka

San Jose StateUniversity

Relations &Their Properties

EquivalenceRelations

Matrices,Digraphs, &RepresentingRelations

c© R. P. Kubelka

Equivalence Classes � Examples

Example

(cont'd)

For example,

[(1, 2)] = {(1, 2), (−1,−2), (2, 4), (−2,−4), (3, 6), . . .} (10)

Remark

When we use the rational numbers�numbers of the form

p/q where p,q ∈ Z and q 6= 0�we're actually dealing with

the equivalence classes discussed in this example because

we're using the fact that

1

2=

−1

−2=

2

4=

−2

−4=

3

6=

−3

−6= . . .

That's why this rather complicated example is important.

Page 21: Math 42, Discrete Mathematics

Math 42,Discrete

Mathematics

Richard P.Kubelka

San Jose StateUniversity

Relations &Their Properties

EquivalenceRelations

Matrices,Digraphs, &RepresentingRelations

c© R. P. Kubelka

Equivalence Relations & Partitions

De�nition

A partition of a set S is a collection of disjoint nonempty

subsets of S whose union is all of S.

In other words, P = {Ai |Ai ⊆ S ∀i ∈ I} is a partition of S if

and only if

1. Ai 6= ∅ ∀i ∈ I;2. Ai ∩Aj = ∅ ∀i 6= j ∈ I; and3.

⋃i∈IAi = S.

Remark

We call the Ai in a partition mutually exclusive and

exhaustive because they are mutually exclusive and

together they exhaust all the possible elements of S.

Page 22: Math 42, Discrete Mathematics

Math 42,Discrete

Mathematics

Richard P.Kubelka

San Jose StateUniversity

Relations &Their Properties

EquivalenceRelations

Matrices,Digraphs, &RepresentingRelations

c© R. P. Kubelka

Partitions

Figure 1: A Partition of a Set

Page 23: Math 42, Discrete Mathematics

Math 42,Discrete

Mathematics

Richard P.Kubelka

San Jose StateUniversity

Relations &Their Properties

EquivalenceRelations

Matrices,Digraphs, &RepresentingRelations

c© R. P. Kubelka

Equivalence Relations & Partitions

Partitions are intimately related to equivalence relations, as

the following theorem shows.

Theorem

Let R be an equivalence relation on a set S. Then the

equivalence classes of R form a partition of S. Conversely,

given a partition {Ai | i ∈ I} of the set S, there is an

equivalence relation R that has the sets Ai, i ∈ I, as itsequivalence classes.

Page 24: Math 42, Discrete Mathematics

Math 42,Discrete

Mathematics

Richard P.Kubelka

San Jose StateUniversity

Relations &Their Properties

EquivalenceRelations

Matrices,Digraphs, &RepresentingRelations

c© R. P. Kubelka

Equivalence Relations & Partitions

Proof.

�⇒"

Assume that R is an equivalence relation on S and Let

P = {[s] | s ∈ S} be the set of its equivalence classes. We

must show that P is a partition of S.

1. For each equivalence class [s] ∈ P, s ∈ [s] by re�exivity,

so [s] 6= ∅.2. Given [s], [t] ∈ P, we must show that either [s] = [t] or

[s] ∩ [t] = ∅.Suppose u ∈ [s] ∩ [t], i.e., [s] ∩ [t] 6= ∅. Then u ∼ s and

u ∼ t. Now if v ∈ [s], then v ∼ s. But symmetry of the

relation gives that s ∼ u, and transitivity then implies

that v ∼ u; transitivity again gives v ∼ t. Hence

[s] ⊆ [t]. A similar argument gives that [t] ⊆ [s], and so

[s] = [t].

Page 25: Math 42, Discrete Mathematics

Math 42,Discrete

Mathematics

Richard P.Kubelka

San Jose StateUniversity

Relations &Their Properties

EquivalenceRelations

Matrices,Digraphs, &RepresentingRelations

c© R. P. Kubelka

Equivalence Relations & Partitions

Proof.

(cont'd)

3.⋃

s∈S[s] ⊆ S since [s] ⊆ S ∀s ∈ S. On the other hand,

for every s ∈ S, s ∈ [s] ⊆⋃

s∈S[s]. Thus⋃

s∈S[s] = S,as needed.

�⇐"

If P = {Ai | i ∈ I} is a partition of S, we must show that P

de�nes an equivalence relation R on S whose equivalence

classes are the Ai.

De�ne the relation R by s ∼ t if and only if s, t ∈ Ai0 for

some i0 ∈ I, i.e., if and only if s and t are in the same piece

of the partition.

Re�exive If s ∈ S, then since⋃

i∈IAi = S, we must

have s ∈ Ai0 for some i0 ∈ I. But since s is inthe same Ai0 as itself, s ∼ s.

Page 26: Math 42, Discrete Mathematics

Math 42,Discrete

Mathematics

Richard P.Kubelka

San Jose StateUniversity

Relations &Their Properties

EquivalenceRelations

Matrices,Digraphs, &RepresentingRelations

c© R. P. Kubelka

Equivalence Relations & Partitions

Proof.

(cont'd)

Symmetric If s ∼ t, then s, t ∈ Ai0 for some i0 ∈ I. Butthen t, s ∈ Ai0 , and so t ∼ s.

Transitive If s ∼ t and t ∼ v, then s, t ∈ Ai0 for some

i0 ∈ I, and t, v ∈ Aj0 for some j0 ∈ I. Thent ∈ Ai0 ∩Aj0 6= ∅. This implies that

Aj0 = Ai0 , so s, v ∈ Ai0 , and thus s ∼ v, as

needed.

Page 27: Math 42, Discrete Mathematics

Math 42,Discrete

Mathematics

Richard P.Kubelka

San Jose StateUniversity

Relations &Their Properties

EquivalenceRelations

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c© R. P. Kubelka

Equivalence Relations & Partitions � Example

Example

Let A = R. And consider

P = {[k+ i/100− 0.005,k+ i/100+ 0.005) |k ∈ Z, i = 0, . . . , 99}(11)

Claim

P is a partition of A = R.

So that means that P gives rise to an equivalence relation ∼.

De�ne x ∼ y if and only if

b100x+ 0.5c = b100y+ 0.5c

This ∼ is the equivalence relation that gives rise to the

partition P.

Page 28: Math 42, Discrete Mathematics

Math 42,Discrete

Mathematics

Richard P.Kubelka

San Jose StateUniversity

Relations &Their Properties

EquivalenceRelations

Matrices,Digraphs, &RepresentingRelations

c© R. P. Kubelka

Equivalence Relations & Partitions � Example

Example

(cont'd)

What's it all about? This is the equivalence relation we get

when we round a number to the nearest hundredth. A similar

argument yields the equivalence relation of rounding to any

speci�ed decimal place.

Page 29: Math 42, Discrete Mathematics

Math 42,Discrete

Mathematics

Richard P.Kubelka

San Jose StateUniversity

Relations &Their Properties

EquivalenceRelations

Matrices,Digraphs, &RepresentingRelations

c© R. P. Kubelka

Equivalence Relations & Partitions � Example

Example

Suppose f : X→ Y is a function. De�ne a relation ∼ on X by

x1 ∼ x2 if and only if f(x1) = f(x2). Show that ∼ is an

equivalence relation.

Proof.

Re�exive For each x ∈ X, f(x) = f(x) since �=� is

re�exive. Thus x ∼ x.

Symmetric If x1 ∼ x2, then f(x1) = f(x2). But �=� is

symmetric, so f(x2) = f(x1), and thus x2 ∼ x1.

Transitive x1 ∼ x2 and x2 ∼ x3, then f(x1) = f(x2) andf(x2) = f(x3). But then f(x1) = f(x3) by the

transitivity of �=�, and so x1 ∼ x3, as needed.

Page 30: Math 42, Discrete Mathematics

Math 42,Discrete

Mathematics

Richard P.Kubelka

San Jose StateUniversity

Relations &Their Properties

EquivalenceRelations

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c© R. P. Kubelka

Equivalence Relations & Partitions � Example

Example

(cont'd)

Remark

Note that the same result�with the same proof�would hold

if ≈ were an equivalence relation on Y and we de�ned ∼ on

X by x1 ∼ x2 if and only if f(x1) ≈ f(x2)

Suppose that f : R→ R is given by f(x) = sin x and that we

de�ne x1 ∼ x2 if and only if f(x1) = f(x2). What is [π/2],i.e., the equivalence class represented by π/2?

[π/2] = {x ∈ R | sin x = sinπ/2 = 1}

= {π/2+ k2π |k ∈ Z}.(12)

Note that x2 ∼ x1 if and only if x2 − x1 = k · 2π for some

k ∈ Z. Does this look familiar?

Page 31: Math 42, Discrete Mathematics

Math 42,Discrete

Mathematics

Richard P.Kubelka

San Jose StateUniversity

Relations &Their Properties

EquivalenceRelations

Matrices,Digraphs, &RepresentingRelations

c© R. P. Kubelka

Equivalence Relations & Partitions � Example

Let A = R and de�ne ∼ on A by

a ∼ b if and only if a− b = k · 2π, (13)

where k ∈ Z.

RemarksI The relation ∼ de�ned in (13) is an equivalence

relation. The proof is similar to the proof that

congruence modulo n is an equivalence relation.

I The relation ∼ is the same relation as the one de�ned by

the sine function above.

Page 32: Math 42, Discrete Mathematics

Math 42,Discrete

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Richard P.Kubelka

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c© R. P. Kubelka

Matrices

De�nition

A matrix is a rectangular array of numbers. A matrix with m

rows and n columns is called an m× n matrix�pronounced

�m by n matrix.�

I The plural of �matrix� is �matrices.� There's no such

thing as a �matrice.�

I A matrix with the same number of rows as columns is

called square.

I Two matrices are equal if they have the same number of

rows and the same number of columns and the

corresponding entries in every position are equal.

Page 33: Math 42, Discrete Mathematics

Math 42,Discrete

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Richard P.Kubelka

San Jose StateUniversity

Relations &Their Properties

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c© R. P. Kubelka

Matrices

A generic matrix looks like

A =

a11 a12 · · · a1na21 a22 · · · a2n...

... · · ·...

am1 am2 · · · amn

, (14)

or A = [aij], for short.

We call aij the (i, j)th entry or element of A. The left-hand

subscript gives the row number, while the right-hand

subscript gives the column number.

Thus aij gives the element that occupies the intersection of

the i-th row and the j-th column.

Remark

Matrices are ubiquitous in science, mathematics, and even

business, but we will be interested only in binary matrices,

matrices whose entries are only 0's and 1's.

Page 34: Math 42, Discrete Mathematics

Math 42,Discrete

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c© R. P. Kubelka

Relations and Matrices

Given a relation R on the set A = {a1, . . . ,an}, we build a

binary matrix M = [mij] as follows:

mij =

{1 if aiRaj

0 if ai�Raj.(15)

Page 35: Math 42, Discrete Mathematics

Math 42,Discrete

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Richard P.Kubelka

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Relations and Matrices

Example

Let A = {2, 3, 4, 5, 6} and suppose R is de�ned by aRb if and

only if a ≡ b mod 3. Then

I R ={(2, 2), (2, 5), (3, 3), (3, 6), (4, 4), (5, 2), (5, 5), (6, 3), (6, 6)}.

I The matrix M associated to R is

M =

1 0 0 1 00 1 0 0 10 0 1 0 01 0 0 1 00 1 0 0 1

. (16)

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I Could we detect Re�exivity in R by looking at M? How?

Note that for Re�exivity, we need aRa for all a ∈ A.But since the elements of A are all numbered, that

means we need aiRai for all i = 1, . . . ,n. And that

will be true exactly when mii = 1 for all i = 1, . . . ,n,i.e., when all the diagonal entries of M are 1's.

I Could we detect Symmetry in R by looking at M? How?

For Symmetry, we need ajRai whenever we have aiRaj,

i.e., we need mji = 1 whenever mij = 1. On the other

hand, if mij = 0 and mji = 1, we would have ai�Rajand ajRai, contradicting symmetry. So we conclude

that R will be symmetric if and only if mji = mij for all

i, j = 1, . . . ,n. A square matrix M is called symmetric

if mji = mij for all i, j = 1, . . . ,n. So R will be

symmetric exactly when its corresponding matrix M is

symmetric.

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Boolean Product of Matrices

De�nition

Let S = {0, 1}, we de�ne two operations on S as follows:

a1 ∧ a2 =

{1 if a1 = a2 = 1

0 otherwise,(17)

a1 ∨ a2 =

{0 if a1 = a2 = 0

1 otherwise.(18)

Remark

Note that if we interpret 1 as T and 0 as F, these are

precisely the operations of disjunction (∧) and conjunction

(∨) we studied in Chapter 1.

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De�nition

Given two n× n binary matrices A and B , the Booleanproduct C = A� B is the n× n binary matrix whose

(i, j)th entry cij is given by:

cij = (ai1 ∧ b1j)∨ (ai2 ∧ b2j)∨ · · ·∨ (ain ∧ bnj) (19)

(See p. 192, Rosen, 8th Ed., for this de�nition.)

Remark

We denote A�A by A[2].

De�nition

Given two n× n binary matrices N and M, we say N ⊆ M if

nij 6 mij for all i, j = 1, . . . ,n.

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Binary Matrices and Transitivity

FactsI Suppose R1 and R2 are two relations on

A = {a1, . . . ,an}, and MR1 and MR2 are their binary

matrices, then MR1 ⊆ MR2 if and only if R1 ⊆ R2.I A relation R on A = {a1, . . . ,an} is transitive, if and

only if M[2]R ⊆ MR.

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Representing Relations with Digraphs

De�nition

A directed graph, or digraph, consists of a set V of

vertices (or nodes) together with a set E of ordered pairs of

elements of V called edges (or arcs). The vertex a is called

the initial vertex of the edge (a,b) and the vertex b is

called the terminal vertex of this edge.

Remark

Note any relation R on a set A = {a1, . . . ,an} immediately

gives rise to a digraph: let V = A and let E = R. This works

because R is already a set of ordered pairs of the elements of

V = A.

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Digraphs

Figure 2: Digraph 1

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Representing Relations with Digraphs

Example

The matrix for the relation R represented by the digraph in

Figure 2 is

MR =

0 0 1 11 1 1 01 0 1 01 0 1 0

(20)

I Is R re�exive? No, since not all the diagonal entries of

MR are 1s: m11 = 0 = m22

I Is R symmetric? No, since MR is not a symmetric

matrix.

I is R transitive? Well, we need to compute M[2]R .

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Representing Relations with Digraphs

Example (cont'd)

The matrix M[2]R for the relation represented by the digraph

in Figure 2 is

NR = M[2]R =

1 0 1 01 1 1 11 0 1 11 0 1 1

(21)

Since n11 = 1 and m11 = 0, R cannot be transitive.

Why is n11 = 1? Because, for example,

m13 ∧m31 = 1∧ 1 = 1. This means that a1Ra3 and

a3Ra1. Transitivity would then imply a1Ra1. But m11 = 0,so a1�Ra1, a contradiction.

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Representing Relations with Digraphs

Example (cont'd)

Remark

Note that n14 = 0 while m14 = 1. What's the signi�cance

of this? The fact that mij = 1 encodes the fact that we can

get from ai to aj in one step, while nij = 0 encodes the

impossibility of getting from ai to aj in exactly two steps.

So here we can get from a1 to a4 in one step but not in two

steps.

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Digraphs

Figure 3: Digraph 2

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Representing Relations with Digraphs

Example

The matrix for the relation R represented by the digraph in

Figure 3 is

MR =

1 0 1 10 1 0 01 0 1 01 0 0 1

(22)

I Is R re�exive? Yes, since all the diagonal entries of MR

are 1s.

I Is R symmetric? Yes, since MR is a symmetric matrix.

I is R transitive? Well, we need to compute M[2]R .

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Representing Relations with Digraphs

Example (cont'd)

The matrix M[2]R for the relation represented by the digraph

in Figure 3 is

NR = M[2]R =

1 0 1 10 1 0 01 0 1 11 0 1 1

(23)

Since n34 = 1 and m34 = 0, R cannot be transitive.

Why is n34 = 1? Because, for example,

m31 ∧m14 = 1∧ 1 = 1. This means that a3Ra1 and

a1Ra4. Transitivity would then imply a3Ra4. But m34 = 0,so a3�Ra4, a contradiction.

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Example (cont'd)

Now suppose we consider the relation represented by the

digraph in Figure 3 but with A = {a, c,d,b}, i.e., with the

elements of A written in a di�erent order.

The new matrix MR will be

MR =

1 1 1 01 1 0 01 0 1 00 0 0 1

(24)

Notice that we can �partition� the matrix MR into blocks as

follows:

MR =

1 1 1 01 1 0 01 0 1 0

0 0 0 1

(25)

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Digraphs

Figure 4: Digraph 3

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Example

The matrix for the relation R represented by the digraph in

Figure 4 is

MR =

1 1 1 10 1 0 10 0 1 10 0 0 1

(26)

I Is R re�exive? Yes, since all the diagonal entries of MR

are 1s.

I Is R symmetric? No, since MR is not a symmetric

matrix.

I is R antisymmetric? Well, we haven't looked at that

yet.

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Example (cont'd)

I Recall that R is antisymmetric if and only if aiRaj and

ajRai together imply that ai = aj, i.e., i = j. In terms

of the matrix MR, this means that if i 6= j and mij = 1,then we must have mji = 0. MR satis�es this

condition, so R is antisymmetric.

I The matrix M[2]R for the relation represented by the

digraph in Figure 4 is

NR = M[2]R =

1 1 1 10 1 0 10 0 1 10 0 0 1

=

1 1 1 10 1 0 10 0 1 10 0 0 1

= MR

(27)

But since NR ⊆ MR, R is transitive.