Propositions and Truth Tables

STATEMENTS A statement (or verbal assertion) is any collection of symbols (or sounds) which is either

true or false, but not both. Statements will usually be denoted by the letters

P , (7, r , . . . The truth or falsity of a statement is called its truth value.

Example 1.1 : Consider the following expressions:

(i) Paris is in England. (iii) Where a re you going?

(ii) 2 + 2 = 4 (iv) P u t the homework on the blackboard.

The expressions (i) and 6) a r e statements; the first is false and the second is true. The expressions (iii) and [iv) a re not statements since neither is either t rue or false.

COMPOUND STATEMENTS Some statements are composite, that is, composed of substatements and various logical

connectives which we discuss subsequently. Such composite statements are called com- pound statements.

Example 2.1: "Roses a r e red and violets a re blue" is a con~pound statement with substatements "Roses a r e red" and "Violets a r e blue".

Example 2.2: "He is intelligent or studies every night" is, implicitly, a compound statement with substatements "He is intelligent" and "He studies every night".

The fundamental property of a compound statement is that its truth value is completely determined by the truth values of its substatements together with the way in which they are connected to form the compound statement. We begin with a study of some of these connectives.

CONJUNCTION, p A q

Any two statements can be combined by the word "and" to form a compound statement called the conjunction of the original statements. Symbolically,

P A 4

denotes the conjunction of the statements p and q, read " p and q".

Example 3.1: Let p be "It is raining" and let q be "The sun is shining".

Then p A q denotes the statement "It is raining and the sun is shining".

The truth value of the compound statement p ~ q satisfies the following property:

[TI If p is true and q is true, then p A q is true; otherwise, 11 A q is false.

In other words, the conjunction of two statements is true only in the case when each sub- statement is true.

PROPOSITIONS AND TRUTH TABLES

Observe that both propositions have the same truth table.

illISCELLANEOUS PROBLEMS 1.11. Let Apq denote p ,;q and let Sp denote -p. Rewrite the following propositions

using A and N instead of A and -. (9 P A -q (iii) -p A (-q A r )

( 4 -(-P A q) (iv) - ( p A -q) A (-q A -7,)

(i) p A -q = p A 'Yq = ApNq

(ii) -(-p q) = -(.Vp A q) = -(ASpq) = NANpq

(iii) -p A (-q A 1.) = N p A (Nq A r) = S p A (ANqr) = ANpALVq~

(iv) -(p A -q) A (-q -r) = -(ApXq) A (ANqSr) = (NApNq) A ( A S q S r ) = ANApNqANqXr

Observe tha t there a r e no parentheses in the final answer when A and .V are used instead of A and -. In fact, i t has been proved that parentheses a re never needed in any proposition using A and N.

1.12. Rewrite the following propositions using A and - instead of A and N.

(i) NAP^ (iii) ApNq (v) Iv'AANpql.

(ii) ANpq (iv) ApAqr (vi) A S p A qNr

(i) NApq = S ( p A q) = -(p A q) (iii) ApNq = Ap(-q) = p A -q

(ii) ANpq = A(-p)q = -p A q (iv) ApAqr = Ap(q i\ 7,) = p A (q A r )

(v) NAASpqr = SAA(-p)qr = NA(-p f i q)r = Nr(-p A q) / r' = -:(-p A q) A r]

(vi) ANpAqSr = ANpAq(-r) = ANp(q A -7) = A(-p)(q A -1.) = -P A (9 A my)

Notice tha t the propositions involving A and N are unraveled from right to left.

Supplementary Problems

STATEMENTS

1.13. Let p be "Marc is rich" and let q be "hlarc is happy". Write each of the following in symbolic form.

(i) Marc is poor but happy.

(ii) Marc is neither rich nor happy.

(iii) Marc is either rich or unhappy.

(iv) Marc is poor or else he is both rich and unhappy.

Algebra of Propositions

TAUTOLOGIES AND CONTRADICTIONS

Some propositions P(p, q, . . . ) ca,ntain only T in the last column of their t ruth tables, i.e. a re true for any truth values of their variables. Such propositions are called tautologies. Similarly, a proposition P(p , (1, . . .) is called a cotlt~.adiction if it contains only F in the last column of its t ruth table, i.e. is false for any truth values of its variables.

Example 1.1: The proposition "p or not p", i.e. p v -p, is a tautology and the proposition " p and not p", i.e, p -11, is a contradiction. This is verified by constructing their t ruth tables'

11 1 -11 p v - p

T F T

F I T T

Since a tautology is always true, the negation of a tautology is always false, i.e. is a contradiction, and vice versa. That is,

Theorem 2.1: If P(p,q , . . .) is a tautology then - P ( p , q , . . . ) is a contradiction, and conversely.

Now let P(j1, q , . . .) be a tautology, and let Pl (p , q, . . .), P,(P, q , . . .), . . . be any propo- sitions. Since P(p , q , . . . ) does not depend upon the particular truth values of its variables p, q, . . ., we can substitute Pl for p, P , for q, . . . in the tautology P(p, q, . . .) and still have a tautology. In other words:

Theorem 2.2 (Principle of Substitution) : If P(p, q, . . .) is a tautology, then P(P,, P,, . . . ) is a tautology for any propositions P,, P,, . . . .

LOGICAL EQUIVALENCE

Two propositions P(p, q, . . . ) and Qjp, q, . . .) are said to be logically equivalelzt, or simply eqzlicnlel~t or equal, denoted by

P ( P , (1, . . = Q ( P , q, . . . ) if they have identical truth tables.

Example 2.1: The t ruth tables of -(]I q i and - p v -q follo~v. , ; , ; - p i - q

T F F T T F F T F T F T F T T F T

F F F T F F T T T

Accordingly, the propositions - ( p A q ) and -11 v - q are logically equivalent:

- ( p A q) -11 V -Q

Set Theory

SETS AND ELEMENTS The concept of a set appears in all branches of mathematics. Intuitively, a set is any

well-defined list or collection of objects, and will be denoted by capital letters A, B, X, Y, . . . . The objects comprising the set are called its elements or menzbe~s and will be denoted by lower case letters a, b, x, y, . . . . The statement "p is an element of A" or, equivalently, " p belongs to A" is written

11 E A The negation of p E A is written p 6! A.

There are essentially two ways to specify a particular set. One way, if i t is possible, is to list its members. For example,

A = {a, e , i, o, u)

denotes the set A whose elements are the letters a, e, i, o, u. Note that the elements are separated by commas and enclosed in braces { 1. The second way is to state those proper- ties which characterize the elements in the set. For example,

B = {x : x is an integer, x > 0}

which reads " B is the set of x such that x is an integer and x is greater than zero," denotes the set B whose elements are the positive integers. A letter, usually x, is used to denote a typical member of the set; the colon is read as "such that" and the comma as "and".

Example 1.1: The set B above can also be written a s B = {1,2,3, . . .I. Observe tha t -6 4 B, 3 E B and .T 4 B.

Example 1.2: The set A above can also be written a s

A = { x : a is a letter in the English alphabet, :c is a vowel)

Observe tha t b 6Z A , e € A and p @ A .

Example 1.3: Let E = {x : x2 - 32 + 2 = 0). In other words, E consists of those numbers which a r e solutions of the equation x2- 3x + 2 = 0, sometimes called the solution set of the given equation. Since the solutions of the equation a re 1 and 2, we could also write E = {I, 2).

Two sets A and B are equal, written A = B, if they consist of the same elements, i.e. if each member of A belongs to B and each member of B belongs to A. The negation of A = B is written A # B.

Example 1.4: Let E = {x : xz - 3x + 2 = 0), F = {2,1) and G = {I, 2,2,1,6/3).

Then E = F = G. Observe that a set does not depend on the way in which i ts elements a r e displayed. A set remains the same if its elements a r e repeated or rearranged.

36 S E T THEORY [CHAP. 5

FINITE AND INFINITE SETS

Sets can be finite or infinite. A set is finite if i t consists of exactly n different elements, where 12 is some positive integer; other~vise i t is infinite.

Example 2.1: Let M be the set of the days of the week. In other words,

M = {Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday] Then M is finite.

Example 2.2: Let Y = { 2 , 4, 6, 8 , . . . ) . Then Y is infinite.

Example 2.3: Let P = { T : r is a river on the earth). Although it may be difficult to count the number of rivers on the earth, P is a finite set.

SUBSETS

A set A is a subset of a set B or, equivalently, B is a s z~perse t of A, written

A C E or B > A

iff each element in A also belongs to B; that is, x € A implies x € B. We also say that A is contailzed in B or B contains A. The negation of A c B is written A $! B or B $ A and states that there is an x E A such that x 6! B.

Example 3.1: Consider the sets

A = { 1 , 3 5 , 7 , . . ) B = {5,10,15,20 , . . . 1 C = { . c : x i s prime, r > 2 ) = {3, 5, 7, 11, ...)

Then C c A since every prime number greater than 2 is odd. On the other hand, B @ A since 10 E B but 10 4 A.

Example 3.2: Let N denote the set of positive integers, Z denote the set of integers, Q denote the set of rational numbers and R denote the set of real numbers. Then

N c Z c Q c R

Example 3.3: The set E = {2,4,6] is a subset of the set F = {6,2,4), since each number 2, 4 and 6 belonging to E also belongs to F. In fact, E = F. In a similar manner it can be shown that every set is a subset of itself.

As noted in the preceding example, A C B does not exclude the possibility that A = B. In fact, we may restate the definition of equality of sets as follows:

Two sets A and B are equal if A c B and E c A.

In the case that A c B but A = B, n7e say that A is a proper subset of B or B contains A properly. The reader should be warned that some authors use the symbol C for a subset and the symbol c only for a proper subset.

The following theorem is a consequence of the preceding definitions:

Theorem 5.1: Let A, B and C be sets. Then: (i) A c A; (ii) if A C B and B C A, then A = B; and (iii) if A C B and B c C, then A c C.

UNIVERSAL AND NULL SETS

In any application of the theory of sets, all sets under investigation a re regarded as subsets of a fixed set. We call this set the universal set or universe o f discourse and denote it (in this chapter) by C.

Example 1.1: In plane geometry, the universal set consists of all the points in the plane.

Example 4.2: In human population studies, the universal set consists of all the people in the world.

CHAP. 51 SET THEORY 37

I t is also convenient to introduce the concept of the empty or null set, that is, a set which contains no elements. This set, denoted by e), is considered finite and a subset of every other set. Thus, for any set A, @ c A c U.

Example 4.3: Let A = {x : x' = 4, z is odd). Then A is empty, i.e. A = @.

Example 4.4: Let B be the set of people in the world who a r e older than 200 years. According to known statistics, B is the null set.

CLASS, COLLECTION, FAMILY Frequently, the members of a set are sets themselves. For example, each line in a set

of lines is a set of points. To help clarify these situations, other words, such as "class", "collection" and "family" are used. Usually we use class or collection for a set of sets, and family for a set of classes. The words subclass, subcollection and subfamily have meanings analogous to subset.

Example 5.1: The members of the class {{2,3), {2), { 5 ,6 ) ) are the sets {2,3), (2) and {5,6).

Example 5.2: Consider any set A. The power set of A, denoted by T(A) or Z A , is the class of all subsets of A. In particular, if A = { a , b, c ) , then

T(A) = {A, { a , b), {a, c), (6, c ) , { a ) , Cb), {c), @)

In general, if A is finite and has n elements, then T ( A ) will have 2n elements.

SET OPERATIONS The union of two sets A and B, denoted by A U B, is the set of all elements which

belong to A or to B: A U B = { x : x E A or x E B )

Here "or" is used in the sense of andlor.

The intersection of two sets A and B, denoted by A n B, is the set of elements which belong to both A and B:

A n B = { x : x E A and x E B )

If A f l B = 0, that is, if A and B do not have any elements in common, then A and B are said to be disjoint or non-intersecting.

The relative co?nplernent of a set 5' with respect to a set A or, simply, the difference of A and B, denoted by A \ B , is the set of elements which belong to A but which do not belong to B:

A \ B = { n : : x E A , x4B)

Observe that A \ B and B are disjoint, i.e. (A \ B) n B = $3.

The absolute co~nplement or, simply, complement of a set A, denoted by Ac, is the set of elements which do not belong to A:

Ac = { x : x E U , x 4 A )

That is, Ac is the difference of the universal set U and A.

Example 6.1: The following diagrams, called Venn diagrams, illustrate the above set operations. Here sets a r e represented by simple plane areas and U , the universal set, by the area in the entire rectangle.

SET THEORY [CHAP. 5

A u B is shaded. A n B is shaded.

A \ B is shaded. AC is shaded.

Example 6.2: Let A = {1 ,2 ,3 ,4 ) and B = {3,4,5,6) where 72 = {I, 2,3, . . .). Then:

A U B = {I, 2, 3, 4, 5, 61 A n B = {3,4)

A\B = {1 ,2 ) Ac = (5, 6 , 7, . . .) Sets under the above operations satisfy various laws or identities which are listed in

Table 5.1 below. In fact we state:

Theorem 5.2: Sets satisfy the laws in Table 5.1.

LAWS OF THE ALGEBRA OF SETS

Idempotent Laws l a . A u A = A lb. A n A = A

Associative Laws 2a. ( A u B ) u C = A u ( B u C ) 2b. ( A n B ) n C = A n ( B n C )

Commutative Laws 3a. A u B = B u A 3b. A n B = B n A

Distributive Laws 4a. A u ( B n C ) = ( A u B ) n ( A u C ) 4b. A n ( B u C ) = ( A n B ) u ( A n C )

Identity Laws 5a. A u @ = A 5b. A n U = A

6a. A u U = U 6b. A n 8 = @

Complement Laws 7a. A uAc = U 7b. AnAc = @

8a. ( A c ) ~ = A 8b. U C = 13, gC = U

De Rlorgan's Laws 9a. ( A u B ) ~ = Acn Bc 9b. ( A nB)c = AcuBc

Table 5.1

Remark: Each of the above laws follows from the analogous logical law in Table 2.1, Page 11. For example,

A n B = { x : x E A and z E B ) = { x : x E B and x E A ) = B n A

CHAP. 51 SET THEORY 3 9

Here me use the fact that if 11 is ,x E A and q is x E B, then p A q is logically equivalent to q A p: p A q = Q A p .

Lastly we state the relationship between set inclusion and the above set operations:

Theorem 5.3: Each of the following conditions is equivalent to A c B:

(i) A n B = A (iii) Bc c A" (v) B u A" = Lr

(ii) A U B = B (iv) A n Bc = $3

ARGUMENTS AND VENN DIAGRAMS

Many verbal statements can be translated into equivalent statements about sets which can be described by Venn diagrams. Hence Venn diagrams are very often used to determine the validity of an argument.

Example 7.1 : Consider the following argument:

S1: Babies are illogical.

SP: Nobody is despised who can manage a crocodile.

S3: Illogical people a r e despised.

S : Babies cannot manage crocodiles.

(The above argument is adapted from Lewis Carroll, Sumbolic Logic; he is also the author of Alice in Wo~zcler la~zd . ) Now by S1, the set of babies is a subset of the set of illogical people:

babies

the set of illogical people is contained in the set of despised people:

Furthermore, by S2 , the set of despised people and the set of people who can manage a crocodile a r e disjoint:

people who can manage

crocodiles

But by the above Venn diagram, the set of babies is disjoint f rom the set of people who can manage crocodiles, or "Babies cannot manage crocodiles" is a consequence of S1, Sz and S3. Thus the above argument,

is valid. S,, sz, s3 + s

S E T THEORY

Solved Problems SETS, ELEMENTS 5.1. Let A = ( x : 3 x = 6). Does A = 2 ?

A is the set which consists of the single element 2, that is, A = ( 2 ) . The number 2 belongs to A; it does not equal A. There is a basic difference between a n element p and the singleton set { p ) .

5.2. Which of these sets are equal: { T , t , s ) , { s , t, r , s ) , { t , s, t , r ) , { s , r , s , t ) ? They are all equal. Order and repetition do not change a set.

5.3. M7hich of the following sets are finite?

(i) The months of the year. (iv) {x : x is an even number) (ii) (1, 2, 3 , . . ., 99, 100) (v) {I, 2 , 3 , . . . 1 (iii) The number of people living on the earth.

The first three sets a re finite; the last two sets a r e infinite.

5.4. Determine which of the follo~ving sets are equal: @, { 0 ) , {PI. Each is different from the other. The set ( 0 ) contains one element, the number zero. The

set contains no elements; i t is the empty set. The set (0) also contains one element, the null set.

5.5. Determine whether or not each set is the null set:

(i) X = { x : x2 = 9, 2x = 41, (ii) Y = { x : x # x), (iii) Z = { x : x + 8 = 8) . (i) There is no number which satisfies both x2 = 9 and 2 2 = 4; hence X is empty, i.e. X = 0.

(ii) We assume tha t any object is itself, so Y is also empty. In fact, some texts define the null set as follows:

$3 = { x : X # X )

(iii) The number zero satisfies z + 8 = 8; hence Z = (0). Accordingly, Z is not the empty set since i t contains 0 . That is, Z f @.

SUBSETS 5.6. Prove that A = { 2 , 3 , 4 , 5 ) is not a subset of B = ( x : x is even).

I t is necessary to show that a t least one element in A does not belong to B. Now 3 E A and, since E consists of even numbers, 3 6? B; hence A is not a subset of B.

5.7. Prove Theorem 5.l(iii): If A C B and B c C, then A c C. We must show t h a t each element in A also belongs to C. Let x E A. Now A c B implies

x E B. But B c C; hence x E C. We have shown t h a t x E A implies x E C, t h a t is, that A c C.

5.8. Find the power set T(S) of the set S = { 1 , 2 , 3 ) .

The power set T ( S ) of S is the class of all subsets of S ; these a re ( 1 , 2 , 3 ) , ( 1 , 2 ) , ( 1 , 3 ) , { 2 , 3 ) , {I}, ( 2 1 , ( 3 ) and the empty set 0. Hence

T ( S ) = { S , 1 1 , 3 ) , { 2 , 3 ) , { 1 , 2 } , { 2 ) , {3),

Note tha t there a re Z 3 = 8 subsets of S.

CHAP. 51 SET THEORY 4 1

5.9. Let V = ( d ) , W = { c ,d ) , X = (a, b , c) , Y = {a , b ) and Z = (a, b , d ) . Determine whether each statement is true or false:

(i) Y c X, (ii) TV - Z, (iii) Z > 17, [iv) V c X, (v) X = W, (vi) W c Y .

(i) Since each element in Y is a member of X, Y c X is true.

(ii) Now a E Z but (1 4 W ; hence W # Z is true.

(iii) The only element in V is d and it also belongs to 2; hence Z > V is true.

(iv) V is not a subset of X since d E V but d 4 X; hence V C X is false.

(v) Now u E X but a 4 W; hence X = W is false.

(vi) W is not a subset of Y since c E 68 but c 4 Y ; hence W c Y is false.

6.10. Prove: If A is a subset of the empty set e), then A = @. The null set @ is a subset of every set; in particular, IZ) c A . But, by hypothesis, A c $3;

hence A = 0.

SET OPERATIONS 5.11. Let U = ( 1 , 2 , . . . , 8,9), A = { 1 , 2 , 3 , 4 ) , B = [2,4,6,8) and C = { 3 , 4 , 5 , 6 ) . Find:

(i) A", (ii) A n C, (iii) (A n C)[, (iv) A U B, (v) B\ C.

(i) Ac consists of the elements in C that are not in A; hence A C = {5,6, 7,8,9).

(ii) A n C consists of the elements in both A and C; hence A n C = {3 ,4} .

(iii) ( A n C)c consists of the elements in C that a re not in A n C . Now by iii), A n C = {3 ,4 ) and so ( A n C ) c = {1,2,5,6,7,8,9) .

(iv) A u B consists of the elements in A or B (or both): hence A L B = {1,2,3,4,6,8) .

(v) B \ C consists of the elements in B which a r e not in C; hence B \ C = {2,8).

5.12. In each Venn diagram below, shade: [i) A U B, (ii) A n B.

(i) A L E consists of those elements which belong to A or B (or both); hence shade the area in A and in B as follows:

A u B is shaded.

(ii) A n B consists of the area tha t is common to both A and B. To compute A n B, first shade A with strokes slanting upward to the right ( / / / I ) and then shade B with strokes slanting downward to the r ight (\\\\), as follows:

Then A n E consists of the cross-hatched area which is shaded below:

SET THEORY

A n B is shaded.

Observe the following:

( a ) A n E is empty if A and E are disjoint.

( b ) A r E = B if B c A .

i c ) A ~ I B = A if A C E .

5.13. In the Venn diagram below, shade: (i) Bc, (ii) ( A L'B)<, (iii) ( B \A)" ( iv) A' n Bc.

( i i BL consists of the elements which do not belong to B: hence shade the area outside B a s follows:

Ec is shaded.

(ii) Firs t shade A u B; then iA 1- BjC is the area outside A IJ E:

AUB i s shaded. (A u E)C is shaded.

(iii, Firs t shade B\A, the area in E which does not lie in A ; then (E\A)c is the area outside E\A.

B \A is shaded. (E \ A)C is shaded.

(ivr Firs t shade AC, the area outside of A, with strokes slanting upward to the right i / l l / ) , and then shade BC with strokes slanting downward to the right ( \ \ \ \ I ; then ACOIE' is the cross-hatched area:

S E T THEORY

AQnd Bc a r e shaded. Acn Bc is shaded.

Observe that ( A L B)c = Acn Bc, a s expected by De Morgan's law.

5.14. In the Venn diagram below, shade (i) A n ( B U C), (ii) ( A n B) U ( A n C).

(i) Firs t shade A with upward slanted strokes, and then shade BUC with downward slanted strokes; now A n (BUC) is the cross-hatched area:

A and B u C a r e shaded. A n ( B u C) is shaded.

(ii) Firs t shade A n B with upward slanted strokes, and then shade A n C with downward slanted strokes; now ( A n 6 ) u ( A n C) is the total area shaded:

A n B and A n C a r e shaded. ( A n B ) u ( A n C) is shaded.

Notice tha t A n ( B u C) = ( A n B) u ( A n C), as expected by the distributive law.

5.15. Prove: B\A = B n A". Thus the set operation of difference can be written in terms of the operations of intersection and complementation.

B\A = { x : x E B , . , ;@A) = {x: x E B , x E A C } = BnAc

44 SET THEORY

5.16. Prove the Distributive Law: A n (B U C) = ( A n B) U (A n C).

A n ( B u C ) ==- {x : x € A ; x E B u C }

= { x : x E A ; x E B or x E C }

1- { x : x E A , x E B ; or x E A , x E C }

= { r e : x E A n B or a E A n C }

= ( A n B ) u ( A n C )

Observe tha t in the third step above we used the analogous logical law

P (9 v r ) = (P A q) v ( P A 4

5.17. Prove: (A \ B) f l B = @.

( A \ B ) n B = { x : x E A \ B , x E B }

= { x : x E A , x 4 B ; z E B }

= F1

The last step follows from the fact tha t there is no element x satisfying z E B and z @ B .

5.18. Prove De Morgan's Law: (A U B)" Ac n Bc.

( A u B ) ~ = { x : x ~ A u B }

= { x : x 4 A , x 4 B )

= {x : x E AC, x E BC}

= ACnBC

Observe that in the second step above we used the analogous logical law

- ( p V q) -p A -q

5.19. Prove: For any sets A and B, A n B c A c A U B.

Let z E A n B; then x E A and x E B. In particular, x E A. Since x E A n B implies x € A , A n B c A. Furthermore, if z E A, then x E A or x E B, i.e. x E A u B. Hence A c A u B. In other words, A n B c A c A u B .

5.20. Prove Theorem 5.3(i): A c B if and only if A n B = A . Suppose A c B. Let x E A ; then by hypothesis, x E B. Hence x € A and x E B, i.e. x E A n B.

Accordingly, A c A n B . On the other hand, i t is always t rue (Problem 5.19) tha t A n B c A. Thus A n B = A .

Now suppose t h a t A n B = A. Then in particular, A C A n B. But i t is always t rue tha t A n B c B. Thus A c A n B c B and so, by Theorem 5.1, A c B.

ARGUMENTS AND VENN DIAGRAMS 5.21. Show that the following argument is not valid by constructing a Venn diagram in

which the premises hold but the conclusion does not hold:

Some students are lazy.

All males are lazy. . . . . . . . . . . . . . . . . . . . . . Some students are males.

Consider the following Venn diagram:

CHAP. 51 SET THEORY

lazy people

students males

Notice tha t both premises hold, but the conclusion does not hold. For an argument to be valid, the conclusion nlust always be t rue whenever the premises a re true. Since the diagram represents a case in which the conclusion is false, even though the premises are true, the argument is false.

I t is possible to construct a Venn diagram in which the preinises and conclusion hold, such as

lazy people

students males

5.22. Show that the following argument is not valid:

All students are lazy.

Nobody svho is wealthy is a student. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Lazy people are not wealthy.

Consider the following Venn diagram:

wealthy people

Now the premises hold in the above diagram, but the conclusion does not hold; hence the argument is not valid.

5.23. For the following set of premises, find a conclusion such that the argument is valid:

S,: All lawyers are wealthy.

S,: Poets are temperamental.

S,: No temperamental person is wealthy. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

By S, , the set of lawyers is a subset of the set of wealthy people: ancl by S,, the set of wealthy people and the set of temperamental people are disjoint. Thus

temperamental people

46 S E T THEORY

By S Y , the set of poets is a subset of the set of temperamental people; hence

Thus the statement "No poet is a lawyer" or equivalently "No lawyer is a poet" is a valid conclusion.

The statements "No poet is wealthy" and "No lawyer is temperamental" are also valid conclusions which do not make use of all the premises.

MISCELLANEOUS PROBLERlS

5.24. Let A = (2, (4 ,5) , 4). Which statements are incorrect and why? ( i 4 5 ) C A, (ii) {4 ,5) E A, (iii) {{4,5}} C A.

The elements of A a re 2, 4 and the set {4 ,5) . Therefore (iii is correct, but (i) is an incorrect statement. Furthermore, (iii) is also a correct statement since the set consisting of the single element {4 ,5$ which beiongs to A is a subset of A.

5.25. Let A = (2, {4,5) , 4) . lTThich statements are incorrect and why?

(il 5 E A, (ii) (5) E A, (iii) { 5 ) c A.

Each statement is incorrect. The elements of A are 2, 1 and the set {4,5}; hence (ii and (ii) are incorrect. There a re eight subsets of A and (5) is not one of them; so (iii) is also incorrect.

5.26. Find the power set T(S) of the set S = (3, {1 ,4 ) ) . Note first tha t S contains two elements, 3 and the set {1 ,4 ) . Therefore T(S) contains

22 = 3 elen~ents: S itself, the empty set (3, and the two singleton sets which contain the elements 3 and { I , 4) respectively, i.e. {3) and {{I, 4)). In other words,

TiS) = {S, {3), {{I, 4)), 0)

Supplementary Problems

SETS, SUBSETS

5.27. Let A = { 1 , 2 ,..., 8,9), B = { 2 , 4 , 6 , 8 ) , C = { 1 , 3 , 5 , 7 , 9 ] , D = {3 ,4 ,5 ) and E = {3 ,5 ) . Which sets can equal X if we are given the following information? (i) X and E are disjoint. (ii) X c D but X # B. (iii) X c A but X' C. (iv) X C C but X # A .

5.28. State whether each statement is t rue or false: (ii Every subset of a finite set is finite, rii) Every subset of an infinite set is infinite.

5.29. Find the power set T(A) of A = ( 1 , 2 , 3 , 4 ) and the power set T(B) of B = {I, {2 ,3 ) , 4 ) .

CHAP. 51 S E T THEORY

5.30. State whether each set is finite or infinite:

(i) The set of lines parallel to the r axis.

(ii) The set of letters in the English alphabet.

(iii) The set of numbers which a re n~ultiples of 5.

(iv) The set of animals living on the earth.

(v) The set of numbers which are solutions of the equation x2' + 2 6 ~ 1 ~ - 17x11 + 7x3 - 10 = 0.

(vi) The set of circles through the origin (0,O).

5.31. State whether each statement is t rue or false:

(i) {1,4,3: = :3,4,1: (iv) C4) E {C4))

(ii) {1,3,1,2,3,2) c {1,2,3} (v) (4) C CC4))

(iii) {1,2} = {2 ,1 ,1 ,2 ,1 ] (vi) 8 c CC43)

SET OPERATIONS

5.32. Let U = {u ,b ,c ,d ,e , f ,g ) , A = {a ,b ,c ,d ,e ) , B = {a,c ,e ,g) and C = {b,e ,f ,g) . Find:

( i ~ A u C (iii) C \ B (v) C C n A ivii) (A \ BC)C

iiir E n A (iv) B C ~ C (vi) (A \ iviii) (A nAc)c

5.33. In the Venn diagrams below, shade (i) W \ V (ii) VCU W ( i i i ~ V n WC (iv) VC\ WC.

( a ) ( b )

5.34. Draw a Venn diagram of three non-empty sets A, B and C so tha t A, B and C have the following properties:

( i ) A c B , CCB, A n C = (29 (iii) A c C , A # C , B n C = @

fii) A c B , C$B, A n C # O (iv) A c ( B n C ) , B c C , C # B , A Z C

5.35. The formula A \ B = A nBc defines the difference operation in terms of the operations of intersection and complement. Find a formula that defines the union of two sets, A u B , in terms of the operations of intersection and complement.

5.36. Prove Theorem 5.3iiii: A C B if and only if A UB = B.

5.37. Prove: If A ? E = (3, then A CBC.

5.38. Prove: Ac \ BC = b' \ A .

5.39. Prove: A c B implies A u (B \ A ) = B.

5.40. (i) Prove: A n (B \ C) = (A n B ) \ ( A ?C).

(ii) Give an example to show tha t A U (B \ C) f (A U B) \ (A C).

-ARGUMENTS AND V E N S DIAGRAMS

5.41. Determine the validity of each argument for each proposed conclusion.

No college professor is wealthy.

Some poets a re wealthy.

(i) Some poets a re college professors.

(ii) Some poets a r e not college professors.

48 SET THEORY

5.42. Deter~nine the validity of each argu111ent fo r each proposed conclusion.

All poets a r e interesting people.

Audrey is an interesting person.

I i I Audrey is a poet.

I ~ i i Audrey is not a poet.

Determine the validity of the a rgu i i i e~~t for each proposed conclusion.

All poets a re poor.

In order to be a teacher, one nus st graduate from college.

No college graduate is poor.

(i) Teachers a r e not poor.

(ii) Poets are not teachers.

(iii) If Marc is a college graduate, then he is not a poet.

Determine the validity of the argument for each proposed conclusion.

A11 mathematicians are interesting people.

Some teachers sell insurance.

Only uninteresting people become insurance salesmen.

(i) Insurance salesnlen a re not mathematicians.

(ii) Some interesting people a r e not teachers.

(iii) Some teachers are not interesting people.

(iv) Some riiathen~aticians a r e teachers.

(v) Some teachers a re not mathematicians.

(vi) If Eric is a mathematician, then he does not sell insurance.

Answers to Supplementary Problems (i) C and E, (ii) D and E, [iii) A, E and D, (iv) None

(i) T, (ii) F

(i) infinite, (ii) finite, (iii) infinite, (iv) finite, (v) finite, (v i ) infinite

(i) T, (ii) T, (iii) T, (iv) T, (v) F, (vi) T

(i) A u C = U (v) CC n A = {a, c, d ) = CL

(ii) B n A = {a, c, e ) (vi) (A \ C)C = { b , e , f, g }

(iii) C \ B = { b , f) (vii) (A \ Bc)c = { b , cl, t', g )

(iv) B C u C = { b , d , e , f , g } (viii) (A nAc)c = U

CHAP. 51 SET THEORY 4 9

W \ V Vcu W Vn Wc Vc \ W c

Observe that V c U W = U and V n T.Vc = O in case ( b ) where V c ti'.

5.34. (i)

@

5.40. (ii)

A L ( E \ C) is shaded. ( A u B) \ (A u C) is shaded.

5.41. (i) fallacy, (ii) valid

5.42. (i) fallacy, (ii) fallacy

5.43. (i) valid, (ii) valid, iiii) valid

5.44. (i) valid, (ii) fallacy, riii) valid, (iv) fallacy, I V ~ valid, (vi) valid

The following Venn diagrams show why (iij and (iv) a r e fallacies:

interesting

insurance

mathematicians 0 teachers u

Product Sets

ORDERED PAIRS An o r d e ~ e d pair consists of two elements, say a and b , in which one of them, say a,

is designated as the first element and the other as the second element. Such an ordered pair is written

(a, b)

Two ordered pairs (a, b ) and (c, d ) are equal if and only if a = c and b = d.

Example 1.1: The ordered pairs ( 2 , 3 ) and (3 ,2 ) are different.

Example 1.2: The points in the Cartesian plane in Fig. 6-1 below represent ordered pairs of real numbers.

Example 1.3: The set {2 ,3} is not an ordered pair since the elements 2 and 3 are not dis- tinguished.

Example 1.4: Ordered pairs can have the same first and second elements, such as (1, I), (4,4) and (5,5).

Remark: An ordered pair (a, b ) can be defined rigorously as follows:

(a, b ) = {{a), (a, b ) )

the key here being that {a} c {a, b J which we use to distinguish a as the first element.

From this definition, the fundamental property of ordered pairs can be proven:

(a, b) = (c, d ) if and only if a = c and b = d

PRODUCT SETS Let A and B be two sets. The p ~ o d u c t set (or Car tes ian product ) of A and B, written

A x B, consists of all ordered pairs (a, b ) where a E A and b E B:

A x B = {(a, b ) : a E A, b E B } The product of a set with itself, say A X A, is sometimes denoted by A2.

Example 2.1: The reader is familiar with the Cartesian plane R2 = R X R (Fig. 6-1 below). Here each point P represents a n ordered pair (u, b) of real numbers and vice versa; the vertical line through P meets the x axis a t a , and the horizontal line through P meets the 7~ axis a t b.

Fig. 6-1 Fig. 6-2

CHAP. 61 PRODUCT SETS

Example 2.2: Let A = ( 1 , 2 , 3 ) and B = {a, h ) . Then

A E = C(1, a ) , (1, b) , (2, a ) , (2, b ) , (3 , a) , (3, b ) )

Since A and B do not contain many elements, i t is possible to represent A ,.( B by a coordinate diagram as shown in Fig. 6-2 above. Here the vertical lines through the points of A and the horizontal lines through the points of B meet in 6 points which represent A X B in the obvious way. The point P is the ordered pair (2, b).

In general, if a finite set A has s elements and a finite set B has t elements, then A x B has s times t elements. If either A or B is empty, then A X B is empty. Lastly, if either A or B is infinite, and the other is not empty, then A x B is also infinite.

PRODUCT SETS I N GENERAL

The concept of a product set can be extended to more than two sets in a natural way. The Cartesian product of sets A, B, C, denoted by A x B x C, consists of all ordered triplets (a, b, c ) where u E A, 2, E B and c E C:

Analogously, the Cartesian product of n sets A,, A,, . . . ,An, denoted by A, X A, X . X An, consists of all ordered n-tuples (a,, a,, . . . ,a , , ) where a, E A,, . . . , an E An:

x A , x . . x An = {(a1, . . . , an ) : a, E A,, . . ., an € 4 1 Here an ordered 71-tuple has the obvious intuitive meaning, that is, it consists of n elements, not necessarily distinct, in which one of then1 is designated as the first element, another as the second element, etc. Furthermore,

( a , . . a ) = ( b , . . . , b ) iff a, = b,, . . . , a,, = b,,

Example 3.1: In three dimensional Euclidean geometry each point represents a n ordered triplet: its x-component, its y-component and its z-component.

Example 3.2: Let A = { a , b} , B = { 1 , 2 , 3 ) and C = { x , y). Then:

A A E A C = { ( a , I , % ) , (a , 1, y), ( a , 2 , ~ ) ~ (a, 2, Y),

( a , 3 , x ) , ( a , 3 , ~ ) , ( b , l , s ) , (b , l , y ) ,

(b , 2, x), (b , 2, .Y), (b , 3 , 2 ) , (b , 3 , ~ ) )

TRUTH SETS OF PROPOSITIONS

Recall that any proposition P containing, say, three variables p, (r and r , assigns a truth value to each of the eight cases below:

T

Let U denote the set consisting of the eight 3-tuples appearing in the table above:

U = {TTT, TTF, TFT, TFF, FTT, FTF, FFT, F F F )

(For notational convenience we have written, say, TTT for (T, T, T).)

52 PRODUCT SETS [CHAP. 6

The truth set of a proposition P, written T(P), consists of those elements of U for which the proposition P is true.

Esample 1.1: Following is the t ruth table of ( p + q ) A ( q + T ) :

?# F F F T T F

T F F F

Accordingly, the t ruth set of (p -+ q) A ( q + T ) is

T ( ( p + q) A ( q + r ) ) = {TTT, FTT, F F T , F F F }

The next theorem shows the intimate relationship between the set operations and the logical connectives.

Theorem 6.1: Let P and Q be propositions. Then:

(i) T ( P + Q) = T(P) n T(Q)

(ii) T ( P \) Q) = T(P) U T ( Q )

(iii) T(-P) = (T(P)).

(iv) P logically implies Q if and only if T(P) c T(Q)

The proof of this theorem follows directly from the definitions of the logical connectives and the set operations.

Solved Problems

ORDERED PAIRS

6.1. Let W = {John, Jim, Tom) and let V = {Betty, Mary}. Find W X V.

TV x V consists of all ordered pairs (a, b) where a € IT' and b EV. Hence,

W x V = {(John, Betty), (John, Mary), (Jim, Betty), (Jim, ilIaryi, (Tom, Betty), (Tom, Mary))

6.2. Suppose (x + y, 1) = (3, T - y). Find r and y.

If ( x - U, 1) = (3, x - y) then, by the fundamental property of ordered pairs,

. r + y = 3 and 1 = z - y

The solution of these simultaneous equations is giverrby .z: = 2, y = 1.

Relations

RELATIONS

A b i n u i 8 ~ i7elotion or, simply, re lu f ion R from a set A to a set B assigns to each pair (a, b) in A r B exactly one of the following statements:

(i) "a is related to b", written a R b (ii) "a is not related to b", written a 4 b

A relation from a set A to the same set A is called a relation in A.

Example 1.1: Set inclusion is a relation in any class of sets. For, given any pair of sets A and B, either A c B or A 2 E.

Example 1.2: Marriage is a relation from the set M of men to the set W of women. For , given any man m E -ll and any woman w E TV, either 771 is married to ~ t . or )?z is not married to tc.

Example 1.3: Order, symbolized by "<", or, equivalently, the sentence "x is less than y", is a relation in any set of real numbers. For, given any ordered pair ( a , b ) of real numbers, either

a < b or a Q b

Example 1.4: Perpendicularity is a relation in the set of lines in the plane. For, given any pair of lines a and b, either a is perpendicular to b or a is not perpendicular to b.

RELATIOKS AS SETS OF ORDERED PAIRS

Now any relation R from a set A to a set B uniquely defines a subset R" of A x B as follows:

R:: = { (a , b ) : a is related to b ) = {(a, b ) : a R b )

On the other hand, any subset R,' of A x B defines a relation R from A to B as follows:

a 1: 2, iff (a ,b) € Rb'

In view of the correspondence between relations R from A to B and subsets of A X B, we redefine a relation by

A relation R from A to B is a subset of A x B.

Example 2.1: Let R be the following relation from A = {I, 2, 3) to B = {a , b):

R = ( ( 1 , a), (1, b ) , ( 3 , ~ ) ) a

Then 1 R a , 2 @ b , 3 R a, and 3 $ b. The relation R is dis- played on the coordinate diagram of A X B on the right.

b @ 1 2 3

Example 2.2: Let R be the following relation in W = {a , b , c]

R = ( ( a , b), (a , c) , (c, 4 , ( c , b ) )

Then a $ a , u R b , c R c and c $ a .

CHAP. 71 RELATIONS 5 9

Example 2.3: Let A be any set. The i d e x t i t y relation in A, denoted by 1 or P A , is the set of all pairs in A X A with equal coordinates:

A, = { ( a , a ) : a E A ) The identity relation is also called the diagonal by virtue of i ts position in a coordi- nate diagram of A x A.

INVERSE RELATION Let R be a relation from A to B. The inverse of R, denoted by R-l, is the relation

from B to A which consists of those ordered pairs which when reversed belong to R:

R-I = { ( b , a ) : (a, b) E R )

Example 3.1 : Consider the relation R = { (1 ,2 ) , (1 ,3 ) , ( 2 , 3 ) )

in A = {1,2 ,3} . Then R-' = ( (2 , I ) , (3 ,1) , ( 3 , 2 ) )

Observe tha t R and R-I are identical, respectively, to the relations < and > in A, i.e.,

( a , b ) E R iff a < b and ( d , b ) E R P 1 iff a > b

Example 3.2: The inverses of the relations defined by

"X is the husband of y" and "x is taller than y"

a r e respectively

"x is the wife of y" and ''3: is shorter than y"

EQUIVALENCE RELATIONS A relation R in a set A is called ~eflexive if aRa, i.e. (a , a) E R, for every a E A.

Example 4.1: (i) Let R be the relation of similarity in the set of triangles in the plane. Then R is reflexive since every triangle is similar to itself.

(ii) Let R be the relation < in any set of real numbers, i.e. (a , b ) E R iff a < b. Then R is not reflexive since a 4 a f o r any real number a.

A relation R in a set A is called symnzetric if whenever a R b then b R a, i.e. if (a, b ) E R implies ( b , a ) E R.

Example 4.2: ii) The relation R of similarity of triangles is symmetric. For if triangle ct is similar to triangle 13, then ,8 is similar to a.

(ii) The relation R = {(I, 3) , (2 ,3 ) , (2 ,2) , ( 3 , l ) ) in A = {I, 2 ,3 ) is not symmetric since (2 ,3 ) E R but ( 3 , 2 ) 4 R.

A relation R in a set A is called transitive if whenever a R b and b Rc then aRc, i.e. if (a, b) E R and ( b , c ) E R implies (a , c) E R.

Example 4.3: (i) The relation R of similarity of triangles is transitive since if triangle ct is similar to /3 and /3 is similar to y, then ol is similar to y.

(ii) The relation R of perpendicularity of lines in the plane is not transitive. Since if line a is perpendicular to line b and line b is perpendicular to line c , then a is parallel and not perpendicular to c.

A relation R is an equivalence relafio?~ if R is (i) reflexive, (ii) symmetric, and (iii) transitive.

Example 4.4: By the three preceding examples, the relation R of similarity of triangles is a n equivalence relation since it is reflexive, symmetric and transitive.

60 RELATIONS [CHAP. 7

PARTITIONS

A parti t ion of a set X is a subdivision of X into subsets which are disjoint and whose union is X, i.e. such that each a € X belongs to one and only one of the subsets. The subsets in a partition are called cells.

Thus the collection {A,,A,, . . . ,A, , , ) of subsets of X is a partition of X iff:

(i) X = A , U A , U - . . U A , , , ; (ii) for any Ai,Aj, either A i = A j or A i n A j = @

Example 5.1 : Consider the following classes of subsets of X = {I, 2, . . . ,8,9):

(i) [{I, 3,51, {2,6), {4,8,931

!ii) [{I, 3,51, {2,4,6,8), {5,7,91]

(iii) [{I, 3,51, {2,4,6,8}, (7,911

Then (i) is not a partition of X since 7 E X but 7 does not belong to any of the cells. Furthermore, (ii) is not a partition of X since 5 E X and 5 belongs to both {I, 3,5) and {5,7,9). On the other hand, (iii) is a partition of X since each element of X belongs to exactly one cell.

EQUIVALENCE RELATIONS AND PARTITIONS

Let R be an equivalence relation in a set A and, for each a E A , let [a], called the equivale?zce class of A, be the set of elements to which a is related:

[a] = {X : (a, x) E R )

The collection of equivalence classes of A, denoted by A l R, is called the quotient of A by R:

A I R = {[a]: a E A j

The fundamental property of a quotient set is contained in the following theorem.

Theorem 7.1: Let R be an equivalence relation in a set A. Then the quotient set A I R is a partition of A. Specifically,

(i) aE[a] , for every a E A ;

(ii) [a] = [b] if and only if (a, b) E R;

(iii) if [a] # [b], then [a] and [b] are disjoint.

Example 6.1: Let R, be the relation in Z, the set of integers, defined by x = y (mod 5)

which reads "x is congruent to y modulo 5" and which means tha t the difference x - y is divisible by 5 . Then R5 is a n equivalence relation in Z. There a r e exactly five distinct equivalence classes in Z I R5 :

Observe tha t each integer x which is uniquely expressible in the form x = 5q + r where 0 5 r < 5 is a member of the equivalence A, where r is the remainder. Note t h a t the equivalence classes a r e pairwise disjoint and tha t

Z = A, u A, u A, u A , u A,

CHAP. 71 RELATIONS

Solved Problems RELATIONS 1 Let R be the relation < from A = {1,2,3,4) to B = {1,3,5), i.e. defined by

"x is less than y".

(i) Write R as a set of ordered pairs.

(ii) Plot R on a coordinate diagram of '4 x B. (iii) Find the inverse relation R-'.

(i) R consists of those ordered pairs (a, b) E A A E for which a < b; hence

R = {(1,3) , 11,5), 12,3), (2,5), (3,5), (4,51j

(ii) R is sketched on the coordinate diagram of A X E a s shown in the figure.

(iii) The inverse of R consists of the same pairs a s a re in R hut in the reverse order: hence

R-I = {13,lI, ( .? , I ) , (3 ,2) , (5,2), (5 ,3 ) , 15,8)]

Observe tha t R-I is the relation >, i.e. defined by ''7 is greater than y".

7.2. Let R be the relation from E = :2,3,4,5) to F = (3,6,7,10) defined by "x: divides y".

(i) Write IZ as a set of ordered pairs. ( i i ) Plot R on a coordinate diagram of E x F. (iii) Find the inverse relation R-'.

(i) Choose from the sixteen ordered pairs in E X F those in which the first element divides the second; then 10

R = {(2,61, (2,101, (3,3), (3,6), (5,101) 7

(ii) R is sketched on the coordinate diagram of E X F as 6

shown in the figure. 3

(iii) To find the inverse of R, write the elements of R but in reverse order. 2 3 4 5

R-I = :16,2), l10,2), (3,3), (6,3), (10, 5 ) )

Let 144 = {a, b, c, d ) and let R be the relation in ;12 consisting of those points which are displayed on the d

coordinate diagram of M x M on the right. C

(i) Find all the elements in '$1 which are related to Z?, that is, {x : ( r , b) E R J . b

(ii) Find all those elements in M to which d is re- a lated, that is, {x : (d, x) E R ) .

(iii) Find the inverse relation R-'. a b c d

ii) The horizontal line through b contains all points of R in which b appears as the second element: (a, b), ( b , b ) and (cl, b). Hence the desired set is {a, b,d).

iii) The vertical line through (1 contains all the points of R in which d appears as the first element: ( d , a) and ( t l , h ) . Hence { u , b\ is the desired set.

(iii) Firs t \vrite R as a set of ordered pairs, and then write the pairs in reverse order:

R = ',(a, b), (b, a ) , (6, b), (b, 4 , (c, c ) , (d , (0, id, b)l

R-' = {(b, a), (a, b), (b, b), (d, b) , (c, c), (a, d), (b, cl):

62 RELATIONS [CHAP. 7

EQUIVALENCE RELATIONS

7.4. Let R be the relation 6 in N = {1,2,3, . . .), i.e. (a, b) E R iff a ' b. Determine whether R is (i) reflexive, (ii) symmetric, (iii) transitive, (iv) an equivalence relation.

(i) R is reflexive since a'a for every a € N.

(ii) R is not symmetric since, for example, 3 6 5 but 5 3, i.e. (3,5) E R but (5,3) 4 R.

(iii) R is transitive since a 6 b and b 6 c implies a 6 c.

(iv) R is not an equivalence relation since i t is not symmetric.

7.5. Let R be the relation ( 1 (parallel) in the set of lines in the plane. Determine whether R is (i) reflexive, (ii) symmetric, (iii) transitive, (iv) an equivalence relation. (Assume that every line is parallel to itself.)

(i) R is reflexive since, by assumption, a 1 1 a f o r every line a.

(ii) R is symmetric since if a ( 1 p then p 1 1 a, i.e. if the line a is parallel to the line /3 then /3 is parallel to a.

iiii) R is transitive since if n 1 j p and /3 J J y then oc I) y.

(iv) R is an equivalence relation since it is reflexive, symmetric and transitive.

7.6. Let W = {1,2,3,4). Consider the following relations in W:

Rl = {(I, Z), (4,317 (2, Z), (2, 1)) (3 , l ) )

R, = {(2,2), (2,3), (3,211

R, = {(1,3))

Determine whether each relation is (i) symmetric, (ii) transitive.

(i) Now a relation R is not symmetric if there exists an ordered pair (a, b ) E R such that (b, a ) B R. Hence:

R, is not symmetric since (4,3) E R1 but (3,4) 4 R1,

R3 is not symmetric since (1,3) E R3 but (3, l ) 4 R3.

On the other hand, R, is symmetric.

(ii) A relation R is not transitive if there exist elements a, b and c, not necessarily distinct, such tha t

(a, b) E R and ( b , c ) E R but (a, c) 4 R

Hence Rl is not transitive since

(4,3) E R1 and (3, l ) E R1 but ( 4 , l ) 4 R1

Furthermore, R, is not transitive since

(3,2) E R, and (2,3) E R2 but (3,3) 4 R2

On the other hand, R3 is transitive.

7.7. Let R be a relation in A. Show that:

(i) R is reflexive iff A c R; (ii) R is symmetric iff R = R-l.

(i) Recall tha t A = {(a, a ) : a E A ) . Thus R is reflexive iff (a, a) E R fo r every a E A iff A c R.

(ii) Suppose R is symmetric. Let (a, b) E R, then ( b , a) E R by symmetry. Hence (a , b) E R-1, and so R c R-1. On the other hand, let (a, b) E R-1; then ( b , a) E R and, by symmetry, (a, b) E R. Thus R-1 c R, and so R = R-1.

Now suppose R = R-1. Let (a, b) E R; then (b, a) E R-1 = R. Accordingly R is synln~etric.

CHAP. 71 RELATIONS 6 3

7.8. Consider the relation R = ((1, I), (2 ,3) , ( 3 , 2 ) ) in X = { 1 , 2 , 3 ) . Determine whether or not El is ( i) reflexive, (ii) symmetric, (iii) transitive. (i) R is not reflexive since 2 E X but 12,2) 4 R. (ii) R is synlmetric since R-1 = ((1, I), (3,2), (2,3)) = R. (iii) R is not transitive since (3 ,2 ) E R and 12,3) E R but (3,3) 4 R.

7.9. Let N = [1 ,2 ,3 , . . . ), and let R be the relation = in N X N defined by

(a, b ) ( c , d ) iff a d = bc

Prove that R is an equivalence relation. For every (a, b ) E N X N, (a, b ) - (a, b ) since ab = ba; hence R is reflexive.

Suppose (a , b ) = ( c , d). Then ad = bc, which implies that c b = da. Hence (c, d) (a, b ) and so R is symmetric.

Now suppose ia ,b) - (c,d) and (c,d) = (e,f). Then ad = bc and cf = de. Thus

iatl)(cf) = (bc)(de)

and, by cancelling from both sides, at' = be. Accordingly, ( a , b ) - (e, f ) and so R is transitive.

Since R is reflexive, symmetric and transitive, R is an equivalence relation.

Observe that if the ordered pair (a, b ) is written a s a fraction a , then the above relation R is, h a - c .

in fact, the usual definition of equality between fractions, i.e. 7 - - lff ad = bc. d

7.10. Prove Theorem 7.1: Let R be an equivalence relation in a set A. Then the quotient set A IT: is a partition of A. Specifically, (i) a E [ a ] , for every a E A ; (ii) [a] = rbl if and only if ( a , b ) E R; (iii) if [ a ] + Ibl, then la] and [bl are disjoint.

Proot o f (1). Since R is reflexwe, (a, a ) E R for every a E A and therefore a € [a].

Proof ot n). Suppose (a, b) E R. We want to show tha t [a] = [ b ] . Let x E [b] ; then ( b , x) E R. But by hypothes~s (a, b) E R and so, by transitivity, (a, x) E R. Accordingly x € [a]. Thus [bl c [a]. To prove that a c L b ] , we observe that (a, b ) E R implies, by symmetry, tha t (b, a ) E R. Then by a similar argument, n e obtain [a] ~ [ b l . Consequently, [a] = [b].

On the other hand, if [a] = [b], then, by ( I ) , b E [b] = [a] ; hence (a, b ) E R.

Proof ot ( n ~ i . We prove the equivalent contrapositive statement

if 'a1 bl # @ then [a] = [bl

If [a] r b ] + $3, then there exists an element x E A with .c E [a] n jbl. Hence (a, x) E R and (b, x) E R. By symn~etry, (x, b) E R and by transitivity, (a, b) E R. Consequently by (ii), [a] = [b l .

PARTITIONS 7.11. Let X = { a , b , c , d, e, f , g ) , and let:

(i) A , = { a , c , e ) , A,= { b ) , A,= { d , g ) ; (ii) B l = (a, e , g ) , B, = { c , d ) , B , = { b , e , f ) ; (iii) C l = ( a , b , e , g ) , C ,= { c ) , C, = { d , j ) ;

(iv) Dl = { a , b , c, d , e , f , g l .

Which of { A , , A , ,A , ) , {B,, B,, B , ) , JC,, C,, C,), {Dl) are partitions of X? (i) {A , , A,, A ,] is not a partition of X since t E X but f does not belong to either A l , A>, or A,. (ii) { E l , E? , E , ] IS not a partition of X slnce e E X belongs to both B1 and B,. (iii) {C,, C2, C?] is a partition of X since each element in X belongs to exactly one cell, i.e.

X = C, d C, u C3 and the sets are pairwise disjoint.

(iv) {D, i IS a partition of X.

64 RELATIONS [CHAP. 7

Find all the partitions of X = {a, b , c, d ) . Note first tha t each partition of S contains either 1, 2, 3, or 4 distinct sets. The partitions

a re as follo\vs: (1) [{a, b, c, dl]

There a re fifteen different partitions of X .

Supplementary Problems RELATIONS

7.13. Let R be the relation in A = {2,3,4,5) defined by "x and y a re relatively prime", i.e. "the only common divisor of x and y is 1".

(i) Write R a s a set of ordered pairs. (ii) Plot R on a coordinate diagram of A X A. (iii) Find R-1.

7.14. Let 'V = {1,2,3, . . .) and let R be the relation in N defined by z + 2y = 8, i.e., R = {(rr,y) : x , y E N , r + 2 y = 8 )

(i) Write R a s a set of ordered pairs. (ii) Find R-1.

7.15. Let C = {I, 2 ,3,4,5} and let R be the relation in C consisting of the points displayed in the following coordinate diagram of C x C:

(i) State whether each is t rue or false: (a) 1 R 4, (b) 2 R 5, (c? 3 $ 1, (d) 5 $ 3.

(ii) Find the elements of each of the following subsets of C: ( a ) Cn : 3 R x), (b) {x : (4, z) E RS, (c) (x : (x, 2) 6? R}, (d) {x : x R 5).

7.16. Consider the relations < and 5 in N = { 1 , 2 , 3 , . . .}. Show tha t < U A = 6 where 1 is the diagonal relation.

EQUIVALENCE RELATIONS

7.17. Let TY = {I , 2,3,4}. Consider the following relations in W:

RI = {(I, 11, i l , 2 ) } R4 = {( I , l), ( 2 , 2 ) , (3,311

R2 = {(1,1), (2,3), ( 4 , l ) ) R, = T.Ti x W

R3 = {(1,3), (2 ,4)) = Q) Determine whether or not each relation is reflexive.

CHAP. 71 RELATIONS

7.18. Determine whether or not each relation in Problem 7.17 is symmetric.

7.19. Determine whether or not each relation in Problem 7.17 is transitive.

7.20. Let R be the relation I of perpendicularity in the set of lines in the plane. Determine whether R is ( i ) reflexive, (ii) symmetric, tiii) transitive, (iv) a n equivalence relation.

7.21. Let N = {I , 2,3, . . . } and let r be the relation in N X N defined by

(a, b) - (c , d ) iff a + d = b + c

(i) Prore r is an equivalence relation. (ii) Find the equivalence class of (2,5), i.e. [(2,5)].

7.22. Prove: If R and S a r e equivalence relations in a set X, then R n S is also a n equivalence relation in X.

7.23. (i) Show tha t if relations R and S are each reflexive and symmetric, then R U S is also reflexive and symmetric.

(ii) Give an example of transitive relations R and S for which R U S is not transitive.

PARTITIONS

7.24. Let W = {I, 2,3,4,5,6) . Determine whether each of the following is a partition of TV:

(i) IC1,3,51, {2,4), C3,6) I (iii) [{I, 51, {2}, C41, { I , 5}, {3,61]

I 51, {21, {3, Q l (iv) [{I, 2,3,4,5,6)1

7.25. Find all partitions of V = {1,2, 3).

7.26. Let [ A , , A , , . . . ,A , , , ] and [B,, B2, . . ., B,] be partitions of a set X. Show tha t the collection of sets

[A, n B, : i = l , ..., m, j=l,.. .,n] is also a partition (called the c ~ o s s par t i t ion) of X.

Answers to Supplementary Problems R = R - 1 = f ~ ( 2 , 3 ) , (3, 2), (2,5), (5,2), (3,4), (4,3), (3,5), (5,3), (4 ,5) , (5,4)}

R = {(2,3), (4,2), R-' = {(3,2), (2,4), (1,6))

(i) T, F, F , T. (ii) (a) {1,4,51, ( b ) 8 , ( c ) '12,3,41, ( 4 (31

R, is the only reflexive relation.

R4, R5 and R, are the only symmetric relations.

All the relations are transitive.

(i) no, tii) yes, iiii) no, (iv) no

(ii) [(2, 5)l = {(a, b) : a + 5 = b + 2, a , b E N )

= {(a, a + 3) : a E N l = {(I , 4), (2,5), (3,6), (4,7), . . .I

(ii) R = { \ I , 21) and S = {(2,3))

(i) no, (iii no, (iii) yes, (iv) yes

[{l, 2,3)], [{I:, C2, 3)], [{21, (1,311, [{3}, {1,2)1 and [{l}, {2:, (311

Functions

DEFINITION OF A FUNCTION

Suppose that to each element of a set A there is assigned a unique element of a set B; the collection, f , of such assignments is called a functioz (or mapping) from (or on) A into B and is written

f : ~ + B or A $ B

The unique element in B assigned to a E A by f is denoted by J(a), and called the inzage of a under f or the value of f a t a. The dovlzain of f is A, the co-domain B. The ?.a?lge of J , denoted by f [ A ] is the set of images, i.e.,

flA] = { f (a ) : a € A )

Example 1.1: Let f assign to each real number its square, tha t is, for every real number 2 let f (x) = x2. Then the image of -3 is 9 and so we may write f(-3) = 9 or f : -3 + 9.

Example 1.2: Let f assign to each country in the world its capital city. Here the domain of f is the set of countries in the world; the co-domain is the list of cities of the world. The image of France is Paris, t h a t is, f \France) = Paris.

Example 1.3: Let A = {a , b, c , d ) and B = {a, b, c}. The assignments

a - b , b - t c , c - c and d + b

define a function f from A into B. Here f ( a ) = b, f(b) = c, f (c ) = c and f(c1) = b. The range of f is {b, c ) , that is, f [A] = {b , c ) .

Example 1.4: Let R be the set of real numbers, and let f : R + R assign to each rational number the number 1, and to each irrational number the number -1. Thus

1 if .?: is rational f ( x ) = -1 if x is irrational

The range of f consists of 1 and -1: f [ R ] = {I , -1).

Example 1.5: Let A = {a, b , c , d ) and B = {x, y, 2) . The following diagram defines a function f : A - , B .

Here f(a) = y, f(b) = N , f(c) = x and f(d) = y. Also f [A] = B, tha t is, the range and the co-domain a re identical.

CHAP. 81 FCNCTIONS

GRAPH OF A FUNCTION

To each function f : A -+ B there corresponds the relation in A X B given by

((a, f ( a ) ) : a E A J

We call this set the graph of f . Two functions f : A + B and g : A -+ B are defined to be equal, written f = y, if f ( a ) = g(a) for every a € A , tha t is, if they have the same graph. Accordingly, we do not distinguish between a function and its graph. The negation of f = g is written J = y and is the statement:

there exists an a E A for which f ( a ) + g(a)

A subset J of A x B, i.e. a relation from A to B, is a function if it possesses the follo\~ling property:

[F] Each a E A appears as the first coordinate in exactly one ordered pair (a, b) in f . Accordingly, if (a, b) E f, then j(a) = b.

Example 2.1: Let f : A -+ B be the function defined by the diagr'am in Example 1.5. Then the graph of f is the relation

{ ( a , Y), ( b , x), (G, x ) , ( d . u)S

Example 2.2: Consider the following relations in A = {1,2,3] :

f = C(1 ,3 ) , ( 2 , 3 ) , ( 3 , l ) ) cl = { ( I , 2 ) , ( 3 , l ) I Ft = { ( I , 31, ( 2 , I ) , ( 1 , Z), ( 3 , l ) I

f is a function from A into A since each member of A appears a s the first coordi- nate in exactly one ordered pair in f ; here f ( 1 ) = 3, f ( 2 ) = 3 and f ( 3 ) = 1 . g is not a function from A into A since 2 E A is not the first coordinate of any pair in g and so g does not assign any image to 2 . Also h is not a function from A into A since 1 E A appears as the first coordinate of two distinct ordered pairs in h , (1,3) and ( 1 , 2 ) . If h is to be a function i t cannot assign both 3 and 2 to the element 1 E A.

Example 2.3: A real-valued function f : R + R of the form f ( r ) = a r t b (or: defined by y = a.c + b )

is called a linear fuxctio?z: its graph is a line in the Cartesian plane R2, The graph of a linear function can be obtained by plotting ( a t least) two of its points. For example, to obtain the graph of %(.I.) = 2x - 1, set up a table with a t least two values of x and the corresponding values of f ( x ) a s in the adjoining table. The line through these points, ( - 2 , - 5 ) , (0, - 1 ) and (2,3), is the graph of f as shown in the diagram.

Graph of f(x) = 2 x - 1

68 FUNCTIONS [CHAP. 8

A real-valued function f' : R + R of the fonn

f ( x ) = ( I ~ T ) ~ f alxn-l + . . . + a,,-,:c + a, is called a poly?zo?i~iul f~c?~c t ion . The graph of such a function is sketched by plotting various points and drawing a smooth continuous curve through them.

Consider, for example, the function f i x ) = .c' - 2.1: - 3. Set up a table of values for s and then find the corresponding values fo r f ( x ) a s in the adjoining table. The diagram shows these points and the sketch of the graph.

Graph of f ( x ) = xL 22.1: - 3

COMPOSITION FUNCTION

Consider now functions f : A + B and g : B + C illustrated below:

Let a E A; then its image f(a) is in B, the domain of g. Hence we can find the image of f(a) under the function g, i.e. g(f(a)). The function from A into C which assigns to each a € A the element g(f(a)) E C is called the composition or product of f and g and is denoted by g 0 f . Hence, by definition,

( g O f ) ( a ) = g(f(a))

Example 3.1 : Let f : A + E and g : B - C be defined by the following diagrams:

A f B g C

We compute ( g o t ') : A + C by its definition:

( g o f ) ( a ) = s ( f ( a ) ) = g ( u ) = t ( g o f ) ( b ) = u ( f ( b ) ) = g ( z ) = r

i u O f ) ( c ) = u ( f ( c ) ) = u ( u ) = t

Notice tha t the con~position function g is equivalent to "following the arrows" from A to C in the diagrams of the functions J and g.

CHAP. 81 FUNCTIONS 6 9

Example 3.2: Let R be the set of real numbers, and let f : R + R and g : R -+ R be defined a s follows:

f (2) = x2 and g ( x ) = z + 3

Then ( f 0 g ) ( 2 ) = f ( g ( 2 ) ) = f ( 5 ) = 25

( s O f I ( 2 ) = s ( f ( 2 ) ) = g(4) = 7 Observe tha t the product functions f o g and g 0 f are not the same function. We compute a general formula fo r these functions:

( f o g ) ( z ) = f ( g ( z ) ) = f ( x + 3 ) = (.T + 3)2 = x2 + 6% + 9

( g o f ) ( x ) = g ( J ( x ) ) = g(x2) = z1 + 3

ONE-ONE AND ONTO FUNCTIONS

A function f : A -, B is said to be one-to-one (or: one-one or 1-1) if different elements in the domain have distinct images. Equivalently, f : A + B is one-one if

f (a) = f (a') implies a = a'

Example 4.1: Consider the functions f : A + B, g : B + C and h : C + D defined by the follow- ing diagram:

A f B B C h D

Now f is not one-one since the two elements a and c in its domain have the same image 1. Also, g is not one-one since 1 and 3 have the same image y. On the other hand, h is one-one since the elements in the domain, x, y and z, have distinct images.

A function f : A + B is said to be onto (or: f is a function from A onto B or f maps A onto B) if every Z, E B is the image of some a E A. Hence f : A + B is onto iff the range of f is the entire co-domain, i.e. f [ A ] = B.

Example 4.2: Consider the functions f, g and h in the preceding example. Then f is not onto since 2 E B is not the image of any element in the domain A. On the other hand, both g and h a r e onto functions.

Example 4.3: Let R be the set of real numbers and let f : R + R, g : R + R and h : R + R be defined a s follows:

f ( ~ ) = 22, g(x ) = 2 3 - x and h ( z ) = x2

The graphs of these functions follow:

f ( x ) = 2" g ( x ) = 2 3 - x h ( x ) = x2

FUNCTIONS [CHAP. 8

The function f is one-one; geometrically, this means tha t each horizontal line does not contain more than one point of f. The function g is onto; geometrically, this means t h a t each horizontal line contains a t least one point of g. The function h is neither one-one or onto; fo r h(2) = h ( - 2 ) = 4, i.e. the two elements 2 and -2 have the same image 4, and h[R] is a proper subset of R; fo r example, -16 4 h[R] .

INVERSE AND IDENTITY FUNCTIONS In general, the inverse relation f P 1 of a function f c A x B need not be a function.

However, if f is both one-one and onto, then f P 1 is a function from B onto A and is called t,he inverse function.

Example 5.1: Let A = {a, b, c ) and B = { r , s, t } . Then

f = {(a , s), ( b , t ) , (c, 4) is a function from A into B which is both one-one and onto. This can easily be seen by the following diagram of f:

Hence the inverse relation

f-l = ( ( 5 , a ) , ( t , b ) , ( r , 4) is a function from B into A. The diagram of f-I follows:

Observe tha t the diagram of f-I can be obtained from the diagram of f by revers- ing the arrows.

For any set A , the function f : A + A defined by f(x) = x, i.e. which assigns to each element in A itself, is called the i d e n t i t y function on A and is usually denoted by 1, or simply 1. Note that the identity function 1, is the same as the diagonal relation: 1, = A,. The identity function satisfies the follo~ving properties:

Theorem 8.1: For any function f : A -+ B,

Theorem 8.2: If f : A + B is both one-one and onto, and so has an inverse function f -' : B + A, then

f - ' o f = l , and f o f P 1 = l g

The converse of the previous theorem is also true:

Theorem 8.3: Let f : A + B and g : B -+ A satisfy

g o f = 1, and f o g = 1,

Then f is both one-one and onto, and g = f - ' .