RELATIONS: THE SECOND TIME AROUND - Ralph p grim- 123docz.net

Section.

1. (a) {(I,l )j(2,2)~(3,3),( 4,4),( 1,2),(2,1 ),(2,3),(3,2)}

(b) ((1,1),(2,2),(3,3),( 4,4),(1,2)}

3. (a) Let it,/2,/s E with II{n) = n 1, 12(n) = 5n, and 13(n) = 4n lin.

(b) Let 9}' 92, 93 E F with 9l(n) = 3, 92(n) = lIn, and 13(n) = sin n.

4. (a) The relation R on the set A is (i) reflexive ifVx E A (x,x) E 1(.

(ii) symmetric ifVx,y E A [(x,y) E 'R = } (y,x) E 'R]

(iii) transitive ifVx,y,z E A [(x,y),(y,z) E R = } (x,z) E 'R]

(iv) antisymmetric if Vx, yEA [(x, y), (y, x) En=} x = y], (b) The relation 'R on the set A is

(i) not reflexive if 3x E A (x, x) ¢ 'R

(ii) not symmetric if 3x, yEA [(x, y) E n A (y, x) ¢ R]

(iii) not transitive if 3x,y,z E A [(x,y),(y,z) E R/\ (x,z) ¢'R]

(iv) not antisymmetric if yEA {(x, y),(y,x) E R /\ x "I

5. (a) reflexive, antisymmetric:, b:ansitive (b) transitive

(d) (f)

x E A,

transitive

so (x, x) E n

ofthese are if ,R2 are (x, y), ('II, E Rl n'R2 then (x, y), ('II, z) E 'Rb 'R.'},1 so (x, z) E 'R(1 ) 'R2 (transitive property) and (x, z) E

'R1 n for symmetric and are oUU.u.c:.w.

8. (a) For aU x E A, (x, x) E 'R1 , 'R'}, ~ 'R1 U 'R.2 , so if either 'Rl or 'R.2 is reflexive, then

10.

U 'R'}, is reflexive.

(b) (i) H x, 'II E A and (x, 'II) E R.l U'R'},1 assume without loss of generality, that (x, y) E 'R.t • (x, y) E 'Rl and symmetric:::::::} (y, x) E 'R1 ::=:} ('II, E 'Rl U'R21 80 'R.1 U 'R.'J, is symmetric.

False: A = {1,2}, R.l = {(I, 1),(1,2)}, R'J = {(2, In. (1, (2,1) E

'R.1 U 'R2, and 1 f:. 2, 80 'R.1 U 'R'}, is not antisymmetric.

(iii) False: Let A = {1,2,3},'R1 = {(I, 1), (1,2)},'R'}, = {(2,3)}. Then ,2),(2,3) E R.l U 'R.Z1 but (1,3) ¢ n1 U'R.21 so n1 U R'}, is not transitive.

(a) False: Let A = {1,2} and n = {(1,2),(2, I)}.

(b) (i) Reflexive: True

(ii) Symmetric: False. Let A = {I,2}, Rl = {(I, I)}, Rz = {(I, 1), (1, 2)}.

(iii) Alltisymmetric & Transitive: False. Let A = {I, 2}, Rl = {(I, 2)},

R2 = {(I, 2), (2, I)}.

(ii) Symmetric: False. Let A = {1,2}, 'Rl = {(I, 2)}, 1(.z = {(I, 2), (2, In.

(iii) Antisymmetric: True

(iv) Transitive: False. Let A = {1,2}, 'Rl = {(l, 2), (2, In, 'R2 = {(I, 1),(1,2), (2,1), (2, 2)}.

(d) True (a)

(d) 211 (g)

(b) (24)(26) = 210 (24)(25) = 29

12.

Since 5880 -

2)(k+l), we

For n = p~pipg there are (5 1)(3 + 1)(6 1) = (6)(4)(7) = 168 positive integer divisors, so IAI = 168.

13. There may exist an element a E such for b) nor (b, a) E'Rão There are n ordered of form (x, x), x E each of (n:! - n)/2 sets {(x, y), (y,x)} of ordered pairs where x, yEA, x =1= y, one element is chosen. This results in a maximum value of n + (n2 - n)/2 = (n2 + n)/2.

The number of antisymmetric relations that can have this size is 2{n2-n)/2.

15. r - n counts the elements in 'R of the form ( a, b), a =1= b. Since 'R is symmetric, r - n is even.

16. (a) xRy if x < y.

(b) For example, suppose that R satisfies conditions (ii) and (iii). Since R =1= 0, let (x,y) E R, for X,Y E A. Since R is symmetric, it follows that (y,x) E R. Then by the transitive property we have (x, x) E R (and (y, y) E R). But if (x, x) E R the relation R is not i:rreflexive.

(e) 2(n2-n); 2",2 _ 2(2(n2-nằ

17. (a) G) (~l) + G) en + (~) en

(b) G) (~) + G) (;1) + G) (;1)

18. (a) Let Al = I-l(x), A2 = j-l(y), and As = j-l(z). Then R = (AI X AI) U (Az X A2) u

(As x As), so 11(,1 = 102 102 = 225.

(b) ni + ni + n~ n~

xE E E E 0

(a, d) E ('R,1 0 E 'R,3 some c E

,(b, E R2,(c,d) E for some b E B, c E C ==> b) E 'R,1I

(a, E 'R1 (;I (;I (;I 0 0 (Ka 0

4. (8.) R1o('R2 U ::::::R1o{(w, (W, (x,6),(y, (y,

= {(1,4),(1,5),(3,4),(3, (2,6),(1,6)) (R.! 0 R.2 ) U CR.! 0 R.a)

= {{I, (3, (2,6), (1, 4), (1, U ,4), (1,5), 5)}

= ((1,4), (1,5), (1, 6), (2,6), (3, 4), (3, 5)}

(b) R1 0 ('Rz n 1'i3) = R1 ° {( w, 5)} = {(I, (3,5))

(R1 0 R2 ) n (R! 0 'R3 ) :::::: {(I, 5), (3, 5), (2, 6), (1, 4), (1, 6)} n {(I, 4), (1, 5), (3,4), (3, 5)} =

{(1,4), 5),(3,5)}.

'Rl 0 ('R.a n 'R3 ) = 'Rz ° {(m,3), (m, = {(1,3), (1,4)}

CRI ° 'R2 ) n CR1 o'R.a) = {(I, 3), (1, 4)} n {(I, 3), (1, 4)} = {(I, 3), (1, 4)}.

6. (a) (x,z) E 1'il 0 (n2 U 1'il ) ¢=.} for some y E B,(x,y) E 1'il,(y,z) E n'}, U n3 ¢=.} for some 11 E B, ({ x, 11) E 1'i1, (y, z) E 'R'},) or ôx, y) E 'Rh (y, z) E 'R3) ==> (x, z) E 'Rl ° 'R'}, or

(x, z) E 'R1 o'R3 -¢=::} (x, z) E ('R! o'R2)U('R1 o'Ra), so 'Rl O('R20'R3 ) ('R} o'R2)U('R1 o'Ra).

For the opposite inclusion, (x,z) E CR.l o'R2)U('R1 oRa):=::} (x,z) E 'R1 oR'). or (x,z) E

'R.l o'R3 • Assume without loss generality that. (x, z) E RIO 'R2 • Then there exists an element y E B 80 that (x, y) E'R1 and (y, z) E R2 • But (y, z) E Ra :=::} (y, z) E 'R2 U'R31 so (x, z) E Rl ° (Ra U R3 ), and the result follows.

(b) proof here is similar to that in part (a). To show that the inclusion can be proper, let = B = C = {I, 2, 3} with Rl = {(I, 2), (1, In, R2 = {(2, 3)}, R3 = {(I, a)}.

Then 'R1 ° ('R2 01(.3) = R1 00 = 0, but (Rl ° 1(.a) 0 (nl 0 R3 ) = {(1,3)}.

1. This follows by the Pigeonhole Principle. Here the pigeons are the 2,,2 + 1 integers between

o and 2n2, inclusive, and the pigeonholes are the 2",2 relations on A.

8. Let S = {(1,1),(1,2),(1,4)} and T == {(2,1),(2,2),(1,4)}.

9. Here there are two choict'.s for each "ii,l ::; i < 6. For each pair "ij, "ji, 1 ::; i < j 6, are two choices, there are (36 - = 15 such pairs. Consequently are (26)(215) = 221 such matrices.

10. For o in E the

) 1

row '-'V.llU,l.u,uof

is O.

12. Men) = 0, yEA (x, ¢ R = 0. if f-

then yEA where xRy. Hence (x, y) E n and R 0.

m = I, we have M(nl) = }d(R) = [1\d('R)]1, so the is true in this

case. Assuming the truth the statement for m = k we have M(Rk) = {1\d(1<..)]k. Now

consider m = k + 1. M(1<..k+l) = ]1,.f(R 0 Rk) = M('R) . M(1<..It:) (from Exercise 11)

= 1\;[(1<..) . [M(R)]k = [M(R)Jk+l, Consequently this result is true for all m ;;::: 1 by the Principle Mathematical

13. (a) n {:::::;:::} (x, x) E R, for all x E A ~ mxx = 1 in M = (mij)nxro for all x E A {=} In :5 M.

(b) 'R synunetric '¢::=} [\lx,y E A (x,V) E ~ (y,x) E R) -¢=> [Vx,y E A mx'll = 1 in M ~ my a: = 1 in M] ~ M= Mt'l".

14.

10! PROGRAM MAY BE USED DETERMINE IF' A RELATION

20! ON A SET OF SIZE N, WHERE N <= 20,

EQUIVALENCE WE ASSUME OF

40! GENERALITY THAT ELEMENTS 1,2,3, ...

50!

60 80 90 100

110 120 130 140 150 160 170 180 190 200 210 220 230 240 250 260 270 280

=0 320

330 NEX.T 1

360 370

NOT TRANSITIVE"

+ Z = 3 THEN

184

d f

(b)

16. (a)

(a, b),

0 1 0 0 1 0 1 0 1 1 1 0 (c) 0 1 0 0 0 0 1vl(R) =: .

0 1 0 0 1 1 (e) 1 1 0 1 0 0

(/) 0 0 0 1. 0 0

rows are

e), (e, In

o 1 0 0 o 0

0 0 0 0 1 0 M(n) = 0 0 0 0 0 0

o 1 1 0 0 0 0 0 0 o 0 0 0 o 0 o 0 0

n= a),(a,b),(b,a),(c, c), (d, e), (e, d), I), (I, d), (e, 1), (I, e)}

1 1 0 0 0 0 1 0 0 0 0 0 Men) = 0 0 0 1 0 0 0 0 1 0 1 1 0 0 o 1 0 1 0 0 0 1 1 0 (iv) n = ((b,a),(b,c), b), (b, e), (c, d), (c, d)}

o 0 0 0 o 0

1 0 1 0 1 0 M(n) = 0 1 0 1 0 0 0 0 o 0 0 0 0 0 0 1 0 0

o 0 o 0 o 0

18. x),(w,v), x), y),(w,z),(x,z),

(b) x), v),

2t.

22.

225; (25)(210) = 21.5

2211; (25 )(210 ) = 215

'Rl : 'R2 :

1 1 0 0 0 1 1 1 0 0

0 0 1 1ã0 1 1 1 0 0

0 0 1 1 0 0 0 0 1 1

0 0 0 0 1 0 0 0 1 1

(b) Given an equivalence relation n.. on a finite set A, list the elements of A so that elements in the same cell of the partition (See Section 7.4.) are adjacent. resulting relation matrix will then have square blocks of 1 's along the diagonal (from upper left to lower right).

24. (~); G); (;)

25.

When n = 2, we

( c) Let a on where = m. Let he the directed graph Q"'l!>VIv~a.I11C:U

'R,. .- each component of G is a directed cycle Ci on mi vertices, with 1 z ~ k.

ml m2 + ... + mAl = m .. ) The smallest power n loops appea.:r is , for t = min{mil1 i ~

Let s = lcm(ml' m2,'" j mk). Then 'R,.'f'S = the identity (equality) relation on A and

= 'R, all r E smallest power of n n is 8 1.

27. (~) = 703 => n = 38

7.3

/, 18

6" /9

2 3

1 .

3. For all a E A,bE B,an1a and llR',lb so (a,b)'R(a, b), and'R is reflexive. Next

{a,b)'R.(c,d),(c,d)'R(a,b)::=:;:} a'R1c,c'Rla and b'R2d,d'R2b:::=:} a = c,b = d =;. (a,b) =

(c, d), so 'R is antisymmetric. Finally, (a, b )'R,( c, d), (c, d)'R( e, f) ~ a'Rt c, e'R,l e and b'R2d, d'R2f ::::::::} a'Rl e, b'R'2f :=:} (a, II )n( e, f), and n is transitive. Consequently, n is a partial order.

4. No. Let A = B = {1,2} with each of n I l n2 the usual "is less than or equal to" relation.

Then n is a partial order but it is not a total order for we cannot compare (1,2) and (2,1).

5. 0 < {I} < {2} < {3} < {1,2} < {1,3} < {2,3} < {1,2,3}. (There are other possibilities.)

6. (a) (a) (b) (e) (d) (e)

(a) 1 1 1 1 1

(b) 0 1 0 1 1

M(n):= (e) 0 0 1 1 1

(d) 0 0 0 1 1

0 0 0 0 1

(b)

7. (b) 3 < 2 < 1 < 4 or 3 < 1 < 2 < 4.

8. Suppose X, yEA and both are least elements. xRy since X is a least element, and y'Rx since y is a least element. \Vith'R antisymmetric we have x = y.

9. Let x, y be greatest bounds. Then x'Ry since x a lower bound Y IS a greatest lower bound. By similar reasoning yR.x. Since R. is antisymmetric, x = y. [The proof lub is similar.}

10. U = {1,2,3,4}. A be the collection of all proper subsets of U, partially ordered under set inclusion. Then {1,2,3}, {l,2,4}, {1,3,4}, and {2,3,4} are aU maximal elements.

11. Let U = {l,2}, A = P(U), and 'R the inclusion relation. Then (A, R.) a paset but not a total order. Let B = {0, {I}}. Then (B x B) n R. is a total order.

12. For all vertices x,y E A,x =I- y, there is either an edge (x,y) or an edge (y,x), but not both. In addition, if (x,y),(y,z) are edges in G then (x,z) is an edge in G. Finally, at every vertex of the graph there is a loop.

13. n + (;)

14. n + (~)

15. (a) The n elements of A are arranged along a vertical line. For if A = {all a2,'"

where at R.a2 ~ R. ... 'R.an , then the diagram can be drawn as

(b) n!

16. (a) Let a E A with a minimal. Then for x E A, xRa ::=::? x = a. So if M(R) is the relation matrix for R, the column under 'a' has all O's except for the one 1 for the ordered pair (a,a).

17.

(b) Let bE A, with b a greatest element. Then the column under 'b' in M(R) has alII's.

If c E A and c is a least element, then the row of M(R) determined by 'e' has alII's.

lub glb lub glb lub glb

(a) {1,2} 0 (c) {1,2} 0 (e) {l,2,3} 0

(b) {1,2,3} 0 (d) {1,2,3} {l}

18. (a) (i) Only one such upper bound - {1,2,3}. (ii) Here the upper bound has the form {I, 2, 3, x} where x E U and 4 ::; x < 7. Hence there are four such upper bounds. (iii) There are (~) upper bounds of B that contain five elements from U.

(b) (~) + G) (~) (~) + (1) = 24 = 16

(d) One - namely 0 (e) glb =0

each a E Z it follows aRa because a - a = an even nonnegative integer. .... u,,""'""ov IS h, c E Z with anb o1?e

a- b = some mE

b - c = 2n, for some n E

a - e = (a - b) + (b - c) = + n), where m n E a1?c and n

transitive. Finally, that a1?a and bRa for some a, b E Then a - b and b - a are both nonnegative integers. Since this can only occur for a - b = b - a, we find [a1?b A bRa] => a = h, so antisymmetric.

192

'VA04,'Vll 'R is a order for it not a total

example, 2,3 E Z and we have neither 2'R3 nor 3n2, because neither nor 1, respectively, a nonnegative even integer.

20. (8.) For all (a, b) E A, a = a and b 0, so Ca, b)n(a, b) and the relation is reflexive. If d) E with (a, d) (e, b), if a e we that

(a, b)'R(e, d)::} a < c, ( c, d)'R.( u, b) "* c < a,

and we obtain a < a. Henc£) we have a = c.

And now we find that

b )n( c, d) "* b ::; d, and ( c, d)n( u, ::} d $; b,

so b = d. Therefore, (u, b )'R( c, d) and (c, d)'R( a, b) => ( a, b) = (e, d), so the relation antisymmetric. Finally, (',ol1sider (a, b), d), (e, f) E A with (u, b )R( e, d) (e, d)'R( e, f).

Then

(i) a < c, or (ii) a = c and b ::; d; and

(iY c < e, or (ii)' c = e and d :5 f.

Consequently,

(i)" a < e or {iiY' a = e and b f - so, (a, b)R(e,f) and the relation is transitive.

The preceding shows that R is a partial order on A.

b) & c) There is only one minimal element -- namely, (0,0). This is also the least element for this partial order.

The element (1,1) is the only maximal element for the partial order. It is also the greatest element.

d) This partial order is a total order. We find here that (0, O)'R(O, 1)'R(1, O)'R(l, 1).

::n. (a) The refiexive, antisymmetric, and transitive properties are established as in the vious exercise.

e!eDlent (2,2)

(o,O)n(O,l)'R(Oj O)R(l, 1 )'R(1, 2)'R(2, 0)'R(2, 1 )1(.(2, 2).

22. 1

23.

193

(A, a total order, then x, 'II E xRy oryRx. FbI' 'II} = y and glb{x, y} = x. Consequently, (A, R) is a lattice.

24. is finite, A has a maximal element, by x, 'II (x =f. y) are both maximal elements, since x, yRlub{x, y}, thell1ub{x, y} must equal either x or y. Assume

lub{x, = x. Then yRx, 80 'II a maximal A a

maximal element x. Now for each a E A, a ¥ x, if lub{ a, x} f. x, then we contradict x being a maximal element. Hence aRx for a E A, so x is the greatest eu:~m.:lnt

[The proof for the least element is similar.]

25. (a) a (b) a (c) c (d) e (e) z (f) e (g) v

26.

(A, R) is a lattice with z only minimal) element.

a) 5 d) 10

greatest (and only maximal) element and a the least (and

b) and c) n + 1

e) and f) n + (n - 1) + ... + 2 + 1 = n(n + 1)/2.

21. Consider the vertex paqbrc, 0 :$ a < m, 0 :$ b < n, 0 :$ c < k. There are mnk such vertices; eaeh determines three edges - going to the vertices pa+1qOrC,paqb+1rc,pa(lrc+1.

This accounts for 3mnk edges.

Now consider the vertex pmqbrc, 0 :$ b < n, 0 :$ c < k. There are nk of these vertices;

each detemines two edges - going to the vertices pmqb+lrc, pffiqbrc+l. This accounts for 2nk edges. And similar arguments for the vertices paqflrC(O a < m,O :$ c < k) and pCqbrk(O :5 a < m,O :$ b < n) account for 2mk and 2mn edges, respectively.

Finally, each of the k vertices p'mqnrc, 0 S C < k, determines one edge (going to pfflqnr c+1) and 50 these vertices account for k new edges. Likewise, each of the n vertie.es pm (lrk ,

o :$ b < n, determines one edge (going to pmqHlrk), and so these vertices account for n new edges. Lastly, each of the m vertices paqnrk, 0 :$ a < m, determines one edge (going to pl1l+1'i'rk) and these vertie,es account for m new edges.

The preced,mJ; '''.U.U'-'''''. of edges as (m+n+k )+2( mn+mk+nk )+3mnk.

28. a) . 3. There are 4 . 2 = 8 for totally

29. are2ã7=

30. all 1

194

Now let E = (eij )mlo., F =

all 1 :5i :5 m, 1 ~ j :5 n,

)mxn be (O,l)-matrices, with E ~ F and F :5 E. Then,

~ lij :5 =} - ,so E = -- and

"precedes" relation is antisymmetric.

Finally, that - = (fij)mxnl G = are (0,

with E F and F < G. Then, for all 1 :::; i ~ 1 :5 j :::; ằ, eij :5 Iii and lij :5 gij =}

:5 Uij, so E ~. and the "precedes" relation is transitive.

In so much as the "precedes" relation is reflexive, antisymmetric, and transitive, it this a partial order - making into a poset.

Section 7.4

1. (a) Here the collection AI, A:h Aa provides a partition of

(b) Although A = Al U A2 U A3 U~, we have At n A2 0, so the collection All A:h Aa) A.t does not provide a partition for A.

2. (a) There are three choices for placing 8 - in either At, A2! or Aa. Hence there are three partitions of A for the conditions given.

(b) There are two possibilities with 7 E At, and two others with 8 E AI. Hence there are four partitions of A under these conditions.

(c) If we place 7,8 in the same cell for a partition we obtain three of the possibilities. If not, there are three choices of cells for 7 and two choices of cells for 8 - and six more partitions that satisfy the stat,ed restrictions. In total - by the rules of sum and product - there are 3 + (3)(2) = 3 + 6 = 9 such partitions.

3. n = {(I, 1), (1, 2), (2,1), (2,2), (3, 3), (3,4), (4,3), (4,4), (5, 5)}.

4. (a) [1] = {1,2} == [2]; [3] = {3}

(b) = {1,2} U {3} U {4, U {fi}.

5. 'R, is not transitive since 1 R2, 2R3 but 1"1.3.

Y) E X

(Xl, Yl)'R.(X2dJ2) :::::::::;} Xl = X2

). )n.(X2'