378 Relations Solutions for Chapter 16. Section 16.1 Exercises. 3. Let A = {0,1,2,3,4,5}. Write out the relation R that expresses on A.

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1 378 Reltions 16.7 Solutions for Chpter 16 Section 16.1 Exercises 1. Let A = {0,1,2,3,4,5}. Write out the reltion R tht expresses > on A. Then illustrte it with digrm. 2 1 R = { (5,4),(5,3),(5,2),(5,1),(5,0),(4,3),(4,2),(4,1), (4,0),(3,2),(3,1),(3,0),(2,1),(2,0),(1,0) } Let A = {0,1,2,3,4,5}. Write out the reltion R tht expresses on A. Then illustrte it with digrm. R = { 2 1 (5,5),(5,4),(5,3),(5,2),(5,1),(5,0), (4,4),(4,3),(4,2),(4,1),(4,0), 3 0 (3,3),(3,2),(3,1),(3,0), (2,2),(2,1),(2,0),(1,1),(1,0),(0,0) } The following digrm represents reltion R on set A. Write the sets A nd R. Answer: A = {0,1,2,3,4,5}; R = {(3,3),(4,3),(4,2),(1,2),(2,5),(5,0)} 7. Write the reltion < on the set A = Z s suset R of Z Z. This is n infinite set, so you will hve to use set-uilder nottion. Answer: R = {(x, y) Z Z : y x N} 9. How mny different reltions re there on the set A = { 1,2,3,4,5,6 }? Consider forming reltion R A A on A. For ech ordered pir (x, y) A A, we hve two choices: we cn either include (x, y) in R or not include it. There re 6 6 = 36 ordered pirs in A A. By the multipliction principle, there re thus 2 36 different susets R nd hence lso this mny reltions on A. 11. Answer: 2 ( A 2 ) 13. Answer: 15. Answer: (mod 3) Section 16.2 Exercises 1. Consider the reltion R = {(, ),(, ),(c, c),(d, d),(, ),(, )} on the set A = {,, c, d}. Which of the properties reflexive, symmetric nd trnsitive does R possess nd why? If property does not hold, sy why. This is reflexive ecuse (x, x) R (i.e., xrx )for every x A. It is symmetric ecuse it is impossile to find n (x, y) R for which (y, x) R. It is trnsitive ecuse (xr y yrz) xrz lwys holds. 3. Consider the reltion R = {(, ),(, c),(c, ),(, c)} on the set A = {,, c}. Which of the properties reflexive, symmetric nd trnsitive does R possess nd why? If property does not hold, sy why.

2 Solutions for Chpter This is not reflexive ecuse (, ) R (for exmple). It is not symmetric ecuse (, ) R ut (, ) R. It is not trnsitive ecuse cr nd Rc re true, ut crc is flse. 5. Consider the reltion R = { (0,0),( 2,0),(0, 2),( 2, 2) } on R. Sy whether this reltion is reflexive, symmetric nd trnsitive. If property does not hold, sy why. This is not reflexive ecuse (1,1) R (for exmple). It is symmetric ecuse it is impossile to find n (x, y) R for which (y, x) R. It is trnsitive ecuse (xr y yrz) xrz lwys holds. 7. There re 16 possile different reltions R on the set A = {, }. Descrie ll of them. (A picture for ech one will suffice, ut don t forget to lel the nodes.) Which ones re reflexive? Symmetric? Trnsitive? Only the four in the right column re reflexive. Only the eight in the first nd fourth rows re symmetric. All of them re trnsitive except the first three on the fourth row. 9. Define reltion on Z y declring xr y if nd only if x nd y hve the sme prity. Sy whether this reltion is reflexive, symmetric nd trnsitive. If property does not hold, sy why. Wht fmilir reltion is this? This is reflexive ecuse xrx since x lwys hs the sme prity s x. It is symmetric ecuse if x nd y hve the sme prity, then y nd x must hve the sme prity (tht is, xr y yrx). It is trnsitive ecuse if x nd y hve the sme prity nd y nd z hve the sme prity, then x nd z must hve the sme prity. (Tht is (xr y yr z) xr z lwys holds.) The reltion is congruence modulo Suppose A = {,, c, d} nd R = {(, ),(, ),(c, c),(d, d)}. Sy whether this reltion is reflexive, symmetric nd trnsitive. If property does not hold, sy why. This is reflexive ecuse (x, x) R for every x A. It is symmetric ecuse it is impossile to find n (x, y) R for which (y, x) R. It is trnsitive ecuse (xr y yrz) xrz lwys holds. (For exmple (R R) R is true, etc.) 13. Consider the reltion R = {(x, y) R R : x y Z} on R. Prove tht this reltion is reflexive nd symmetric, nd trnsitive.

3 380 Reltions Proof. In this reltion, xr y mens x y Z. To see tht R is reflexive, tke ny x R nd oserve tht x x = 0 Z, so xrx. Therefore R is reflexive. To see tht R is symmetric, we need to prove xr y yrx for ll x, y R. We use direct proof. Suppose xr y. This mens x y Z. Then it follows tht (x y) = y x is lso in Z. But y x Z mens yrx. We ve shown xr y implies yrx, so R is symmetric. To see tht R is trnsitive, we need to prove (xr y yrz) xrz is lwys true. We prove this conditionl sttement with direct proof. Suppose xr y nd yr z. Since xr y, we know x y Z. Since yrz, we know y z Z. Thus x y nd y z re oth integers; y dding these integers we get nother integer (x y) + (y z) = x z. Thus x z Z, nd this mens xrz. We ve now shown tht if xr y nd yrz, then xr z. Therefore R is trnsitive. 15. Prove or disprove: If reltion is symmetric nd trnsitive, then it is lso reflexive. This is flse. For counterexmple, consider the reltion R = {(, ),(, ),(, ),(, )} on the set A = {,, c}. This is symmetric nd trnsitive ut it is not reflexive. 17. Define reltion on Z s x y if nd only if x y 1. Sy whether is reflexive, symmetric nd trnsitive. This is reflexive ecuse x x = 0 1 for ll integers x. It is symmetric ecuse x y if nd only if x y 1, if nd only if y x 1, if nd only if y x. It is not trnsitive ecuse, for exmple, 0 1 nd 1 2, ut is not the cse tht 0 2. Section 16.3 Exercises 1. Let A = {1,2,3,4,5,6}, nd consider the following equivlence reltion on A: R = {(1,1),(2,2),(3,3),(4,4),(5,5),(6,6),(2,3),(3,2),(4,5),(5,4),(4,6),(6,4),(5,6),(6,5)}. List the equivlence clsses of R. The equivlence clsses re: [1] = {1}; [2] = [3] = {2,3}; [4] = [5] = [6] = {4,5,6}. 3. Let A = {,, c, d, e}. Suppose R is n equivlence reltion on A. Suppose R hs three equivlence clsses. Also Rd nd Rc. Write out R s set. Answer: R = {(, ),(, ),(c, c),(d, d),(e, e),(, d),(d, ),(, c),(c, )}. 5. There re two different equivlence reltions on the set A = {, }. Descrie them ll. Digrms will suffice. Answer: R = {(, ),(, )} nd R = {(, ),(, ),(, ),(, )} 7. Define reltion R on Z s xr y if nd only if 3x 5y is even. Prove R is n equivlence reltion. Descrie its equivlence clsses. To prove tht R is n equivlence reltion, we must show it s reflexive, symmetric nd trnsitive. The reltion R is reflexive for the following reson. If x Z, then 3x 5x = 2x is even. But then since 3x 5x is even, we hve xrx. Thus R is reflexive.

4 Solutions for Chpter To see tht R is symmetric, suppose xr y. We must show yrx. Since xr y, we know 3x 5y is even, so 3x 5y = 2 for some integer. Now reson s follows: 3x 5y = 2 3x 5y + 8y 8x = 2 + 8y 8x 3y 5x = 2( + 4y 4x). From this it follows tht 3y 5x is even, so yrx. We ve now shown xr y implies yrx, so R is symmetric. To prove tht R is trnsitive, ssume tht xr y nd yrz. (We will show tht this implies xrz.) Since xr y nd yrz, it follows tht 3x 5y nd 3y 5z re oth even, so 3x 5y = 2 nd 3y 5z = 2 for some integers nd. Adding these equtions, we get (3x 5y) + (3y 5z) = 2 + 2, nd this simplifies to 3x 5z = 2( + + y). Therefore 3x 5z is even, so xrz. We ve now shown tht if xr y nd yrz, then xrz, so R is trnsitive. We ve now shown tht R is reflexive, symmetric nd trnsitive, so it is n equivlence reltion. The completes the first prt of the prolem. Now we move on the second prt. To find the equivlence clsses, first note tht [0] = {x Z : xr0} = {x Z : 3x 5 0 is even} = {x Z : 3x is even} = {x Z : x is even}. Thus the equivlence clss [0] consists of ll even integers. Next, note tht [1] = {x Z : xr1} = {x Z : 3x 5 1 is even} = {x Z : 3x 5 is even} = { x Z : x is odd }. Thus the equivlence clss [1] consists of ll odd integers. Consequently there re just two equivlence clsses {..., 4, 2,0,2,4,...} nd {..., 3, 1,1,3,5,...}. 9. Define reltion R on Z s xr y if nd only if 4 (x+3y). Prove R is n equivlence reltion. Descrie its equivlence clsses. This is reflexive, ecuse for ny x Z we hve 4 (x + 3x), so xrx. To prove tht R is symmetric, suppose xr y. Then 4 (x+3y), so x+3y = 4 for some integer. Multiplying y 3, we get 3x +9y = 12, which ecomes y+3x = 12 8y. Then y + 3x = 4(3 2y), so 4 (y + 3x), hence yrx. Thus we ve shown xr y implies yrx, so R is symmetric. To prove trnsitivity, suppose xr y nd yrz. Then 4 (x + 3y) nd 4 (y + 3z), so x +3y = 4 nd y+3z = 4 for some integers nd. Adding these two equtions produces x + 4y + 3z = 4 + 4, or x + 3z = y = 4( + y). Consequently 4 (x + 3z), so xrz, nd R is trnsitive. As R is reflexive, symmetric nd trnsitive, it is n equivlence reltion. Now let s compute its equivlence clsses. [0] = {x Z : xr0} = {x Z : 4 (x + 3 0)} = {x Z : 4 x} = {... 4,0,4,8,12,16...}

5 382 Reltions [1] = {x Z : xr1} = {x Z : 4 (x + 3 1)} = {x Z : 4 (x + 3)} = {... 3,1,5,9,13,17...} [2] = {x Z : xr2} = {x Z : 4 (x + 3 2)} = {x Z : 4 (x + 6)} = {... 2,2,6,10,14,18...} [3] = {x Z : xr3} = {x Z : 4 (x + 3 3)} = {x Z : 4 (x + 9)} = {... 1,3,7,11,15,19...} 11. Prove or disprove: If R is n equivlence reltion on n infinite set A, then R hs infinitely mny equivlence clsses. This is Flse. Counterexmple: consider the reltion of congruence modulo 2. It is reltion on the infinite set Z, ut it hs only two equivlence clsses. 13. Answer: m A 15. Answer: 15 Section 16.4 Exercises 1. List ll the prtitions of the set A = {, }. Compre your nswer to the nswer to Exercise 5 of Section There re just two prtitions {{},{}} nd {{, }}. These correspond to the two equivlence reltions R 1 = {(, ),(, )} nd R 2 = {(, ),(, ),(, ),(, )}, respectively, on A. 3. Descrie the prtition of Z resulting from the equivlence reltion (mod 4). Answer: The prtition is {[0],[1],[2],[3]} = { {..., 4,0,4,8,12,...},{..., 3,1,5,9,13,...}, {..., 2,2,4,6,10,14,...}, {..., 1,3,7,11,15,...} } 5. Answer: Congruence modulo 2, or sme prity. Section 16.5 Exercises 1. Write the ddition nd multipliction tles for Z 2. + [0] [1] [0] [0] [1] [1] [1] [0] [0] [1] [0] [0] [0] [1] [0] [1] 3. Write the ddition nd multipliction tles for Z 4. + [0] [1] [2] [3] [0] [0] [1] [2] [3] [1] [1] [2] [3] [0] [2] [2] [3] [0] [1] [3] [3] [0] [1] [2] [0] [1] [2] [3] [0] [0] [0] [0] [0] [1] [0] [1] [2] [3] [2] [0] [2] [0] [2] [3] [0] [3] [2] [1] 5. Suppose [],[] Z 5 nd [] [] = [0]. Is it necessrily true tht either [] = [0] or [] = [0]?

6 Solutions for Chpter The multipliction tle for Z 5 is shown in Section In the ody of tht tle, the only plce tht [0] occurs is in the first row or the first column. Tht row nd column re oth heded y [0]. It follows tht if [] [] = [0], then either [] or [] must e [0]. 7. Do the following clcultions in Z 9, in ech cse expressing your nswer s [] with 0 8. () [8] + [8] = [7] () [24] + [11] = [8] (c) [21] [15] = [0] (d) [8] [8] = [1]