Question

[Partial Orders - Six Easy Pieces] A binary relation is R is said to be antisymmetric if (x,y) ER & (y,x) ER = x=y. For example, the relations on the set of numbers is antisymmetric. Next, R is a partial order if it is reflexive, antisymmetric and transitive. Here are several problems about partial orders. (a) Let Ss{a,b} be a set of strings. Let w denote the length of the string w, i.e. the number of occurrences of letters (a or b) in w. For example, abba = 4 P .Is R a partial order? Define a binary relation Ron Sas (x, y) ER Give all the details. (b) Let R be a binary relation on RxR: V(a,b), (c,d)€ RxR, ((a,b),(c,d))ER = [(a<c) (a = c &bsd)]. Prove that R is a partial order. m +n is even (C) Let R be a binary relation on Z: Wm, neZ (m,n) ER Is R a partial order on Z ? Prove. (d) Which of the following relations is an equivalence relation; explain why? R = {(0,0),(1,1),(1,2),(2,1),(2,2),(3,3)} R = {(0,0),(0,1),(0,2), (2,2),(1,0),(1,2),(2,0),(1,1),(3,3)}

Answer 1

a) Let Wis a string. Then, (w/=/w/, Bo (w/< (w). Thus (ww, w) ER, Ris reflexive. Let wis are strings such that (WB), (3,W) ER. - Then, wield and 1115 [wl. => [W) = 131. So R is antigymmetric ro Let wist are strings such that (W,3), (st) ER. Then, Iwl5ls his ltl.. So IWL 314151+/- 2 2.990 I => 1W1s 1413) (wt) ER 0 => - Se R is transitive 3) )) 2 2 Hence, Ris a partial orders of bad s 20 7 6=(26) = Sed b) Let (a, b) EIRXIR, then (9,6)= (ab) ? aza and b=b) (10) > a=a and beb. to rol so ((@6), (a,b)) E R. So, Ris refdexire & bet (96), (sd) EIRXIR such ((a,b), (Cd), (d), (96)) ER. will LA the Then acc or aac & bsd and С<a c =a 2 Sь. Now, acc & c&a cannot happened simultaneously, nor a-c and one of a ce so, aac & bed and c=a & dab. > azc & bed & dsb > a=c & b=d. (a,b) = (ed). So, Ris antisymmetric. or dat on cca.

Let (a, b), (d), (ef) EIRXIR such that ((a,b), (ed), ((Sd), (ef)) ER. Then, acc on aac &e bed and a ece on coefdsf. If acc & cke. happened to Then, aae. so (ab, (ef) ER. If acc & cae & def. nontor Then, acc=e* ace. dona pristos so (@b), (ef)) ER. 1 ja If a=e&bs d and cre, 1+1 = 1813 Then, a=cce ace it w) 161-10 th Sg (@b.cence w Vit Na If a=c & bsd and cae & dsf. Then, azcze » aze & bsdef = befant 1 x 17 (a. so (Carb), Cef)), ER D E Thus, for all cases, (a,b), (sd)), ((sd), Cef) R3 (cab); (ef)er. So, R is transitive. Thus, Ris a partial order, re, ich Let MEZ. Then, mtm=2m is even so, (mm) ER. So, Ris reflexive. s

Let 3,5€ 21. Then, 3+5=8=5+3 implies (3,5), (5,3) € 'R but 375. So Ris not antisymmetric. - Hence, Ris not a partial order. » Given, - Ri= {(0.0), (11), (12), (3,1), (22), (3,3)} so for set = {0, 1, 2, 3}, for ata, (a a) Era. So R is reflexive. If azb, then @b) ER, a baser, Nows for afb, only (22), (31)ER. so, R, is symmebric. Now for any (a,b), (b, c) € R. » (@C) ER. like, (12), (3)ER & CLIDER, (2), (32) fR & (2) ER. (3), (II)ER & (21) ER. So R is transitive. Hence, Ris reflexive, symmetric and transitive and thus R is an equivalence relation. Given, R2 = {(00) (21), (0.2), (22), (1.0), (12), (2,0) (1), (2,3)} since (L2) ER2 but (31) 4R2. so R2 is not symmetric. Hence, R2 is not an equivalence relation.