Question

4. Let S = {1,2,3). Define a relation R on SxS by (a, b)R(c,d) iff a <c and b <d, where is the usual less or equal to on the integers. a. Prove that R is a partial order. Is R a linear order? b. Draw the poset diagram of R.

Answer 1

Hence proved and the poset diagram is drawn, if you still unsatisfied with the diagram drawn by me then you can comment below and let me know to draw in detail to you.

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U Answer helt S= {1,2,3]. Define a relation R on sxs by (a,b) R6 d) iff acc and bed, where & is the usual less (2) equal to on the integers. a) Prove that R is a partial order. Is Ra linear order? Given, s= {1, 2, 3} Now, sxs = {2,6 / aes, bes} le sxs = {(), (2),0,3),(2,2),2,3), (3,3)} since the Rise on sxs defined as (a, b) R(gd) iff as c and bed => R = {(0.0, 01), (1,0,) 60,0,3), (0.1),2,2)(,1),2,3) (6, 1), (3,3), (C, 2), (2),(0,2), (1, 3), (1, 2), 6.,»), 621123) (2.2), 6,3), (0, 3), 31, 3), (C3), 2,»)) ((,3,2,3) (, 3), (3)), (2,2), (6,2)), (3,2), (2,3)), (8,296,3)), (2, 3), (2,3)),(2,3), (3,3), (3, 3), (3, 3)) }. Now, for Reflexive & for every value of (a,b) € sxS aça and bab => (a,b) R@,b) =) R is reflexive.

yera le Can Anti-symmetric & since (a,b) R(,d) and (c,d) R (a,b) > acc and bed & cea and deb By the above relations we can write as a=c and bad => (a,b) = (cd) .: R is anti- symmetus Transitive , - Now, (a,b) R(ed) and (c,d) R (e. f) => asc and bed and cee and def as aseg bsf => (a, R6) Ref) R is transitive. .: R is a partial order since it satisfies all (Reflexive, Anti-symmetic, Transitive). Verification for R is not linear onder (connerity) : Now, take (1,3) & (3, 1) both e sxs . =) 123 8 3 2 1 this part is wrong .: R is not linear order (connesity). cze Unear

b) Draw the poset diagram of R. Ambiente R= Sxs = {(1), (0,2), (, 3), (2, 2), (2, 2),(33)} Co