p.609, icon at Example 2 # 2. Let A be the set of all points in the plane with the origin removed. That is, Define a relation on A by the rule: A = {(x, y)|x, y = R} - {(0,0)}. (a, b)R(c, d) ↔ (a, b) and (c, d) lie on the same line through the origin. (a) Prove that R is an equivalence relation. (b) Describe the equivalence classes arising from the equivalence relation R in part (a). (c) If A is replaced by the entire plane, is R an equivalence relation?

1.A known relationship mode of R (U, F), U= (A, B, C, D, E, G); F={AB->C, D->EG,C->A, BE->C, BC->D, CG->BD, ACD->B, CE->AG}; find the minimum cover of F,then use the Normal Form decomposition algorithm to decompose this mode to 3NF, andcheck if the decomposition has the nonadditive join property.

Let R be the relation on the set {a,b,c,d} containing the ordered pairs (a,a), (a,c), (b,b), (b,c), (b,d), (c,b), (c,c), (d,a) and (d,d). Find R².

List all 3-tuples in the relation {(a,b,c)| abc = 6} where a, b and c are all positive integers.

Let R₁ and R₂ be the relations on X={a,b,c}, R₁={ (b,a), (c,b), (a,c) } R₂={ (a,c), (b,c)} R₁-¹= R₂ ° R₁=

a) Define a relation R on R (the set of all real numbers) as follows: For all x, y E R, xRy → x < y. Is R reflexive, symmetric and transitive?

1) Consider the following relation instance of R1(A, B, C) contains the tuples {(a1,b1.c1). (a2,b1,c1), (a3,b3,c3), (a4,b3,c3), (a5,b5.c5). (a6,b6,c6)} and S1(B, C, D, E) contains the tuples ((b1.c1,d1,e1), (b1.c1.d2.e2),.(b1.c1,d3,e3), (b2.c2,d2.e2), (b3,c3,d4,e4).(b6.c7.d8,e9)) The following SQL query will result in how many tuples? Select From R1 r LEFT OUTER JOIN S1 s (ON r.B= s.B AND r.C=s.C): (The answer will simply be an integer, like 25). Answer: 2) Consider a relation schema R = (A, B, C, D, E, F, G} with the following set of functional dependencies: F = (BG A, CGA, A BC, BC A, BD E, EF, CF B) The decomposition of R into R1 = (A, B, C, G), R2 = {B, D, E), R3 = (B, C, F) and R4 = (E, F) is O In BCNF but Not in 3NF. In 3NF but Not in BCNF. In 3NF and in BCNF. Not in 3NF and Not in BCNF 3)

it says relation room doesnt exist.

1. Let A = {3, 4, 5} and B = {4, 5, 6} and let S be the "divides" relation. That is, for all (x, y) ≤ A × B, xSy ↔ x|y. State explicitly whic ordered pairs are in S and S-1

Question 17 Please provide the answer wisely.