Solution to Let X = {1,2,3,4,5,6,7} and R = {x,y/x–y is divisible by 3} in x. Show … - Sikademy
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Archangel Macsika

Let X = {1,2,3,4,5,6,7} and R = {x,y/x–y is divisible by 3} in x. Show that R is an equivalence relation. b) Let A = {1,2,3,4} and let R = {(1,1), (1,2),(2,1),(2,2),(3,4),(4,3), (3,3), (4,4)} be an equivalence relation on R. Determine A/R. c) Draw the Hasse diagram of lattices, (L1,<) and (L2,<) where L1 = {1, 2, 3, 4, 6, 12} and L2 = {2, 3, 6, 12, 24} and a < b if and only if a divides b.

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a) Let X = \{1,2,3,4,5,6,7\} and R = \{(x,y)|x-y\text{ is divisible by }3\} in X. Let us show that R is an equivalence relation. Since x-x=0 is divisible by 3 for any x\in X, we conclude that (x,x)\in R for any x\in X, and hence R is a reflexive relation. If x,y\in X and (x,y)\in R, then x-y is divisible by 3. It follows that y-x=-(x-y) is also divisible by 3, and hence (y,x)\in R.

We conclude that the relation R is symmetric. If x,y,z\in X and (x,y)\in R,\ (y,z)\in R, then x-y is divisible by 3 and y-z is divisible by 3. It follows that x-z=(x-y)+(y-z) is also divisible by 3, and hence (x,z)\in R. We conclude that the relation R is transitive. Consequently, R is an equivalence relation.

b) Let A = \{1,2,3,4\} and let R = \{(1,1), (1,2),(2,1),(2,2),(3,4),(4,3), (3,3), (4,4)\} be an equivalence relation on R. Let us determine A/R. Taking into account that [a]=\{x\in A\ |\ (x,a)\in R\} and hence [1]=\{1,2\}=[2],\ [3]=\{3,4\}=[4], we conclude that A/R=\{[a]\ |\ a\in A\}=\{[1],[3]\}.

c) Let us draw the Hasse diagram of lattices, (L_1,<) and (L_2,<) where L_1 = \{1, 2, 3, 4, 6, 12\} and L_2 = \{2, 3, 6, 12, 24\} and a < b if and only if a divides b.

Note that a Hasse diagram is a graphical rendering of a partially ordered set displayed via the cover relation of the partially ordered set with an implied upward orientation. A point is drawn for each element of the poset, and line segments are drawn between these points according to the following two rules:

1. If x<y  in the poset, then the point corresponding to x appears lower in the drawing than the point corresponding to y.

2. The line segment between the points corresponding to any two elements  x and  y of the poset is included in the drawing iff  x covers y  or  y  covers x.

In our case, x<y if and only if x|y. Therefore, the Hasse diagrams are the following:

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Question ID: mtid-5-stid-8-sqid-2635-qpid-1105