(1) Consider the following relations on 1, 2, 3, 4, Which...

(1) Consider the following relations on {1, 2, 3, 4}, Which of these relations are equivalence relations? That is which of these relations are reflexive, symmetric, and transitive (a) {(2, 2), (2, 3), (2, 4), (3, 2), (3, 3), (3,4)} (b) {(1, 1), (1, 2), (2, 1), (2,2), (3, 3), (4,4)} (c) {(2, 4), (4, 2)} (d) {(2, 1), (1, 1), (2, 2), (2, 3), (4, 4), (1, 2) (3, 2), (3, 3)} (e) {(1, 1), (2,2), (3,3), (4,4)} (f) {(1, 3), (1,4), (2,3), (2,4), (3,1), (3,4)} (2) Let U={a,b,c,d,e,f,g,h,i,j,k},A={b,c,d,e,f,g},B={a,e,h,k,f}. Find (a)A?B (b)A?B (c)B' (d)A' (e)(A?B)xA (3) (a) How many integers in the set {n? Z|1? n?1100} are divisible by 2 or 5? (b) How many integers in the set {n? Z|1? n?1100} are divisible by 2, 5, or 11?

Transcript

00:01 In this problem we are asked to find out if the given relations are equivalent relations or not.

00:09 Let us denote equivalence with e and we know that a relation is set to be equivalent if it satisfies three properties, that is reflexive which is denoted by r, symmetric which is denoted by s and transitive which is denoted by t.

00:26 So, reflexive means that for every element x of the set, we have x comma x in the relation.

00:35 Symmetric means if x and y are related, then y and x must also be related.

00:41 And transitive means if x and y are related and y and z are related, then it implies that x and z must also be related.

00:53 So now let us begin with the first subpart.

00:57 We have 2 .2 2 .3 2 .4 3.

01:05 3 .3 and 3 .4.

01:13 And the set on which the relation has been created.

01:17 That is the domain is 1 2 3 3 4.

01:20 It can be seen that the element 1 .1 does not belong to the relation are which means that the given relation is not reflexive.

01:33 Therefore, we can conclude that the relation is not equivalent.

01:38 Next, moving towards the second subpart, we have 1 .1, 1 comma 2, 2 .1, 2 .2, 2 .3, 2.

01:51 3 comma 3 and 4 .4.

01:57 So here it can be saying that the reflexive property is clearly satisfied.

02:03 The symmetric property is also satisfied since we have 1 comma 2 and 2 .1 and the transitive property is also satisfied since if we consider 1 .2 and 2 .2 then 1 .2 is there in the relation.

02:19 Likewise if we consider any other pairs we see that the transitive property is satisfied.

02:24 So we have reflexive symmetric and transitive which implies that the given relation is equivalent.

02:31 So this is our answer for subpart b.

02:35 Next moving towards subpart c, in subpart c we have 2 .4 and 4 comma 2.

02:44 It is clear that it is not reflexive since none of the terms, that is none of the ordered pairs 1 .1, 2 .2, 3 comma 3, 3 nor 4 .4 are present here.

02:56 So we can say that the relation is not reflexive which implies that it is not an equal.

03:02 Equivalence relation.

03:04 So this is a final answer for subpart c.

03:06 Next moving towards subart d.

03:08 In subart d we have 2 .1, 1 comma 2 2 .3, 4 .4, 1 .2, 1 .2, 3 .2, 3 .3, 3 .3.

03:23 So here again it is clear that it is reflexive since all the ordered pairs 1 .1, 2 .2, 3, 3 and 4 .4 are present.

03:31 It is also symmetrical.

03:33 Because if we have 2 comma 1 here then we also have 1 comma 2.

03:38 Likewise if we consider 2 .3 then there exists 3 comma 2 as well.

03:43 But it is not transitive because we have 3 .2 and 2 .1 but 3 .1 is not present in the relation.

03:54 So since it is not transitive we can say that the relation is not an equivalence relation...

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