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Problem Define relation 2 a8 follows: For I,V € 2, I -V 15 eVen: This relation is we say' I ~ V if and only if #n equivalence relation [nO need to prove ...

Question

Problem Define relation 2 a8 follows: For I,V € 2, I -V 15 eVen: This relation is we say' I ~ V if and only if #n equivalence relation [nO need to prove this]. Find all of the diflerent equivalence elusses, cuch unique (qquivalence cluxs_ unc show at least % elemnents of4.1 Solution

Problem Define relation 2 a8 follows: For I,V € 2, I -V 15 eVen: This relation is we say' I ~ V if and only if #n equivalence relation [nO need to prove this]. Find all of the diflerent equivalence elusses, cuch unique (qquivalence cluxs_ unc show at least % elemnents of 4.1 Solution



Answers

Show that each of the relation $\mathrm{R}$ in the set $\mathrm{A}=\{x \in \mathbf{Z}: 0 \leq x \leq 12\}$, given by (i) $\mathrm{R}=\{(a, b):|a-b|$ is a multiple of 4$\}$ (ii) $\mathrm{R}=\{(a, b): a=b\}$ is an equivalence relation. Find the set of all elements related to 1 in each case.

In this problem of relation and concerned, we have to show that the religion are in the set. So set is given as a as a set which has element 1234 and five. So this is 1, 2, 3, 4 and five given by. So relations are is given by are as a relation between ordered pair A and B. Such that models of A minus B is even. So this is even number is an equivalence relation. We have to show that this relation are as equivalence relation. And also we have to show that the elements of 135 So we have another set to say elements of 135 are related to each other and another elements Say two and 4 related to each other, but no element of being related to. He had steep. So now we have A 1234 and five. And we have relations A and B. Such that models of A minus B is even so. Now when we take so now we have to check it for the first reflective than symmetric and transitive relation is said to be an equivalence relation if it is reflexive, symmetric and transitive all together at the same time. So first check for reflexive relation is said to be reflective if the ordered pair A. And A. Which is a part of hit here, which is a part of the given set A belongs to are. So this should belongs from our if he belongs to A. The capitalist. So we are taking A what? We 123 and four and five. So when we write any of the value so one minus one will be zero to minus two would be 03 minus three will be zero. So when we write any of the values, so this will give you zero and a minus B. This is the model is so this will be Equals to zero and 0 is an even number. So we can say our is reflexive relation. And now we have to check for symmetric and then we have to check for transitive. So a religion is ready to be symmetric if the ordered pairs say A. B which belongs to us, So this is A. B. Which belongs to our, which implies that we and A should also belongs to us. So now let's take this set. So when we take 13 and five, so this would be one and three. And this implies that the models of 1 -3 should be even and when we write it, this is too and that means the order paper three and one, which means the models of three minus one should be also even. This is again, even. So when we write one in five, this will be against them. This equals to four and five minus model city equals two again four. So we can say that this Set is satisfied. And now when we take the second set that is 2 -4. So again 2 -4 would be equals two. And here when we write 4 -2 would be again a even number. So we can say that both sides are satisfied and which belongs to the given set a. So are as symmetric relations. And now we have to check for transitive relation. A relation is said to be transitive if the ordered pair A. B. Stay here, ordered A. V. Which belongs to our and another ordered pair. That is BNC which belongs to our implies that the ordered pair A. And she should also belongs to from our. So now let's take the order bear 135 years. So we are taking A. Z equals to one and B is equal to three. So one minus three. Again, two models of one minus three is again too and now we are taking C. Is equal to fight, So BNC. So models of 3 -5 is equal to again to and this implies that a minus C. Models of A equals to one and C. Is equal to five which is equals to four again even number. So we can say that this has sat is satisfied. Now when we take the second set we have two and four element. So we have to and food we can write it as So we can write it as 2 -4 which is again two. And now here we can say four and 2. So this will be for my understood again to which is even number. And now this implies that the relation A N C. That is two minus two. Should we should be a even number which is equals to zero. So here we can say that our is here transitive also. So are is transitive also. So hence we can say that hence relations are is and equivalence relation relation. And now we had to check for here, this is the answer. And now we have to check that both the set B and C. Here, I'm showing you. So here both the said B n C r not related to each other. Any of this that is independent of each other. So when we take Here 1, 3, 5, Say 1 -3 is even number one minus five. Models of human -5 is again even number one say three minus type. So this will be three minus five. Again even number. So these are so these are Independent and now when we write 1 -2, So 1 1 -2 is not even number, so this is one, one minus 43 minus 45 minus four is equal to 13 minus two is 11 minus two. Models of one minus two is 15 minus two is three. So these are giving odd number, but when we write to minus four or 4 -2, it even number. So we can say that hence no element of. So this is a reasoning, hence no element of of 135 So this is 1, 3, 5 is so this is a set is related to any element related two. Any element any element of The set which is two and 4. So this is the answer.

In this problem of religion infants. And we have to show that the religion are in the set set is which has the elements 1, 2 and three. So and the relation is given us given by relation is given by this is only two other pair, one and two, and the second one is two and one. So this is the Under repair which have only two pairs that it went 1-1. And we have to show that this relation is symmetric but neither reflexive nor transitive. So first we would So for reflective. So relation is said to be reflexive if ordered pair A and a part of our so ordered pair. Eh and S say when we are putting one, So this should be 11 When we are putting too. They should be too. And when we are putting three, the should be 33 And this will be true for every values of A. Which belongs to the given set A. So here we have three elements 123 but there is no such pair. So we can say our is not reflexive. And now we have to check for here symmetric relation. A relation is said to be symmetric if ordered pair even and a two which is already in relation are which implies that 8 to even should be prison there for every values of even and 82 which belongs to the given, said A. So here we have the give and take 812 and this is a pair. 12 so 12 is in relational. This implies that to anyone should also be present there and this is also present there. So we can see that this relation is a symmetric relation. So R. S symmetric relation are asymmetric. And now we have to check for transitive. So again, if ordered pair X. Y. Which belongs from are they want to? And another order pair. Why end Said which belongs to here. I say this is two and 1. This implies that the ordered pair X. NZ should be also presented there. So that means there should be one and one pair, since this is not present there. So we can say that our is not transitive, so are is not transitive. So hence we can say that we have proved that R. L symmetric, but neither reflexive, not transitive. Hence odd is symmetric, but neither reflexive. Yes, neither reflexive, not the north transitive. So this is the proof.

So here we had our in relation on X Y pairs in the positive 80 years such that if you take two pairs, they're only art. If only a trustee could be policy, these Teoh beast really have to be equal well, to show it looks every we can let a pair a B B inside on the plus, that implies plus B B plus A. Therefore a common be a comedy is an art. We had a common being really itself. Arjun is indeed influx of now efficient symmetry. Let one pair a B being the and let another pair Maybe being in the vault and like a related to be and Phoebe really the really has a trustee would be preceded by the commission. Does that imply we can rearrange this? Get B plus a plus B and then we have plus B equals people? A. Because you can just what? The terms on either side of the coastline. So you have see comedy comma any common be inside our so are indeed now to show transit, let income be comedy being are as well a see quality. So then we have a trustee because people see and people's people see by definition, then apply what we can solve. We want to show relationship a condom, being you come up so we can solve. You're being for a results should be we get e plus the mighty And when we saw you get my and adding to give us a plus B plus C and invited definition of art, that's just be calm. Be calm inside are so are the need chanted so that means or is an equivalent solution that transitive, symmetric and reflexive?

This question, We are asked to show that ah defy as the following is equivalent relation identify on the set off or their pair off positive integer and may be related to CD even only if it eight times the equals B times c So we kind of cross multiplied him and we will shake three properties to to show that it is equivalent in relation. First is reflexive so maybe he has to relate to itself. And you can see here that it it means that we want a B equals to be a But since it and be our positive in hedgerows, they are equals because it competitive, like modification, is committed to so this probably is clear next the symmetry. So if ah may be related to city, we want the reverse to be true. So we want CD to relate to a B s bill and we can do that by just looking at the definition when they be really the two CD. We have this equation, right? This mean that when we we just rearrange them? Basically we swap so perhaps I and commute them. We have cb equal to the as well, right, But This is a definition for for when CD related to a B. All right. Ah, when you shake these kind of probably it is important to arrange the position off off entries in here as exactly the same as your definition. In this case, it doesn't matter because they are indigenes, right? But when they are like metrics or something that cannot commute, then you have to shake carefully. Okay, but in this case is symbol is this probably is clear, then next one is transit E This one we have to do somewhere. So suppose may be related to CD and see the related to E f. Then we have this to equation, right? What we want we won if equals two being you because we won in the end may be related to E f. Right. And we're gonna show this on the next page. Uh, we start with what we know since may be related to city. We have this equation, we multiply f both sigh, and you can see that by the second relation, see if he's equal to the right, and so we shan't it to the here. And then we can divide the boats I so left hands I lived with if and the right hands I become B as required. So this is the definition when maybe we like that to e f. And so it is transitive. We have checked all of three properties. So we have we have shown that this relation is in fact equivalent relation. That is it.


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