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Question: Let \[R=\\{(3,3),(6,6),(9,9),(12,12),(6,12),(3,9),(3,12),(3,6)\\}\] be a relation on \[set\text{ }A=...

Let R=(3,3),(6,6),(9,9),(12,12),(6,12),(3,9),(3,12),(3,6)R=\\{(3,3),(6,6),(9,9),(12,12),(6,12),(3,9),(3,12),(3,6)\\} be a relation on set A=3,6,9,12set\text{ }A=\\{3,6,9,12\\}, The relation is
(1) An equivalence relation
(2) Reflexive and symmetric only
(3) Reflexive and transitive only
(4) Reflexive only

Explanation

Solution

A relation that is reflexive, transitive as well as symmetric is known as an equivalence relation. A relation indicates the relationship between a member of any two sets A and B. First, we should know what is reflexive, symmetric, transitive, and equivalence relation. Then we will prove them one by one and the required answer will be obtained.

Complete step by step answer:
A relation is defined as the set of ordered pairs. If there are two non-empty sets A and then a relation R from A to B is a subset of (A×B)(A\times B).. It can also be written as R(A×B)R\subseteq (A\times B). If we write (a,b)R(a,b)\in R, then we can say that ‘a is related to b’ and we can also write it as a R ba\text{ }R\text{ }b. If we write (a,b)R(a,b)\notin R, then we can say that ‘a is not related to b’.
For reflexive relation:
Let A be any set with any values, then
A relation R on A will be reflexive if (a,a)R(a,a)\in Rfor all aAa\in A.
In the above question, it is given that
set A=3,6,9,12set\text{ }A=\\{3,6,9,12\\} and
R=(3,3),(6,6),(9,9),(12,12),(6,12),(3,9),(3,12),(3,6)R=\\{(3,3),(6,6),(9,9),(12,12),(6,12),(3,9),(3,12),(3,6)\\}
We can see that,
(3,3),(6,6),(9,9),(12,12)R\\{(3,3),(6,6),(9,9),(12,12)\\}\in R
For all 3,6,9,12A\\{3,6,9,12\\}\in A
Which shows that it is a reflexive relation.
For symmetric relation:
Let A be any set with any values, then
A relation R on A will be said to be a symmetric relation if (a,b)R(a,b)\in R and also (b,a)R(b,a)\in R for all a,bAa,b\in A.
In the question, it is given that

& set\text{ }A=\\{3,6,9,12\\} \\\ & R=\\{(3,3),(6,6),(9,9),(12,12),(6,12),(3,9),(3,12),(3,6)\\} \\\ \end{aligned}$$ We can see that $$(3,9)\in R$$but $$(9,3)\notin R$$ for all $$\\{3,6,9,12\\}\in A$$ Hence it is proved that it is not a symmetric relation. For transitive relation: Let A be a set of any values, then A relation R on A will be transitive if $$(a,b)\in R$$and $$(b,c)\in R$$then $$(a,c)\in R$$. In the above question, it is given that $$\begin{aligned} & set\text{ }A=\\{3,6,9,12\\} \\\ & R=\\{(3,3),(6,6),(9,9),(12,12),(6,12),(3,9),(3,12),(3,6)\\} \\\ \end{aligned}$$ We can see that, $$(3,6)\in R$$and $$(6,12)\in R$$ then , $$(3,12)\in R$$for all {$$3,6,9,12\\}\in A$$. Hence ,it is proved that the relation is transitive. So from the above examples, it is proved that the relation R is not a symmetric relation but reflexive and transitive only. **So, the correct answer is “Option 3”.** **Note:** The set of all first coordinates of the elements of relation R is called the domain of R and the set of all second coordinates of the elements of relation R is called the range of R. Any set is a subset of itself. A relation in which none of the elements of the sets are related to each other is known as an empty relation and a Universal relation is one in which every element of set A is related to every element of set B.