Let $A={1,2,3}$. How many minimum number of elements must be added to a relation $R$ containing $(1,2)$ and $(2,3)$ so that it becomes an equivalence relation?
My Attempt
We have, $(1,2)$ and $(2,3)$.
Checking for reflexive relation:
$(1,1)$, $(2,2)$ and $(3,3)$ must be added.
Checking for symmetry
$(2,1)$ and $(3,2)$ must be added.
How should I do further?
On
If you think of the relation as a graph, then you are looking for semi-connected components (connected by lines without arrows). The transitive closure is the union of the complete subgraph of each such component. Since
1, 2, and 3 are in the same component, which has $n^2$ arcs in its complete (sub)graph, where $n$ is the number of nodes. ...
An equivalence relation on $A$ has three properties: reflexivity, symmetry, and transitivity.
Reflexivity means that: "For all $a\in A$, $(a,a)\in R$." You have successfully added the pairs to make the relation reflexive by adding $(1,1)$, $(2,2)$, and $(3,3)$.
Symmetry means that: "If $(a,b)\in R$, then $(b,a)\in R$." You have successfully added the pairs (thus far) to make the relation symmetric by adding $(2,1)$ since $(1,2)$ is in the relation and $(3,2)$ since $(2,3)$ is in the relation.
Transitivity means that: "If $(a,b),(b,c)\in R$, then $(a,c)\in R$." You have not dealt with transitivity in your argument. In particular, there are pairs that you have in your relation of the form $(a,b)$ and $(b,c)$, but $(a,c)$ hasn't been added to the relation.
I teach these relations using dominos. In particular, for reflexivity, you've always got to have the dominos that have the same value on both sides. For symmetry, you're allowed to spin the dominos so that the left becomes the right. For transitivity, you're allowed to put two dominos together if they have the same value on one side (technically, this last description uses both symmetry and transitivity).
I don't want to give a partial answer because any partial answer is actually a full answer.