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**Hint:**First we can take the set \[A = \left\{ {1,2,\left. 3 \right\}} \right.\] then we find the smallest equivalence relation on that set.

By the way, smallest equivalence relation, an equivalence relation on a set of ordered pairs, and one set can be a subset of another.

We have to check the smallest equivalence relation on a set A, whether it satisfies the properties of equivalence relation, they are reflective, symmetric and transitive.

Finally we find the smallest equivalence relation on the set A.

**Complete step-by-step answer:**

It is given that the set \[A = \left\{ {1,2,\left. 3 \right\}} \right.\]

Here first we take the set A and write it for equivalence relation

That is, \[A \times A = \left\{ {(1,1)(1,2)(1,3)(2,1)(2,2)(2,3)(3,1)(3,2)(3,\left. {3)} \right\}} \right.\]

Here the above set is Equivalence Relation for set A

Now, we check if the set \[A \times A\] satisfies the properties are reflexive, symmetric and transitive.

Here we take \[x = y = z = \left\{ {1,2,3} \right\}\]

First we check Reflexive: for all $x \in A$ then $(x,x) \in A \times A$

\[\left\{ {1,2,3} \right\} \in A\], $\left( {1,1} \right)\left( {2,2} \right)\left( {3,3} \right) \in A \times A$

$\therefore $The relation is reflective.

Also, Symmetric: for all $x,{\text{ }}y \in A$ then $\left( {x,{\text{ }}y} \right) \in A \times A = > \left( {y,{\text{ }}x} \right) \in A \times A$

$\left( {1,{\text{ }}2} \right) = > \left( {2,{\text{ }}1} \right) \in A \times A$

$\left( {1,{\text{ }}3} \right) = > \left( {3,{\text{ }}1} \right) \in A \times A$

$\left( {2,{\text{ }}3} \right) = > \left( {3,{\text{ }}2} \right) \in A \times A$ etc.,

Similarly we can write,

$\left( {1,1} \right) = > \left( {1,1} \right) \in A \times A$

$\left( {2,2} \right) = > \left( {2,2} \right) \in A \times A$

$\left( {3,3} \right) = > \left( {3,3} \right) \in A \times A$

$\therefore $ The relation is symmetric.

Transitive: for all $x,y,z \in A$, then $\left( {x,y} \right) \in A \times A,\left( {y,z} \right) \in A \times A = > \left( {x,z} \right) \in A \times A$,

\[\left( {1,2} \right) \in A \times A,\left( {2,3} \right) \in A \times A = > \left( {1,3} \right) \in A \times A\]

\[\left( {2,1} \right) \in A \times A,\left( {1,3} \right) \in A \times A = > \left( {2,3} \right) \in A \times A\]etc.,

Similarly we can write,

\[\left( {1,1} \right) \in A \times A,\left( {1,1} \right) \in A \times A = > \left( {1,1} \right) \in A \times A\]

\[\left( {2,2} \right) \in A \times A,\left( {2,2} \right) \in A \times A = > \left( {2,2} \right) \in A \times A\]

\[\left( {3,3} \right) \in A \times A,\left( {3,3} \right) \in A \times A = > \left( {3,3} \right) \in A \times A\]

$\therefore $The relation is transitive.

Finally we get the relation is equivalence relation and we can write the smallest equivalence relation

\[ \Rightarrow A \times A = \left\{ {(1,1)(2,2)(3,3\left. ) \right\}} \right.\]

The set satisfies the conditions and it is the smallest.

**Note:**Here, the explanation is large but the concept is important to understand the equivalence relation on the set. In this type of question, suppose it is difficult to find the smallest set, write all the possible elements and try to look at it, you get any similarities in all conditions, then the elements are the smallest equivalence relation.

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