Question

Let S be a relation on the set R of all real numbers defined by
\[S=\{(a,b)\in R\times R:{{a}^{2}}+{{b}^{2}}=1\}\]
Prove that S is not an equivalence relation on R.

Answer 1

Hint:-For solving these questions, we would be requiring knowledge about symmetric, reflexive and transitive functions.

Complete step-by-step answer:
A relation is a relationship between sets of values. In math, the relation is between the x-values and y-values of ordered pairs. The set of all x-values is called the domain, and the set of all y-values is called the range.
The types of relations are as follows
Reflexive Relation
A relation is a reflexive relation if every element of set A maps to itself. i.e. for every a \[\in \] A, (a, a) \[\in \] R.
Symmetric Relation
A symmetric relation is a relation R on a set A if (a, b) \[\in \] R then (b, a) \[\in \] R, for all a and b \[\in \] A.
Transitive Relation
If (a, b) \[\in \] R, (b, c) \[\in \] R, then (a, c) \[\in \] R, for all a, b, c \[\in \] A and this relation in set A is transitive.
Equivalence Relation
If and only if a relation is reflexive, symmetric and transitive, it is called an equivalence relation.
As mentioned in the question, we have to prove that S is not an equivalence relation on R.
Now, we can take the following example
\[\left( 1,\text{ }1 \right)\in R\times R\ ~but~\left( 1,1 \right)\notin S\]
 Because from the definition of relation S, we get
\[{{1}^{2}}+{{1}^{2}}=2\ne 1\]
So, as relation S does not contain (1, 1), therefore, it is not a reflexive relation.
As this relation is not reflexive, so it can't be an equivalence relation according to the definition of an equivalence relation.

Note:-The students can make an error if they don’t know about the definitions and the meaning of different types of relations.
Also, it is important to know about the different types of set representations that are the rooster form or the set builder form as without knowing these one could never understand the question properly.