
Which type of relation $ {x^2} = xy $ is:
(A) Symmetric
(B) Reflexive and transitive
(C) Transitive
(D) None of the above
Answer
472.2k+ views
Hint: Here, in the given question, we are given a relation and we need to find which type of relation it is from the given options. We will check for every type of relation individually by using their property which type of relation is given in the question.
Reflexive property-
The reflexive property states that for every real number $ x $ , $ x = x $
Symmetric property-
The symmetric property states that for all real numbers $ x $ and $ y $ , if $ x = y $ , then $ y = x $ .
Transitive property-
The transitive property states that for all real numbers $ x $ , $ y $ and $ z $ , if $ x = y $ and $ y = z $ , then $ x = z $ .
Complete answer:
We have, $ {x^2} = xy $
For reflexive, we have $ xRx $ .
$ \Rightarrow {x^2} = x \cdot x $ , which is true. Hence, the given relation is reflexive.
For symmetric, we have
$ \left( {x,y} \right) \to {x^2} = x \cdot y $
$ \left( {y,x} \right) \to {y^2} = x \cdot y $
Which is not true. Hence, the given relation is not symmetric.
For transitive, we have: $ \left( {x,y} \right) $ , $ \left( {y,z} \right) $ , $ \left( {x,z} \right) $ .
$ {x^2} = xy $ , $ {y^2} = zy $ , $ {x^2} = xz $
$ \Rightarrow x = y $ , $ y = z $ , $ x = z $
$ \Rightarrow x = y = z $
Which is true. Hence, the given relation is transitive.
Thus, the given relation is both reflexive and transitive.
Therefore, the correct option is 2.
Note:
Remember that if a relation is reflexive, symmetric and transitive at the same time it is known as an equivalence relation. To solve this type of question one must know all types of relations and their properties. Representations of types of relations are given below in the table:
Reflexive property-
The reflexive property states that for every real number $ x $ , $ x = x $
Symmetric property-
The symmetric property states that for all real numbers $ x $ and $ y $ , if $ x = y $ , then $ y = x $ .
Transitive property-
The transitive property states that for all real numbers $ x $ , $ y $ and $ z $ , if $ x = y $ and $ y = z $ , then $ x = z $ .
Complete answer:
We have, $ {x^2} = xy $
For reflexive, we have $ xRx $ .
$ \Rightarrow {x^2} = x \cdot x $ , which is true. Hence, the given relation is reflexive.
For symmetric, we have
$ \left( {x,y} \right) \to {x^2} = x \cdot y $
$ \left( {y,x} \right) \to {y^2} = x \cdot y $
Which is not true. Hence, the given relation is not symmetric.
For transitive, we have: $ \left( {x,y} \right) $ , $ \left( {y,z} \right) $ , $ \left( {x,z} \right) $ .
$ {x^2} = xy $ , $ {y^2} = zy $ , $ {x^2} = xz $
$ \Rightarrow x = y $ , $ y = z $ , $ x = z $
$ \Rightarrow x = y = z $
Which is true. Hence, the given relation is transitive.
Thus, the given relation is both reflexive and transitive.
Therefore, the correct option is 2.
Note:
Remember that if a relation is reflexive, symmetric and transitive at the same time it is known as an equivalence relation. To solve this type of question one must know all types of relations and their properties. Representations of types of relations are given below in the table:
| Relation type | Condition |
| Empty Relation | $ R = \phi \subset A \times A $ |
| Universal Relation | $ R = A \times A $ |
| Identity Relation | $ I = \left\{ {\left( {a,a} \right),a \in A} \right\} $ |
| Inverse Relation | $ {R^{ - 1}} = \left\{ {\left( {b,a} \right):\left( {a,b} \right) \in R} \right\} $ |
| Reflexive Relation | $ \left( {a,a} \right) \in R $ |
| Symmetric Relation | $ aRb \Rightarrow bRa,\forall a,b \notin A $ |
| Transitive Relation | $ aRb{\text{ }}and{\text{ }}bRc \Rightarrow aRc\forall a,b,c \in A $ |
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