
If $R=\left\{ \left( 1,3 \right),\left( 4,2 \right),\left( 2,4 \right),\left( 2,3 \right),\left( 3,1 \right) \right\}$ is a relation on the set $A=\left\{ 1,2,3,4 \right\}$. Then, relation R is
A. a function
B. transitive
C. not symmetric
D. reflexive
Answer
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Hint: We explain the condition of relations and its different characteristics like reflexivity, symmetric and transitivity for the given relation. We try to find if the condition $\left( a,a \right)\in R;\forall a\in A$ satisfies for reflexivity. We also find the contradictory examples for the conditions of symmetric and transitivity as $\left( a,b \right)\in R\Rightarrow \left( b,a \right)\in R;\forall a,b\in A$ and \[\left( a,b \right)\in R,\left( b,c \right)\in R\Rightarrow \left( a,c \right)\in R;\forall a,b,c\in A\] respectively.
Complete step by step answer:
It is given that $R=\left\{ \left( 1,3 \right),\left( 4,2 \right),\left( 2,4 \right),\left( 2,3 \right),\left( 3,1 \right) \right\}$ where R be a relation on $A=\left\{ 1,2,3,4 \right\}$.
It cannot be a function as the image of the preimage 2 is both 4 and 3 which goes against definition of function.We have to check the reflexivity, symmetric and transitivity.If we can show one example of contradiction for any characteristics then we can say that the characteristics are not possible.
For reflexivity $\left( a,a \right)\in R;\forall a\in \mathbb{N}$. We can see that, $\left( 2,2 \right),\left( 3,3 \right)\notin R$ but $2,3\in \mathbb{N}$.The relation is not reflexive.
For symmetric $\left( a,b \right)\in R\Rightarrow \left( b,a \right)\in R;\forall a,b\in \mathbb{N}$. We can see that for $2,3\in A$, $\left( 2,3 \right)\in R$ but $\left( 3,2 \right)\notin R$. The relation is not symmetric.
For transitivity \[\left( a,b \right)\in R,\left( b,c \right)\in R\Rightarrow \left( a,c \right)\in R;\forall a,b,c\in \mathbb{N}\]. We can see that for $1,2,3\in A$, $\left( 2,3 \right)\in R$ and $\left( 3,1 \right)\in R$ but $\left( 2,1 \right)\notin R$. The relation is not transitive.
Therefore, the correct option is C.
Note: If a relation satisfies all three relations, then the relation is called Equivalence relation. Equivalence relations are massively used outside of mathematics, as they are simply a means of breaking some set of objects into separate subgroups.
Complete step by step answer:
It is given that $R=\left\{ \left( 1,3 \right),\left( 4,2 \right),\left( 2,4 \right),\left( 2,3 \right),\left( 3,1 \right) \right\}$ where R be a relation on $A=\left\{ 1,2,3,4 \right\}$.
It cannot be a function as the image of the preimage 2 is both 4 and 3 which goes against definition of function.We have to check the reflexivity, symmetric and transitivity.If we can show one example of contradiction for any characteristics then we can say that the characteristics are not possible.
For reflexivity $\left( a,a \right)\in R;\forall a\in \mathbb{N}$. We can see that, $\left( 2,2 \right),\left( 3,3 \right)\notin R$ but $2,3\in \mathbb{N}$.The relation is not reflexive.
For symmetric $\left( a,b \right)\in R\Rightarrow \left( b,a \right)\in R;\forall a,b\in \mathbb{N}$. We can see that for $2,3\in A$, $\left( 2,3 \right)\in R$ but $\left( 3,2 \right)\notin R$. The relation is not symmetric.
For transitivity \[\left( a,b \right)\in R,\left( b,c \right)\in R\Rightarrow \left( a,c \right)\in R;\forall a,b,c\in \mathbb{N}\]. We can see that for $1,2,3\in A$, $\left( 2,3 \right)\in R$ and $\left( 3,1 \right)\in R$ but $\left( 2,1 \right)\notin R$. The relation is not transitive.
Therefore, the correct option is C.
Note: If a relation satisfies all three relations, then the relation is called Equivalence relation. Equivalence relations are massively used outside of mathematics, as they are simply a means of breaking some set of objects into separate subgroups.
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