# Is it true that every relation which is symmetric and transitive is also reflexive ?

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Hint-Make use of the definitions of reflexive , symmetric, transitive functions and solve this.

A relation R on a set A is called reflexive if (a,a) $ \in $ R holds for every element a $ \in $ A .i.e. if set A = {a,b} then R = {(a,a), (b,b)} is reflexive relation.

A relation R on a set A is called symmetric if (b,a) $ \in $ R holds when (a,b) $ \in $ R.i.e. The relation R={(4,5),(5,4),(6,5),(5,6)} on set A={4,5,6} is symmetric

A relation R on a set A is called transitive if (a,b) $ \in $ R and (b,c) $ \in $ R then (a,c) $ \in $ R for all a,b,c $ \in $ A.i.e. Relation R={(1,2),(2,3),(1,3)} on set A={1,2,3} is transitive.

So, from these statements we can say that the given statement is false, Let us try to prove this, Let us prove it by taking a counter example

For example , take a relation R ={(1,1,),(1,2),(2,1),(2,2)} on A={1,2,3}

So, from this example we can clearly infer from the definition that the given relation is symmetric since(1,2),(2,1) $ \in $R and transitive since (1,2)(2,1) $ \in $R and also (1,1) $ \in $R but not reflexive since

(3,3) $ \notin $ R

So, from this we can write that the statement is given is false

Note: Only when the given statement is false, we can prove it by taking a counterexample else if the given statement is true , then we cannot solve it by taking a counter example.

A relation R on a set A is called reflexive if (a,a) $ \in $ R holds for every element a $ \in $ A .i.e. if set A = {a,b} then R = {(a,a), (b,b)} is reflexive relation.

A relation R on a set A is called symmetric if (b,a) $ \in $ R holds when (a,b) $ \in $ R.i.e. The relation R={(4,5),(5,4),(6,5),(5,6)} on set A={4,5,6} is symmetric

A relation R on a set A is called transitive if (a,b) $ \in $ R and (b,c) $ \in $ R then (a,c) $ \in $ R for all a,b,c $ \in $ A.i.e. Relation R={(1,2),(2,3),(1,3)} on set A={1,2,3} is transitive.

So, from these statements we can say that the given statement is false, Let us try to prove this, Let us prove it by taking a counter example

For example , take a relation R ={(1,1,),(1,2),(2,1),(2,2)} on A={1,2,3}

So, from this example we can clearly infer from the definition that the given relation is symmetric since(1,2),(2,1) $ \in $R and transitive since (1,2)(2,1) $ \in $R and also (1,1) $ \in $R but not reflexive since

(3,3) $ \notin $ R

So, from this we can write that the statement is given is false

Note: Only when the given statement is false, we can prove it by taking a counterexample else if the given statement is true , then we cannot solve it by taking a counter example.

Last updated date: 28th Sep 2023

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