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Question: Let S be the set of all points in a plane and let R be a relation in S defined by \(R=\left\\{ \left...

Let S be the set of all points in a plane and let R be a relation in S defined by R=\left\\{ \left( A,B \right):d\left( A,B \right)<2units \right\\} , where d(A,B)d\left( A,B \right) is the distance between the points A and B. Show that R is reflexive and symmetric but not transitive.

Explanation

Solution

Hint:First define set S, which follows relation R. To show that relation R is reflexive, find an example such that relation R is reflexive, find an example such that (a,a)R\left( a,a \right)\in R for aSa\in S . To show that relation R is symmetric, find an example such that (a,b)R\left( a,b \right)\in R and (b,a)R\left( b,a \right)\in R for a,bSa,b\in S . To show that relation R is not transitive, find an example such that (a,b)R\left( a,b \right)\in R and (b,c)R\left( b,c \right)\in R but (a,c)R\left( a,c \right)\notin R for a,b,cSa,b,c\in S .

Complete step-by-step answer:
Let ‘A’ be a set then Reflexivity, Symmetry and transitivity of a relation on set ‘A’ is defined as follows:
Reflexive relation: - A relation R on a set ‘A’ is said to be reflexive if every element of A is related to itself i.e. for every element say (a) in set A, (a,a)R\left( a,a \right)\in R .
Thus, R on a set ‘A’ is not reflexive if there exists an element aAa\in A such that (a,a)R\left( a,a \right)\notin R.
Symmetric Relation: - A relation R on a set ‘A’ is said to be symmetric relation if (a,b)R\left( a,b \right)\in R then (b,a)\left( b,a \right)must be belong to R. i.e. (a,b)R(b,a)R\left( a,b \right)\in R\Rightarrow \left( b,a \right)\in R For all a,bAa,b\in A.
Transitive relation:- A relation R on ‘A’ is said to be a transitive relation If (a,b)R\left( a,b \right)\in R and (b,c)R\left( b,c \right)\in R then (a,c)R\left( a,c \right)\in R.
I.e. (a,b)R\left( a,b \right)\in Rand (b,c)R(a,c)R\left( b,c \right)\in R\Rightarrow \left( a,c \right)\in R.
Given relation R=\left\\{ \left( A,B \right):d\left( A,B \right)<2units \right\\}.
Therefore,
R consists of every pair of points, for which distance between them is less than 2 units.
Check Reflexivity: If (a,a)R\left( a,a \right)\in R , foe aSa\in S , then R is reflexive.
Here, (A,A)R\left( A,A \right)\in R , because distance between point A and point A is zero as point A lies on itself. Therefore, as every point lies on itself, distance between any point and itself is zero, which is obviously less than 2 units.
Therefore, R is reflexive.

Check symmetricity: If (a,b)R\left( a,b \right)\in R for a.bSa.b\in S and (b,a)R\left( b,a \right)\in R , Then R is symmetric.
If (a,b)R\left( a,b \right)\in R , that means the distance between point a and point b is less than 2 units. Its reverse is always true. i.e. distance between point b and a is less than 2 units. This statements always holds
(b,a)R\Rightarrow \left( b,a \right)\in R ( distance between a and b is less than 2 units)\left( \because \text{ distance between a }{ and }\text{ b is less than 2 units} \right)
Hence, if (a,b)R\left( a,b \right)\in R , then (b,a)R\left( b,a \right)\in R .
Therefore, R is symmetric.
Check transitivity: If (a,b)R\left( a,b \right)\in R and (b,c)R\left( b,c \right)\in R for a,b,cSa,b,c\in S , then (a,c)R\left( a,c \right)\in R , then we can say that R is transitive.
If (a.b)R distance between points a and b is less than 2 units.\left( a.b \right)\in R\Rightarrow \text{ distance between points a and b is less than 2 units}\text{.}
And (b,c)R distance between points b and c is less than 2 units\left( b,c \right)\in R\Rightarrow \text{ distance between points b and c is less than 2 units} .
But this does not implies that the distance between a and c is less than 2 units.
i.e. (a,c)R\left( a,c \right)\notin R .
For example:
Let distance between A and B is 1 unit < 2 units and distance between B and C is 1.5 units < 2 units.
That means, (A,B)R\left( A,B \right)\in R and (B,C)R\left( B,C \right)\in R , But, here we can conclude that distance between A and C is (1+1.5=2.5units)\left( 1+1.5=2.5units \right) , which is not less than 2 units.
Hence, (A,C)R\left( A,C \right)\notin R .
Therefore, r is not transitive as (A,B)R and (B,C)R, but (A,C)R\left( A,B \right)\in R\text{ and }\left( B,C \right)\in R,\text{ but }\left( A,C \right)\notin R .
Hence, R is reflexive, symmetric but not transitive.

Note: Be careful that to prove that the given relation is not transitive, we need to just find an example which doesn’t satisfy the condition of transitivity. But to prove that it is reflexive and symmetric. We can’t use an example, instead we need to prove the condition of reflexivity and symmetric for all possible cases.