Question

If R and S are transitive relations on a set A, then prove \[R\cup S\] may not be a transitive relation on A.

Answer 1

Hint: In a reflexive relation, for every \[a\in A,\left( a,a \right)\in R\]. A relation is a symmetric relation R on a set A if \[\left( a,b \right)\in R\] then \[\left( b,a \right)\in R\], for all \[a,b\in A\]. A relation is a transitive relation if \[\left( a,b \right)\in R\], \[\left( b,c \right)\in R\], then \[\left( a,c \right)\in R\] for all \[a,b,c\in A\]. If a relation is reflexive, symmetric and transitive then the relation is said to be equivalence relation. Let us assume a set \[A=\left\{ a,b,c \right\}\]. Now let us assume \[R=\left\{ \left( a,a \right),\left( a,b \right),\left( b,a \right),\left( b,b \right) \right\}\] and \[S=\left\{ \left( b,b \right),\left( b,c \right),\left( c,b \right),\left( c,c \right) \right\}\] because we know that both R and S are transitive. Now let us find \[R\cup S\]. Now from the definition of transitive relation, we can find whether \[R\cup S\]is transitive or not.

Complete step-by-step answer:
Before solving the question, we should know that a relation between two sets is a collection of ordered pairs containing one object from each set. If the object x is from the first set and the object y is from the second set, then the objects are said to be related if the ordered pair \[\left( x,y \right)\] is in the relation. Here the values of x are said to be the domain of the function and the values of y represent the range of the function.
REFLEXIVE RELATION:
A relation is a reflexive relation if every element of set A maps to itself. In a reflexive relation, for every \[a\in A,\left( a,a \right)\in R\].
SYMMETRIC RELATION:
A relation is a symmetric relation R on a set A if \[\left( a,b \right)\in R\] then \[\left( b,a \right)\in R\], for all \[a,b\in A\].
TRANSITIVE RELATION:
A relation is a transitive relation if \[\left( a,b \right)\in R\], \[\left( b,c \right)\in R\], then \[\left( a,c \right)\in R\]for all \[a,b,c\in A\].
EQUIVALENCE RELATION:
If a relation is reflexive, symmetric and transitive then the relation is said to be equivalence relation.

Let us assume a set \[A=\left\{ a,b,c \right\}\].
Now let us assume \[R=\left\{ \left( a,a \right),\left( a,b \right),\left( b,a \right),\left( b,b \right) \right\}\] and \[S=\left\{ \left( b,b \right),\left( b,c \right),\left( c,b \right),\left( c,c \right) \right\}\] because we know that both R and S are transitive.
\[R\cup S=\left\{ \left( a,a \right),\left( a,b \right),\left( b,a \right),\left( b,b \right),\left( b,b \right),\left( b,c \right),\left( c,b \right),\left( c,c \right) \right\}\]
From the definition of transitive relation, we can say that \[\left( a,b \right)\] belongs to \[R\cup S\], \[\left( b,c \right)\] belongs to \[R\cup S\] but \[\left( a,c \right)\] does not belongs to \[R\cup S\].
So, we can prove that \[R\cup S\] may not be a transitive.

Note: Students should have a clear view about the concept of reflexive, symmetric and transitive relations. If a small misconception is there, then students cannot solve this problem. So, this misconception should be avoided. Students should also be able to roster the form of a set in a correct manner. If one cannot write the roster form is written incorrectly, then we cannot get the correct answer.