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

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.