Question

Let $R={(3, 3), (6, 6), (9, 9), (12, 12), (6, 12), (3, 9), (3, 12), (3, 6)}$ be a relation on set $A={3, 6, 9, 12}$. The relation is
A. an equivalence relation.
B. reflexive and symmetric only.
C. reflexive and transitive only.
D. reflexive only.

Answer 1

Hint: We use the definition of reflexive, symmetric and transitive relation to get our desired solution.

Formula used: Let A be a non-empty set and R be a relation on A then (i) R is reflexive if $(a, a)$ \[ \in \]$R$, (ii) R is symmetric if $(a, b)$ \[ \in \]$R$ \[ \Rightarrow \]$(b, a)$ \[ \in \]$R$, (iii) R is transitive if $(a, b), (b, c)$ \[ \in \]$R$ \[ \Rightarrow \]$(a, c)$ \[ \in \]$R$. If R is reflexive, symmetric and transitive then R is an equivalence relation.

Complete step by step solution: Here, $A={3, 6, 9, 12}$, $R={(3, 3), (6, 6), (9, 9), (12, 12), (6, 12), (3, 9), (3, 12), (3, 6)}$.
Here, $(3, 3), (6, 6), (9, 9)$ and $(12, 12)$ \[ \in \] $R$. That means $(a, a)$ \[ \in \]$R$ for all a\[ \in \]A. So, R is reflexive.
Here, $(3, 3), (6, 6), (9, 9)$ and $(12, 12)$ \[ \in \] $R$ but $(6, 12)$ \[ \in \]$R$ \[ \Rightarrow \] $(12, 6)$\[ \notin \] $R$. Similarly, it is for other elements $(3, 9), (3, 12)$ and $(3, 6)$ in R.
Therefore, $(a, b)$ \[ \in \]$R$ \[ \Rightarrow \]$(b, a)$ \[ \notin \]R for all a, b \[ \in \]A.
So, R is not symmetric.
$(3, 3), (6, 6), (9, 9)$ and $(12, 12)$ \[ \in \] $R$
$(3, 6), (6, 12)$ \[ \in \]$R$ \[ \Rightarrow \]$(3, 12)$ \[ \in \]$R$.
Therefore, $(a, b), (b, c)$ \[ \in \]$R$ \[ \Rightarrow \]$(a, c)$ \[ \in \]$R$ for all $a, b, c$ \[ \in \]A.
So, R is transitive.

So, Option ‘C’ is correct

Note: Students often confuse the element $(3, 9)$. Using this element we can reach whether the relation is symmetric or not.