On \[\mathbb{R}\], the relation \['\rho '\] be defined as \[x\rho y\] holds if and only if \[x-y\] is zero or irrational. Then
(a) \['\rho '\] is reflexive and transitive but not symmetric
(b) \['\rho '\] is reflexive and symmetric but not transitive
(c) \['\rho '\] is symmetric and transitive but not reflexive
(d) \['\rho '\] is an equivalence relation.

Hint: For solving this problem we go through the definitions of reflexive, transitive and symmetric relations. Reflexive relation states that for a domain \[\mathbb{R}\] if \[\left( x,y \right)=\left( a,a \right)\] satisfies the relation when \[a\in \mathbb{R}\] then we can say that relation is reflexive.

Complete step-by-step solution
For a domain \[\mathbb{R}\] if \[\left( a,b \right),\left( b,c \right)\] satisfies the relation and also \[\left( a,c \right)\] satisfies the relation then we can say that the relation is transitive.
For a domain \[\mathbb{R}\] if \[\left( a,b \right)\] satisfies the relation and \[\left( b,a \right)\] also satisfies the relation we can say that the relation is symmetric.
If the relation satisfies reflexive, transitive, and symmetric we can say that relation is an equivalence relation.
By using these definitions we can find about the nature of \['\rho '\].
Let us assume that \[x\in \mathbb{R}\] then we can say that
\[\begin{align}
  & \Rightarrow x-x=0 \\
 & \Rightarrow x\rho x \\
\end{align}\]
We know that reflexive relation states that for a domain \[\mathbb{R}\] if \[\left( x,y \right)=\left( a,a \right)\] satisfies the relation when \[a\in \mathbb{R}\] then we can say that relation is reflexive.
Here, \['x'\] satisfies the relation \['\rho '\] for all \['x'\]
So, we can say that \['\rho '\] is reflexive.
Now let us consider \[a,b,c\in \mathbb{R}\].
Let us assume \[\left( a,b \right),\left( b,c \right)\] satisfies the relation \['\rho '\] that is
\[\Rightarrow a-b=0\text{ and }b-c=0\]
From this, we can say that \[a-c=0\]
Similarly, if \[a-b,b-c\] are irrational numbers, so is \[a-c\].
We know that for a domain \[\mathbb{R}\] if \[\left( a,b \right),\left( b,c \right)\] satisfies the relation and also \[\left( a,c \right)\] satisfies the relation then we can say that the relation is transitive.
From this, we can say that the relation \['\rho '\] is transitive.
Now, let us assume that \[a,b\in \mathbb{R}\] and \[\left( a,b \right)\] satisfies \['\rho '\]
So we can write
\[\Rightarrow a-b=0\]
This can also be written as
\[\Rightarrow b-a=0\]
Similarly, if \[a-b\] is an irrational number then \[b-a\] should also be an irrational number.
We can say that \[\left( b,a \right)\] satisfies the relation \['\rho '\].
We know that for a domain \[\mathbb{R}\] if \[\left( a,b \right)\] satisfies the relation and \[\left( b,a \right)\] also satisfies the relation we can say that the relation is symmetric.
So, the relation \['\rho '\] is symmetric.
As the relation \['\rho '\] satisfies reflexive, transitive and symmetric we can say that \['\rho '\] is an equivalence relation.
So, option (d) is the correct answer.

Note: Students will make mistakes in checking the relation by definition. We need to check all possibilities in the given domain. If by any chance anyone value does not satisfy any type of relationship then we are not allowed to say that relation comes under that respective type. Checking the type of relationship using the definition in the general method is best rather than taking examples, because we may miss some cases in examples.