Question

R is a relation on the set Z of integers and it is given by $\left( x,y \right)\in R\Leftrightarrow \left| x-y \right|\le 1$, then R is
(a)Reflexive and transitive
(b)Reflexive and symmetric
(c)Symmetric and transitive
(d)An equivalence relation

Answer 1

Hint: For solving this problem, we consider all options individually. By using the necessary conditions for a set to be reflexive, symmetric and transitive, we proceed for solving the question. If any of the options fails to satisfy the condition, it would be rejected.

Complete step-by-step answer:
The conditions which must be true for a set to be reflexive, transitive and symmetric are:
1)For a relation to be reflexive, $\left( a,a \right)\in R$.
2)For a relation to be symmetric, $\left( a,b \right)\in R\Rightarrow \left( b,a \right)\in R$.
3)For a relation to be transitive, $\left( a,b \right)\in R,\left( b,c \right)\in R\Rightarrow \left( a,c \right)\in R$.
4)For a relation to be equivalence, it should be reflexive, symmetric and transitive.
According to the problem statement, we are given a relation set $\left( x,y \right)\in R\Leftrightarrow \left| x-y \right|\le 1$.
For proving that this set is reflexive, we take $\left( x,x \right)\in R$. Therefore,
$\begin{align}
  & \left| x-x \right|\le 1 \\
 & \left| 0 \right|\le 1 \\
 & \therefore \left( x,x \right)\in R \\
\end{align}$
Hence, this equality is true. So, the given relation is reflexive.
To prove that this set is symmetric, we take $\left( x,y \right)\text{ and }\left( y,x \right)\in R$. Therefore
$\begin{align}
  & \left| y-x \right|\le 1\ldots (1) \\
 & \because \left| x-y \right|\le 1 \\
\end{align}$
Now, takin negative common in equation (1) and then by using mod property of sign change, we get
$\begin{align}
  & \therefore \left| -\left( x-y \right) \right|\le 1 \\
 & \left| x-y \right|\le 1 \\
 & \therefore \left( x,y \right)\in R\Rightarrow \left( y,x \right)\in R \\
\end{align}$
Therefore, the set is symmetric in nature.
Now, for transitive we consider $\left( x,y \right)\text{, }\left( y,z \right)\text{and }\left( z,x \right)\in R$
If we consider first two brackets and invoke relation, we get
$\begin{align}
  & \left| x-y \right|\le 1\ldots (2) \\
 & \left| y-z \right|\le 1\ldots (3) \\
\end{align}$
From these two equations, no conclusive evidence supports the existence of (z, x). hence, the set is not transitive.
Therefore, option (b) is correct.

Note: Students must remember all the necessary conditions for proving a set a reflexive, symmetric and transitive. All the options should be conclusively determined to give a final answer. If some particular operation yields the result, then only we can signify the relation to be reflexive, symmetric and transitive.