Question

Let S be the set of all real numbers and let R be a relation on S, defined by $a\text{ }R\text{ }b\text{ }\Leftrightarrow \text{ }\\!\\!|\\!\\!\text{ a-b }\\!\\!|\\!\\!\text{ }\le \text{1}$. Then R, is
(a) reflexive and symmetric but not transitive
(b) reflexive and transitive but not symmetric
(c) symmetric and transitive but not reflexive
(d) an equivalence relation

Answer 1

Hint:Here, we will use the definitions of reflexive, symmetric and transitive relations to check whether the given relations are reflexive, symmetric or transitive.

Complete step-by-step answer:
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 (x, y) is in the relation.
A relation R is reflexive if each element is related itself, i.e. (a, a) $\in$ R, where a is an element of the domain.
A relation R is symmetric in case if any one element is related to any other element, then the second element is related to the first, i.e. if (x, y) $\in$ R then (y, x) $\in$ R, where x and y are the elements of domain and range respectively.
A relation R is transitive in case if any one element is related to a second and that second element is related to a third, then the first element is related to the third, i.e. if (x, y) $\in$ R and (y, z) $\in$ R then (x, z) $\in$ R.
Here, the given relation is:
$a\text{ }R\text{ }b\text{ }\Leftrightarrow \text{ }\\!\\!|\\!\\!\text{ a-b }\\!\\!|\\!\\!\text{ }\le \text{1}$
Let us take an ordered pair (a, a).
|a-a|=0, which is always less than 1.
So, R is reflexive.
Now, let us suppose that (a, b) $\in$ R.
For R to be symmetric, we will have to show that (b, a) $\in$ R.
$\begin{aligned} & |a-b|\le 1 \\\ & \Rightarrow |-\left( b-a \right)|\le 1 \\\ & \Rightarrow |b-a|\le 1 \\\ \end{aligned}$
So, (b, a) $\in$ R.
This means that R is symmetric.
Now, let us suppose that a-b $\in$ R and b-c $\in$ R, then R would be transitive if a-c $\in$ R.
Since, (a, b) $\in$ R and (b, c) $\in$ R. So, we can write:
$|a-b|\le 1$
$\Rightarrow -1\le a-b\le 1..........\left( 1 \right)$
And, $|b-c|\le 1$
$\Rightarrow -1\le b-c\le 1..........\left( 2 \right)$
On adding equations (1) and (2), we get:
$\begin{aligned} & -1-1\le a-b+b-c\le 1+1 \\\ & \Rightarrow -2\le a-c\le 2 \\\ \end{aligned}$
It means that $|a-c|\notin R$.
This implies that R is not transitive.
Therefore, R is reflexive and symmetric but not transitive.
Hence, option (a) is the correct answer.

Note: Students should note here that for a relation to be a particular type of relation, i.e. reflexive, symmetric or transitive, all its elements must satisfy the required conditions. If we are able to find even a single counter example, i.e. if any of the elements doesn’t satisfy the criteria, then we can’t proceed further.