Answer
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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.
$\begin{align}
& |a-b|\le 1 \\
& \Rightarrow |-\left( b-a \right)|\le 1 \\
& \Rightarrow |b-a|\le 1 \\
\end{align}$
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:
$\begin{align}
& -1-1\le a-b+b-c\le 1+1 \\
& \Rightarrow -2\le a-c\le 2 \\
\end{align}$
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.
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.
$\begin{align}
& |a-b|\le 1 \\
& \Rightarrow |-\left( b-a \right)|\le 1 \\
& \Rightarrow |b-a|\le 1 \\
\end{align}$
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:
$\begin{align}
& -1-1\le a-b+b-c\le 1+1 \\
& \Rightarrow -2\le a-c\le 2 \\
\end{align}$
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.
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