
Let R be the relation on the set N defined by {(x, y) |x, y $\in $ N, 2x+y=41}. Then R is
(a) Reflexive
(b) Symmetric
(c) Transitive
(d) None of these
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:
{(x, y) |x, y $\in $ N, 2x+y=41}
We will take an element (x, x) and check whether it belongs to R or not.
If (x, x) $\in $ R, then we can write:
$\begin{align}
& 2x+x=41 \\
& \Rightarrow 3x=41 \\
& \Rightarrow x=\dfrac{41}{3} \\
\end{align}$
Since, it is given that R is defined on the set N, but here the value of x does not belong to N.
Therefore, (x, x) $\notin $ R.
So, R is not reflexive.
Now, take an element (x, y) $\in $ R.
Therefore, 2x+y=41.
This does not imply that 2y+x=41.
It means that (y, x) $\notin $ R.
So, R is not symmetric.
Now, take elements (x, y) and (y, z) $\in $ R.
So, we have:
$\begin{align}
& 2x+y=41.........\left( 1 \right) \\
& 2y+z=41..........\left( 2 \right) \\
\end{align}$
On multiplying equation (1) by 2 and subtracting it from equation (2), we get:
$\begin{align}
& 2y+z-\left( 4x+2y \right)41-82 \\
& \Rightarrow 2y+z-4x-2y=-41 \\
& \Rightarrow z-4x=-41 \\
\end{align}$
So, (x, z) $\notin $ R.
This implies that R is not transitive.
Hence, option (d) 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:
{(x, y) |x, y $\in $ N, 2x+y=41}
We will take an element (x, x) and check whether it belongs to R or not.
If (x, x) $\in $ R, then we can write:
$\begin{align}
& 2x+x=41 \\
& \Rightarrow 3x=41 \\
& \Rightarrow x=\dfrac{41}{3} \\
\end{align}$
Since, it is given that R is defined on the set N, but here the value of x does not belong to N.
Therefore, (x, x) $\notin $ R.
So, R is not reflexive.
Now, take an element (x, y) $\in $ R.
Therefore, 2x+y=41.
This does not imply that 2y+x=41.
It means that (y, x) $\notin $ R.
So, R is not symmetric.
Now, take elements (x, y) and (y, z) $\in $ R.
So, we have:
$\begin{align}
& 2x+y=41.........\left( 1 \right) \\
& 2y+z=41..........\left( 2 \right) \\
\end{align}$
On multiplying equation (1) by 2 and subtracting it from equation (2), we get:
$\begin{align}
& 2y+z-\left( 4x+2y \right)41-82 \\
& \Rightarrow 2y+z-4x-2y=-41 \\
& \Rightarrow z-4x=-41 \\
\end{align}$
So, (x, z) $\notin $ R.
This implies that R is not transitive.
Hence, option (d) 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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