Question

Let R be the relation on the set N of natural numbers be defined as (x, y) $\in$ R if and only if ${{x}^{2}}-4xy+3{{y}^{2}}=0$ for all x, y $\in$ N. Then R is
(1) Reflexive
(2) Symmetric
(3) Transitive
(4) 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:
We will take an element (x, x) and check whether it belongs to R or not.
as (x, y) $\in$ R if and only if ${{x}^{2}}-4xy+3{{y}^{2}}=0$ .
Let us take an element (x, x) and check whether it belongs to R or not.
If (x, x) belongs to R, then ${{x}^{2}}-4\times x\times x+3\times {{x}^{2}}$ must be equal to 0.
We see that:
$\begin{aligned} & {{x}^{2}}-4\times x\times x+3\times {{x}^{2}} \\\ & =-3{{x}^{2}}+3{{x}^{2}}=0 \\\ \end{aligned}$
So, (x, x) $\in$ R.
So, R is reflexive.
Now, take an element (x, y) $\in$ R.
Then ${{x}^{2}}-4xy+3{{y}^{2}}=0$.
This does not imply that ${{y}^{2}}-4yx+3{{x}^{2}}=0$.
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:
$\begin{aligned} & {{x}^{2}}-4xy+3{{y}^{2}}=0.........\left( 1 \right) \\\ & {{y}^{2}}-4yz+3{{z}^{2}}=0.........\left( 2 \right) \\\ \end{aligned}$
On multiplying equation (2) by 3 and subtracting it from equation (1), we get:

& {{x}^{2}}-4xy+3{{y}^{2}}-\left( 3{{y}^{2}}-12yz+9{{z}^{2}} \right)=0 \\\ & \Rightarrow {{x}^{2}}-4xy+3{{y}^{2}}-3{{y}^{2}}+12yz-9{{z}^{2}}=0 \\\ & \Rightarrow {{x}^{2}}-4xy+12yz-9{{z}^{2}}=0 \\\ \end{aligned}$$ So, (x, z) $\notin $ R. This implies that R is not transitive. Therefore, R is reflexive but neither symmetric nor 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.