Question

Let S be the set of all real numbers and let R be a relation on S, defined by $\text{a R b }\Leftrightarrow {{\text{a}}^{2}}+{{b}^{2}}=1$ . Then R, is
(a) symmetric but neither reflexive nor transitive
(b) reflexive but neither symmetric nor transitive
(c) transitive but neither reflexive nor symmetric
(d) none of these

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:
$\text{a R b }\Leftrightarrow {{\text{a}}^{2}}+{{b}^{2}}=1$
To check reflexivity let (a,a) $\in$ R, then
${{a}^{2}}+{{a}^{2}}=2{{a}^{2}}\ne 1$ for all real numbers except $a=\dfrac{1}{\sqrt{2}}$ .
This implies that (a, a) $\notin $ R.
So, R is not reflexive.
Now, suppose that (a, b) $\in $ R, i.e. ${{a}^{2}}+{{b}^{2}}=1$ .
We can also write it as ${{b}^{2}}+{{a}^{2}}=1$, which means that (b, a) $\in $ R.
This means that R is symmetric.
Now, let us take an ordered pair (a, b) $\in $ R and (b, c) $\in $ R.
This means that ${{a}^{2}}+{{b}^{2}}=1.........\left( 1 \right)$
And, ${{b}^{2}}+{{c}^{2}}=1.........\left( 2 \right)$
On subtracting equation (2) from equation (1), we get:
$\begin{align}
  & {{a}^{2}}+{{b}^{2}}-\left( {{b}^{2}}+{{c}^{2}} \right)=1-1 \\
 & \Rightarrow {{a}^{2}}+{{b}^{2}}-{{b}^{2}}-{{c}^{2}}=0 \\
 & \Rightarrow {{a}^{2}}-{{c}^{2}}=0 \\
\end{align}$
So, (a, c) $\notin $ R.
This implies that R is not transitive.
Therefore, R is symmetric but neither reflexive 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.