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Any relation P is defined in set \[{{R}_{0}}\] of non-zero real numbers by following ways:
$xPy\Leftrightarrow x{}^\text{2}\text{ }+\text{ }y{}^\text{2}=1$

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Last updated date: 20th Jul 2024
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Answer
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Hint: Type of question is based on the Relation. As in this question we had to find out the any relation that ‘P’ possess that can be Reflexive, symmetric or transitive or all three, from the condition given i.e. $xPy\Leftrightarrow x{}^\text{2}\text{ }+\text{ }y{}^\text{2}=1$. As to find the relation we do not have any formula, rather we had to find the possible case if there is any to prove the relation wrong, if it is proven wrong then there is no relation, otherwise there will be.

Complete step-by-step solution:
As we know that reflexive relation is the one in which every element maps to itself. For example, consider a set A = {1, 2,}. Now, the reflexive relation will be \[R\text{ }=\text{ }\left( 1,\text{ }1 \right),\text{ }\left( 2,\text{ }2 \right),\]. Hence, a relation is reflexive \[if:\text{ }\left( a,\text{ }a \right)\]. And a set A is called symmetric if \[\left( b,a \right)\text{ }\in \text{ }R\] holds when \[\left( a,b \right)\in \text{ }R\] i.e. relation \[R=\left\{ \left( 4,5 \right),\left( 5,4 \right),\left( 6,5 \right),\left( 5,6 \right) \right\}\] on set \[A=\left\{ 4,5,6 \right\}\] is symmetric and A relation R on set A is called Transitive if \[xRy\]and \[yRz\]implies\[xRz,\forall x,y,z\in A\]. As the relation\[R=\left\{ \left( 1,2 \right),\left( 2,3 \right),\left( 1,3 \right) \right\}\] on set \[A=\left\{ 1,2,3 \right\}\] is transitive.
Moving ahead with the question to find the relation of ‘P’ with respect to the given condition i.e. $xPy\Leftrightarrow x{}^\text{2}\text{ }+\text{ }y{}^\text{2}=1$
First we will check for reflexive relation; then let \[3\in {{R}_{0}}\] as 3 is non-zero real number; than
Then for reflexive $(x,y)=(x,x)$ then for the given condition if 3 satisfies then it is a reflective relation otherwise not. So;
$\begin{align}
  & xPy\Leftrightarrow x{}^\text{2}\text{ }+\text{ }y{}^\text{2}=1 \\
 & x,y\in 3 \\
 & 3{}^\text{2}\text{ }+\text{ 3}{}^\text{2}=1 \\
 & 9+9\ne 1 \\
\end{align}$
As we can clearly say that it does not satisfy the condition, hence ‘P’ will not be a reflexive relation for the given condition.
Now we will check for symmetric relation;
Now Let us assume \[a,b\in {{R}_{0}}\]. In this way\[\left( a,\text{ }b \right)\in P~\] ;
$\begin{align}
  & xPy\Leftrightarrow x{}^\text{2}\text{ }+\text{ }y{}^\text{2}=1 \\
 & a{}^\text{2}\text{ }+\text{ b}{}^\text{2}=1 \\
\end{align}$
It can also be written as$\text{ b}{}^\text{2}+a{}^\text{2}=1$, means \[\left( a,\text{ }b \right)\in P~\] then \[\left( b,a \right)\in P~\] will also be there which is the condition of symmetric.
So, P is a symmetric relation.
Now moving ahead to check transitive relation, let the \[\left( \dfrac{1}{2},\text{ }\dfrac{\sqrt{3}}{2} \right)\in P~\] as both \[\dfrac{1}{2}\] and \[\dfrac{\sqrt{3}}{2}\] belongs to non-zero real numbers. So when assumed condition applied to given function we will have;
\[\begin{align}
  & xPy\Leftrightarrow x{}^\text{2}\text{ }+\text{ }y{}^\text{2}=1 \\
 & \left( \dfrac{1}{2},\text{ }\dfrac{\sqrt{3}}{2} \right)\in P~ \\
 & \Rightarrow \left( \dfrac{1}{2} \right){}^\text{2}\text{ }+\left( \dfrac{\sqrt{3}}{2} \right){}^\text{2}=1 \\
 & \Rightarrow \dfrac{1}{4}\text{ }+\dfrac{3}{4}=1 \\
 & \Rightarrow 1=1 \\
\end{align}\]
And if we take \[\left( \dfrac{\sqrt{3}}{2},-\dfrac{1}{2} \right)\in P~\]
\[\begin{align}
  & xPy\Leftrightarrow x{}^\text{2}\text{ }+\text{ }y{}^\text{2}=1 \\
 & \left( \dfrac{\sqrt{3}}{2},-\dfrac{1}{2} \right)\in P~ \\
 & \Rightarrow \text{ }\left( \dfrac{\sqrt{3}}{2} \right){}^\text{2}+\left( -\dfrac{1}{2} \right){}^\text{2}=1 \\
 & \Rightarrow \dfrac{3}{4}+\dfrac{1}{4}\text{ }=1 \\
 & \Rightarrow 1=1 \\
\end{align}\]
But when we will take \[\left( \dfrac{1}{2},-\dfrac{1}{2} \right)\in P~\]then;
\[\begin{align}
  & xPy\Leftrightarrow x{}^\text{2}\text{ }+\text{ }y{}^\text{2}=1 \\
 & \left( \dfrac{1}{2},-\dfrac{1}{2} \right)\in P~ \\
 & \Rightarrow \text{ }\left( \dfrac{1}{2} \right){}^\text{2}+\left( -\dfrac{1}{2} \right){}^\text{2}=1 \\
 & \Rightarrow \dfrac{1}{4}+\dfrac{1}{4}\text{ }=1 \\
 & \Rightarrow \dfrac{1}{8}\ne 1 \\
\end{align}\]
Hence it is not transitive Relation.
So ‘P’ is only a symmetric Relation.

Note: As mentioned we do not have any formula to find the relation that might exist in a function, but if we know exactly what does the relation mean then through taking different examples as we did in our case.