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Question: Let \[P = \\{ (x,y)/{x^2} + {y^2} = 1,x,y \in R\\} \]then P is \(1)\)Reflexive \(2)\)Symmetric ...

Let P=(x,y)/x2+y2=1,x,y∈RP = \\{ (x,y)/{x^2} + {y^2} = 1,x,y \in R\\} then P is
1)1)Reflexive
2)2)Symmetric
3)3)Transitive
4)4)Anti-symmetric

Explanation

Solution

First, we will need to know about the relations on reflexive, transitive, symmetric, and anti-symmetric.
After that, we can easily solve the given problem as P is defined as P=(x,y)/x2+y2=1,x,y∈RP = \\{ (x,y)/{x^2} + {y^2} = 1,x,y \in R\\} where x,yx,y are in real numbers
We will use the definitions of the given functions to check whether the options hold or not and hence we deduce the result.

Complete step by step answer:
Now, we have been given a relation and we have to find whether the relation is reflexive, symmetric, transitive, or a combination of these which is also known as the equivalent relation definition.
1)1)Reflexive
Staring with the reflexive relation, which is every element in the reflexive is mapped to itself.
That is, (x,x)∈R(x,x) \in R.
Apply this relation on the given function, P=(x,y)/x2+y2=1,x,y∈RP = \\{ (x,y)/{x^2} + {y^2} = 1,x,y \in R\\} then we get, P=(x,x)/x2+x2=1,x∈RP = \\{ (x,x)/{x^2} + {x^2} = 1,x \in R\\}
Further solving we get; x2+x2=1⇒2x2=1{x^2} + {x^2} = 1 \Rightarrow 2{x^2} = 1but which is not the reflexive because function gets x2=12{x^2} = \dfrac{1}{2}which is not x∉Rx \notin R and thus option 1)1)Reflexive incorrect.
2)2)Symmetric
Symmetric relations are those for which if (x,y)∈R(x,y) \in R then (y,x)∈R(y,x) \in R
From the given that we have P=(x,y)/x2+y2=1,x,y∈RP = \\{ (x,y)/{x^2} + {y^2} = 1,x,y \in R\\} then this function can be rewritten as P=(x,y)/x2+y2=1,x,y∈R⇒P=(y,x)/y2+x2=1,y,x∈RP = \\{ (x,y)/{x^2} + {y^2} = 1,x,y \in R\\} \Rightarrow P = \\{ (y,x)/{y^2} + {x^2} = 1,y,x \in R\\} just changing the position of (x,y)∈R(x,y) \in Rto (y,x)∈R(y,x) \in R.
Hence, we get P=(y,x)/y2+x2=1,y,x∈R⇒(y,x)∈RP = \\{ (y,x)/{y^2} + {x^2} = 1,y,x \in R\\} \Rightarrow (y,x) \in R
Therefore, option 2)2)Symmetric is correct.
3)3)Transitive
Transitive are those elements in which if (x,y)∈R(x,y) \in R and (y,z)∈R(y,z) \in Rthen (x,z)∈R(x,z) \in Rmust be held.
Which is impossible because we have, the functions P=(x,y)/x2+y2=1,x,y∈RP = \\{ (x,y)/{x^2} + {y^2} = 1,x,y \in R\\} and P=(y,z)/y2+z2=1,y,z∈RP = \\{ (y,z)/{y^2} + {z^2} = 1,y,z \in R\\} while equating these values we don’t have the chance to get x2+z2=1{x^2} + {z^2} = 1 , and thus option 3)3)Transitive is incorrect.
4)4)Anti-symmetric
Since R is symmetric, so R will not the antisymmetric because antisymmetric is absolute cannot go on both ways like the symmetric which is (x,y)∈R(x,y) \in R or (y,x)∉R(y,x) \notin Rwhenever x≠yx \ne yand unless x=yx = y like the symmetric.
Hence option 4)4)Anti-symmetric is incorrect.

So, the correct answer is “Option 2”.

Note: It is important to remind that xRyxRy which means a is related to b with the help of the given real number R or it can be also written in the form of (x,y)∈R(x,y) \in R
These types of given questions are solved using examples and counterexamples to verify the relation is correct or not.
Since if the relation satisfies the symmetric, transitive, and also reflexive then it is called the equivalence relation.
If symmetric property holds then anti-symmetric will be not hold unless x=yx = y