Let P = { (x, y) | x^2 + y^2 = 1, x, y in mat | vedclass

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(B) relation $P$ on a set $A$ is reflexive if $(x, x) \in P$ for all $x \in A$. Here,$x^2 + x^2 = 1 \Rightarrow 2x^2 = 1 \Rightarrow x = \pm \frac{1}{

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Symmetric

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(B) relation $P$ on a set $A$ is reflexive if $(x, x) \in P$ for all $x \in A$. Here,$x^2 + x^2 = 1 \Rightarrow 2x^2 = 1 \Rightarrow x = \pm \frac{1}{\sqrt{2}}$. Since this is not true for all $x \in \mathbb{R}$,$P$ is not reflexive.$A$ relation $P$ is symmetric if $(x, y) \in P \Rightarrow (y, x) \in P$. Given $x^2 + y^2 = 1$,it follows that $y^2 + x^2 = 1$. Thus,if $(x, y) \in P$,then $(y, x) \in P$. Hence,$P$ is symmetric.$A$ relation $P$ is transitive if $(x, y) \in P$ and $(y, z) \in P \Rightarrow (x, z) \in P$. Let $(x, y) = (1, 0)$ and $(y, z) = (0, 1)$. Both are in $P$ because $1^2 + 0^2 = 1$ and $0^2 + 1^2 = 1$. However,$(x, z) = (1, 1)$ is not in $P$ because $1^2 + 1^2 = 2 
eq 1$. Thus,$P$ is not transitive.

Let $P = \{ (x, y) | x^2 + y^2 = 1, x, y \in \mathbb{R} \}$. Then $P$ is:

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