P - x - y - x - y - R - x 2- y 2-1- then P isA- transitiveB- reflexiveC- equivalenceD- symmetric

The correct option is D symmetric P = { (x, y) : x, y ϵ R, x^2 + y^2 = 1 } (a,a) ∉ P, (a,b) ϵ P ⇒ (b,a) ϵ P ∴R is symmetric ...

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Symmetric Relations

P = x , y : x...

Question

$P = {(x, y) : x, y ϵ R, x^{2} + y^{2} = 1}, t h e n P i s$

reflexive

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symmetric

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transitive

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equivalence

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