Relation

Question

For $a,b\epsilon Real$ define $aRb$ iff $a^{2}+b^{2}>2$.Is it Reflexive, Symmetric or Transitive?

Comments

(-1,3) and (3,-1) are there but (-1,-1) are not so not transitive, also not reflexive, it will be symmetric

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not reflexive, because {$(x,x) |\forall x$ such that $|x|\leq 1$} is not in relation

Answer 1

not reflexive, because {(x,x)|∀x such that |x|≤1} is not in relation

not transitive becoz  

for :  a²+b²>2  and b²+c²>2 . Adding the two, we get  a²+c² + 2b² > 4

for transitive relation to hold, we should have  a²+c² > 2. Assuming that transitive relation holds, we can say a²+c²  = 2 + ε

thus, 2 + ε + 2b² > 4

 b² > (2 - ε ) / 2 .... clearly  'ε' can take value >0 and make RHS max value ~ 1.  But, b² can have value in range[0,1). This will prove our assumption a²+c² > 2  to be false. Hence, transitive relation does not hold.

Symmetric relation holds as  a²+b²>2  and   b² + a² > 2 both holds true.