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Question
Let S be a relation on the set R of all real numbers defined by
S = {(a, b) ∈ R × R : a2 + b2 = 1}
Prove that S is not an equivalence relation on R.
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Solution
We observe the following properties of S.
Reflexivity :
Let a be an arbitrary element of R. Then,
a ∈ R
⇒ a2+a2 ≠ 1∀ a ∈ R
⇒ (a, a) ∉ S
So, S is not reflexive on R.
Symmetry: Let (a, b) ∈ R
⇒ a2 + b2= 1
⇒ b2+ a2= 1
⇒ (b, a) ∈ S for all a, b ∈ R
So, S is symmetric on R.
Transitivity :
Let (a, b) and (b, c) ∈ S
⇒ a2+b2=1 and b2+c2=1
Adding the above two, we get
a2+c2=2−2b2≠1 for all a, b, c ∈ R
So, S is not transitive on R.
Hence, S is not an equivalence relation on R.
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