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.

The given relation is not reflexive as (1,1)∈R×R but (1,1)∈S since 12+12=2=1.As this relation is not reflexive so it can't be an equivalance relation.

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Let S be a relation on the set R of all real numbers defined b

Question

Let $S$ be a relation on the set $R$ of all real numbers defined by $S = {(a, b) \in R \times R : a^{2} + b^{2} = 1}$ . Prove that $S$ is not an equivalence relation on $R$ .

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Harper Discussed

Let

S

be a relation on the set

R

of all real numbers defined by

S = {(a, b) \in R \times R : a^{2} + b^{2} = 1}

. Prove that

S

is not an equivalence relation on

R

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14 mins ago

Text solutionVerified

The given relation is not reflexive as

(1, 1) \in R \times R

but

(1, 1) \neq \in S

since

1^{2} + 1^{2} = 2 \neq = 1

As this relation is not reflexive so it can't be an equivalance relation.

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Question Text	Let $S$ be a relation on the set $R$ of all real numbers defined by $S = {(a, b) \in R \times R : a^{2} + b^{2} = 1}$ . Prove that $S$ is not an equivalence relation on $R$ .
Answer Type	Text solution:1
Upvotes	150

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.

Text solutionVerified

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Let $S$ be a relation on the set $R$ of all real numbers defined by $S = {(a, b) \in R \times R : a^{2} + b^{2} = 1}$ . Prove that $S$ is not an equivalence relation on $R$ .