JEE Exam > JEE Questions > Let C = {(a, b): a2 + b2 = 1; a, b ∈ R} a rel...
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Let C = {(a, b): a2 + b2 = 1; a, b ∈ R} a relation on R, set of real numbers. Then C is
a)Equivalence relation
b)Reflexive
c)Transitive
d)Symmetric
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
Let C = {(a, b): a2 + b2 = 1; a, b ∈ R} a relation on R, set of real n...
Correct Answer :- D
Explanation:- Check for reflexive
Consider (a,a)
∴ a2+a2
=1 which is not always true.
If a=2
∴ 22+22
=1⇒4+4=1 which is false.
∴ R is not reflexive ---- ( 1 )
Check for symmetric
aRb⇒a2+b2=1
bRa⇒b2+a2 =1
Both the equation are the same and therefore will always be true.
∴ R is symmetric ---- ( 2 )
Check for transitive
aRb⇒a2+b2=1
bRc⇒b2+c2=1
∴ a2+c2=1 will not always be true.
Let a=−1,b=0 and c=1
∴ (−1)2+02=1,
= 02+12
=1 are true.
But (−1)2+12
=1 is false.
∴ R is not transitive ---- ( 3 )
Most Upvoted Answer
Let C = {(a, b): a2 + b2 = 1; a, b ∈ R} a relation on R, set of real n...
For symmetry condition, replace a by b and b by a. if equation remain same then it's symmetry
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Community Answer
Let C = {(a, b): a2 + b2 = 1; a, b ∈ R} a relation on R, set of real n...
Equivalence Relation:
An equivalence relation on a set is a relation that is reflexive, symmetric, and transitive. Let's check if the relation C satisfies these properties.
Reflexive:
A relation is reflexive if each element is related to itself. In this case, we need to check if (a, b) is related to (a, b) for all (a, b) ∈ R.
Let's substitute a = a and b = b in the given relation: a^2 - b^2 = 1
(a^2 - b^2) = (a^2 - b^2) simplifies to 0 = 0, which is true for all real numbers.
Since the relation holds for all (a, b) ∈ R, the relation C is reflexive.
Symmetric:
A relation is symmetric if whenever (a, b) is related to (c, d), then (c, d) is related to (a, b). In this case, we need to check if whenever a^2 - b^2 = 1, then b^2 - a^2 = 1.
Let's assume a^2 - b^2 = 1. Rearranging the equation, we get b^2 = a^2 - 1.
Now, let's substitute b^2 = a^2 - 1 in the original relation: (a^2 - 1) - a^2 = 1
This simplifies to -1 = 1, which is not true.
Since the relation does not hold for all (a, b) ∈ R, the relation C is not symmetric.
Transitive:
A relation is transitive if whenever (a, b) is related to (c, d) and (c, d) is related to (e, f), then (a, b) is related to (e, f). In this case, we need to check if whenever a^2 - b^2 = 1 and c^2 - d^2 = 1, then a^2 - b^2 = 1.
Let's assume a^2 - b^2 = 1 and c^2 - d^2 = 1. Rearranging the equations, we get b^2 = a^2 - 1 and d^2 = c^2 - 1.
Now, let's substitute b^2 = a^2 - 1 and d^2 = c^2 - 1 in the original relation:
(a^2 - 1) - (c^2 - 1) = 1
Simplifying, we get a^2 - c^2 = 1
This is true for all (a, b), (c, d), and (e, f) ∈ R.
Since the relation holds for all (a, b), (c, d), and (e, f) ∈ R, the relation C is transitive.
Conclusion:
Since the relation C satisfies the properties of being reflexive and transitive, but not symmetric, it is not an equivalence relation. Therefore, the correct answer is option 'A'.
An equivalence relation on a set is a relation that is reflexive, symmetric, and transitive. Let's check if the relation C satisfies these properties.
Reflexive:
A relation is reflexive if each element is related to itself. In this case, we need to check if (a, b) is related to (a, b) for all (a, b) ∈ R.
Let's substitute a = a and b = b in the given relation: a^2 - b^2 = 1
(a^2 - b^2) = (a^2 - b^2) simplifies to 0 = 0, which is true for all real numbers.
Since the relation holds for all (a, b) ∈ R, the relation C is reflexive.
Symmetric:
A relation is symmetric if whenever (a, b) is related to (c, d), then (c, d) is related to (a, b). In this case, we need to check if whenever a^2 - b^2 = 1, then b^2 - a^2 = 1.
Let's assume a^2 - b^2 = 1. Rearranging the equation, we get b^2 = a^2 - 1.
Now, let's substitute b^2 = a^2 - 1 in the original relation: (a^2 - 1) - a^2 = 1
This simplifies to -1 = 1, which is not true.
Since the relation does not hold for all (a, b) ∈ R, the relation C is not symmetric.
Transitive:
A relation is transitive if whenever (a, b) is related to (c, d) and (c, d) is related to (e, f), then (a, b) is related to (e, f). In this case, we need to check if whenever a^2 - b^2 = 1 and c^2 - d^2 = 1, then a^2 - b^2 = 1.
Let's assume a^2 - b^2 = 1 and c^2 - d^2 = 1. Rearranging the equations, we get b^2 = a^2 - 1 and d^2 = c^2 - 1.
Now, let's substitute b^2 = a^2 - 1 and d^2 = c^2 - 1 in the original relation:
(a^2 - 1) - (c^2 - 1) = 1
Simplifying, we get a^2 - c^2 = 1
This is true for all (a, b), (c, d), and (e, f) ∈ R.
Since the relation holds for all (a, b), (c, d), and (e, f) ∈ R, the relation C is transitive.
Conclusion:
Since the relation C satisfies the properties of being reflexive and transitive, but not symmetric, it is not an equivalence relation. Therefore, the correct answer is option 'A'.
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Let C = {(a, b): a2 + b2 = 1; a, b ∈ R} a relation on R, set of real numbers. Then C isa)Equivalence relationb)Reflexivec)Transitived)SymmetricCorrect answer is option 'A'. Can you explain this answer?
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