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Let R be a relation on N (set of natural numbers) such that (m, n) R (p, q)mq(n + p) = np(m + q). Then, R is
- a)An Equivalence Relation
- b)Only Reflexive
- c)Symmetric and reflexive.
- d)Only Transitive
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
Let R be a relation on N (set of natural numbers) such that (m, n) R (...
(m, n) R (p, q) <=> mq(n + p) = np(m + q)
For all m,n,p,q € N
Reflexive:
(m, n) R (m, n) <=> mn(n + m) = nm(m + n)
⇒ mn2 + m2n = nm2 + n2m
⇒ mn2 + m2n = mn2 + m2n
⇒ LHS = RHS
So, (m, n) R (m, n) exists.
Hence, it is Reflexive
Symmetric:
Let (m, n) R (p, q) exists
mq(n + p) = np(m + q) --- (eqn1)
(p, q) R (m, n) <=> pn(q + m) = qm (p + n)
⇒ np(m + q) = mq(n + p)
⇒ mq(n + p) = np(m + q)
This equation is true by (eqn1).
So, (p, q) R (m, n) exists
Hence, it is not symmetric.
Transitive:
Let (m, n) R (p, q) and (p, q) R (r, s) exists.
Therefore,
mq(n + p) = np(m + q) --- (eqn1)
ps(q + r) = qr (p + s) --- (eqn2)
We cannot obtain ms(n+r) = nr(m+s) using eqn1 and eqn2.
So, ms(n + r) ≠ nr(m + s)
Therefore, (m, n) R (r, s) doesn’t exist.
Hence, it is transitive.
For all m,n,p,q € N
Reflexive:
(m, n) R (m, n) <=> mn(n + m) = nm(m + n)
⇒ mn2 + m2n = nm2 + n2m
⇒ mn2 + m2n = mn2 + m2n
⇒ LHS = RHS
So, (m, n) R (m, n) exists.
Hence, it is Reflexive
Symmetric:
Let (m, n) R (p, q) exists
mq(n + p) = np(m + q) --- (eqn1)
(p, q) R (m, n) <=> pn(q + m) = qm (p + n)
⇒ np(m + q) = mq(n + p)
⇒ mq(n + p) = np(m + q)
This equation is true by (eqn1).
So, (p, q) R (m, n) exists
Hence, it is not symmetric.
Transitive:
Let (m, n) R (p, q) and (p, q) R (r, s) exists.
Therefore,
mq(n + p) = np(m + q) --- (eqn1)
ps(q + r) = qr (p + s) --- (eqn2)
We cannot obtain ms(n+r) = nr(m+s) using eqn1 and eqn2.
So, ms(n + r) ≠ nr(m + s)
Therefore, (m, n) R (r, s) doesn’t exist.
Hence, it is transitive.
Most Upvoted Answer
Let R be a relation on N (set of natural numbers) such that (m, n) R (...
The answer is b as (a,b)R(a,b) givesab(a+b)=ba
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Community Answer
Let R be a relation on N (set of natural numbers) such that (m, n) R (...
Equivalence Relation:
To determine whether relation R is an equivalence relation, we need to check whether it satisfies three properties: reflexive, symmetric, and transitive.
Reflexive Property:
A relation is reflexive if every element is related to itself. In other words, for all a in N, (a, a) R (a, a) should hold true.
Let's check if R satisfies the reflexive property:
(a, a) R (a, a)
aa(a - a) = aa(0) = 0
aa(a - a) = 0
0 = 0
Since 0 = 0, we can see that R is reflexive.
Symmetric Property:
A relation is symmetric if for all a, b in N, if (a, b) R (b, a), then (b, a) R (a, b).
Let's check if R satisfies the symmetric property:
If (a, b) R (b, a), then ab(b - a) = ba(a - b) should hold true.
ab(b - a) = ba(a - b)
ab(b - a) = -ba(b - a)
ab(b - a) + ba(b - a) = 0
(b - a)(ab + ba) = 0
Since (b - a)(ab + ba) = 0, we can see that R is symmetric.
Transitive Property:
A relation is transitive if for all a, b, c in N, if (a, b) R (b, c), then (a, c) R (c, b).
Let's check if R satisfies the transitive property:
If (a, b) R (b, c), then ab(b - c) = bc(c - b) should hold true.
ab(b - c) = bc(c - b)
ab(b - c) = -bc(b - c)
ab(b - c) + bc(b - c) = 0
(b - c)(ab + bc) = 0
Since (b - c)(ab + bc) = 0, we can see that R is transitive.
Conclusion:
Since R satisfies all three properties: reflexive, symmetric, and transitive, we can conclude that R is an equivalence relation. Therefore, the correct answer is option 'C' - Symmetric and reflexive.
To determine whether relation R is an equivalence relation, we need to check whether it satisfies three properties: reflexive, symmetric, and transitive.
Reflexive Property:
A relation is reflexive if every element is related to itself. In other words, for all a in N, (a, a) R (a, a) should hold true.
Let's check if R satisfies the reflexive property:
(a, a) R (a, a)
aa(a - a) = aa(0) = 0
aa(a - a) = 0
0 = 0
Since 0 = 0, we can see that R is reflexive.
Symmetric Property:
A relation is symmetric if for all a, b in N, if (a, b) R (b, a), then (b, a) R (a, b).
Let's check if R satisfies the symmetric property:
If (a, b) R (b, a), then ab(b - a) = ba(a - b) should hold true.
ab(b - a) = ba(a - b)
ab(b - a) = -ba(b - a)
ab(b - a) + ba(b - a) = 0
(b - a)(ab + ba) = 0
Since (b - a)(ab + ba) = 0, we can see that R is symmetric.
Transitive Property:
A relation is transitive if for all a, b, c in N, if (a, b) R (b, c), then (a, c) R (c, b).
Let's check if R satisfies the transitive property:
If (a, b) R (b, c), then ab(b - c) = bc(c - b) should hold true.
ab(b - c) = bc(c - b)
ab(b - c) = -bc(b - c)
ab(b - c) + bc(b - c) = 0
(b - c)(ab + bc) = 0
Since (b - c)(ab + bc) = 0, we can see that R is transitive.
Conclusion:
Since R satisfies all three properties: reflexive, symmetric, and transitive, we can conclude that R is an equivalence relation. Therefore, the correct answer is option 'C' - Symmetric and reflexive.
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Let R be a relation on N (set of natural numbers) such that (m, n) R (p, q)mq(n + p) = np(m + q). Then, R isa)An Equivalence Relationb)Only Reflexivec)Symmetric and reflexive.d)Only TransitiveCorrect answer is option 'C'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Let R be a relation on N (set of natural numbers) such that (m, n) R (p, q)mq(n + p) = np(m + q). Then, R isa)An Equivalence Relationb)Only Reflexivec)Symmetric and reflexive.d)Only TransitiveCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let R be a relation on N (set of natural numbers) such that (m, n) R (p, q)mq(n + p) = np(m + q). Then, R isa)An Equivalence Relationb)Only Reflexivec)Symmetric and reflexive.d)Only TransitiveCorrect answer is option 'C'. Can you explain this answer?.
Let R be a relation on N (set of natural numbers) such that (m, n) R (p, q)mq(n + p) = np(m + q). Then, R isa)An Equivalence Relationb)Only Reflexivec)Symmetric and reflexive.d)Only TransitiveCorrect answer is option 'C'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Let R be a relation on N (set of natural numbers) such that (m, n) R (p, q)mq(n + p) = np(m + q). Then, R isa)An Equivalence Relationb)Only Reflexivec)Symmetric and reflexive.d)Only TransitiveCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let R be a relation on N (set of natural numbers) such that (m, n) R (p, q)mq(n + p) = np(m + q). Then, R isa)An Equivalence Relationb)Only Reflexivec)Symmetric and reflexive.d)Only TransitiveCorrect answer is option 'C'. Can you explain this answer?.
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