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?

Question

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&#8203;a)An Equivalence Relationb)Only Reflexivec)Symmetric and reflexive.d)Only TransitiveCorrect answer is option 'C'. Can you explain this answer?

Accepted Answer

(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.

Answer

The answer is b as (a,b)R(a,b) givesab(a+b)=ba

Answer

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.

Answer

Jdude

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