Test: Types of Relations (April 16) - JEE MCQ

To be reflexive, a line must be perpendicular to itself, but which is not true. So, R is not reflexive
For symmetric, if  l₁ R l₂ ⇒ l₁ ⊥ l₂.
⇒  l₂ ⊥ l₁ ⇒ l₁ R l₂ hence symmetric
For transitive,  if l₁ R l₂ and l₂ R l₃
⇒ l₁ R l₂  and l₂ R l₃  does not imply that l₁ ⊥ l₃ hence not transitive.

Reflexive : A relation is reflexive if every element of set is paired with itself. Here none of the element of A is paired with themselves, so S is not reflexive.
Symmetric : This property says that if there is a pair (a, b) in S, then there must be a pair (b, a) in S. Since there is no pair here in S, this is trivially true, so S is symmetric.
Transitive : This says that if there are pairs (a, b) and (b, c) in S, then there must be pair (a,c) in S. Again, this condition is trivially true, so S is transitive.

The relation { } ⊂ A x A on a is surely not reflexive. However, neither symmetry nor transitivity is contradicted. So { } is a transitive and symmetry relation on A.

(a, b) R (c, d) <=> a + d = b + c
Reflexive:
(a, b) R (a, b) <=> a + b = b + a
This is true for all (a, b) € N x N
Hence, it is reflexive.
Symmetric:
Let (a, b) R (c, d) <=> a + d = b + c
a + d = b + c
c + b = d + a (same)
By the above equation,
(c, d) R (a, b)
Hence,
(c, d) R (a, b) <=> c + b = d + a
Hence, it is symmetric
Transitive:
Let (a, b) R (c, d) <=> a + d = b + c, --- eqn 1
(c, d) R (e, f) <=> c + f = d + e --- eqn 2
for all a, b, c, d, e, f € N
eqn 1 : a + d = b + c
⇒  a - b = c - d
eqn 2 : c + f = d + e
⇒ c - d = e - f
So, a-b = e-f
⇒ a + f = b + e
⇒ (a, b) R (e, f)
Hence, it is transitive.
This is an equivalence relation

 R be the relation in the set {1, 2,3, 4] given by R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3, 3), (3, 2)}
it is seen that (a, a) ∈ R for every a ∈ {1, 2, 3, 4}
so,R is reflexive.
it is seen that (a, b) = (b, a) ∈ R 
because, (1, 2)∈ R but (2, 1) ∉ R
so, R is not symmetric.
it is seen that (a, b), (b, c) ∈ R ⇒ (a, c) ∈ R for all a, b, c ∈ {1, 2, 3, 4}.
so, R is transitive.
Hence, R is reflexive and transitive but not symmetric.

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 )

Let there be a natural number n,
We know that n divides n, which implies nRn.
So, Every natural number is related to itself in relation R.
Thus relation R is reflexive .

Let there be three natural numbers a,b,c and let aRb,bRc
aRb implies a divides b and bRc implies b divides c, which combinedly implies that a divides c i.e. aRc.
So, Relation R is also transitive .

Let there be two natural numbers a,b and let aRb,
aRb implies a divides b but it can't be assured that b necessarily divides a.
For ex, 2R4 as 2 divides 4 but 4 does not divide 2 .
Thus Relation R is not symmetric .

(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.

It is equivalence relation...
as a triangle is congruent to itself that means (T₁, T₁) exist in relation which implies it is reflexive.
And also if T₁ is congruent to T₂ then T₂ is also congruent to T₁ as simple that means (T₁,T₂) and (T₂,T₁) both belongs to relation ...which implies it is symmetric.
And is T1 is congruent to T₂ and T₂ is congruent to T₃, then T₁ is also congruent to T₃, congruency rule that means (T₁,T₂), (T₂,T₃) and (T₁,T₃) belongs to relation which implies it transitive.
this relation is reflexive, symmetric, and transitive, hence it is equivalence relation.

A = {1,2,3}
B = {(2,3)} is not reflexive or symmetric on A but it is transitive
 ∵ if (a,b) exists but (b,c) does not exist then (a,c) does not need to exist and the relation is still transitive.