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Let R = {(P, Q) : OP = OQ , O being the origin} be an equivalence relation on A. The equivalence class [(1, 2)] is
Correct Answer : a
Explanation : A = {(x,y) : x^{2} + y^{2} = 5}
=> (1,2) {(1)^{2} + (4)^{2} = 5}
=> {1 + 4 = 5}
=> {5 = 5}
Let a relation T on the set R of real numbers be T = {(a, b) : 1 + ab < 0, a, ∈ R}. Then from among the ordered pairs (1, 1), (1, 2), (1, 2), (2, 2), the only pair that belongs to T is________.
Since T is a set of real number for it's given ordered pairs we have a condition provided that is 1 + ab is less than zero. so in the given option c if we put order of a and b then it satisfy our given condition
For real number x and y, we write _{x}R_{y} ⇔ xy + √2 is an irrational number. Then the relation R is:
_{x}R_{y} => x  y + √2 is an irrational number.
Let R is a binary relation on real numbers x and y.
Clearly, R is reflexive relation
As _{x}R_{x} iff x – x +√2 = √2 ,which is an irrational number.
Here R is not symmteric if we take x =√2 and y =1 then x – y + √2 is an irrational
number but y – x + √2 = 1, which is not irrational number
Now, R is transitive iff for all (x, y) ∈ R and (y, z) ∈ R implies (x, z) ∈ R
But here R is not transitive as we take x = 1, y = 2√2, z=√2
Given, _{x}R_{y} => x  y + √2 is irrational ............1
and _{y}R_{z} => y  z + √2 is irrational ............2
Add equation 1 and 2, we get
(x  y + √2) + (y  z + √2)
= x  z + √2 = 1, which is not an irrational
Let R = {(1, 3), (4, 2), (2, 4), (2, 3), (3, 1)} be a relation on the set A = {1, 2, 3, 4}. The relation R is
Let R={(1,3),(4,2),(2,4),(2,3),(3,1)} be a relation on the set A={1,2,3,4}, then
(a) Since (2,3) ∈ R but (3,2) ∈/ ′R, so R is not symmetric.
(b) Since (1,3) ∈ R and (3,1) ∈/ R but (1,1) ∈/ R, so R is not transitive:
(c) Since (1,1) ∈/ R, so R is not reflexive.
(d) Since (2,4) ∈ R and (2,3) ∈ R, so R is not a function.
If A = {1, 3, 5, 7} and we define a relation R = {(a, b), a, b ∈ A: a  b = 8}. Then the number of elements in the relation R is
Clearly there is no pair in set A whose difference is 8 or 8.
so D is the correct option.
If A = {1, 3, 5, 7} and define a relation, such that R = {(a, b) a, b ∈ A : a + b = 8}. Then how many elements are there in the relation R
Number of relations would be 4 as.. 1 + 7,7 + 1, 3 + 5, 5 + 3 all are equal to 8
In the set N x N, the relation R is defined by (a, b) R (c, d) ⇔ a + d = b + c. Then R is
(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, ab = ef
⇒ a + f = b + e
⇒ (a, b) R (e, f)
Hence, it is transitive.
This is an equivalence relation
If A = {1, 2, 3, 4} and B = {1, 3, 5} and R is a relation from A to B defined by(a, b) ∈ element of R ⇔ a < b. Then, R = ?
A = {1, 2, 3, 4}
B = {1, 3, 5}
(a, b) ∈ element of R ⇔ a < b for all a ∈ A, b ∈ B
(a, b) pairs satisfying the condition of R are:
(1, 3), (1, 5), (2, 3), (2, 5), (3, 5), (4, 5)
So,
R = {(1, 3), (1, 5), (2, 3), (2, 5), (3, 5), (4, 5)}
Let R be the relation on the set {1, 2, 3, 4} given by R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3), (3,3), (3,2)}. then R is
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.
Let A = {1, 2, 3, 4, 5, 6, 7}. P = {1, 2}, Q = {3, 7}. Write the elements of the set R so that P, Q and R form a partition that results in equivalence relation.
P{1, 2} union Q {3, 7} union R{} = A {1, 2, 3, 4, 5, 6, 7}
therefore R = {4, 5, 6}
Let R be a relation on set A of triangles in a plane.
R = {(T_{1}, T_{2}) : T_{1}, T_{2} element of A and T_{1} is congruent to T_{2}} Then the relation R is______
It is equivalence relation...
as a triangle is congruent to itself that means (T_{1}, T_{1}) exist in relation which implies it is reflexive.
And also if T_{1} is congruent to T_{2} then T_{2} is also congruent to T_{1} as simple that means (T_{1},T_{2}) and (T_{2},T_{1}) both belongs to relation ...which implies it is symmetric.
And is T1 is congruent to T_{2} and T_{2} is congruent to T_{3}, then T_{1} is also congruent to T_{3}, congruency rule that means (T_{1},T_{2}), (T_{2},T_{3}) and (T_{1},T_{3}) belongs to relation which implies it transitive.
this relation is reflexive, symmetric, and transitive, hence it is equivalence relation.
Which one of the following relations on set of real numbers is an equivalence relation?
If a = b  then b = a so this is symmetric as well as reflexive and if a = b and b = c then c = a then it is transitive as well so it is an equivalence relation.
Let C = {(a, b): a^{2} + b^{2} = 1; a, b ∈ R} a relation on R, set of real numbers. Then C is
Correct Answer : D
Explanation: Check for reflexive
Consider (a,a)
∴ a^{2}+a^{2}
=1 which is not always true.
If a=2
∴ 2^{2}+2^{2}
=1⇒4+4=1 which is false.
∴ R is not reflexive  ( 1 )
Check for symmetric
aRb⇒a^{2}+b^{2}=1
bRa⇒b^{2}+a^{2} =1
Both the equation are the same and therefore will always be true.
∴ R is symmetric  ( 2 )
Check for transitive
aRb⇒a^{2}+b^{2}=1
bRc⇒b^{2}+c^{2}=1
∴ a^{2}+c^{2}=1 will not always be true.
Let a=−1,b=0 and c=1
∴ (−1)^{2}+0^{2}=1,
= 0^{2}+1^{2}
=1 are true.
But (−1)^{2}+1^{2}
=1 is false.
∴ R is not transitive  ( 3 )
Let A = {1, 2, 3, 4} and B = {x, y, z}. Then R = {(1, x), (2, z), (1, y), (3, x)} is
Let a set of A = (1, 2, 3, 4) and set B (x, y, z) so. set A of all elements in set B then the relation of A to B
Let R be a relation on N, set of natural numbers such that m R n ⇔ m divides n. Then R is
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 .
If R be a relation “less than” from set A = {1, 2, 3, 4} to B = {1, 3, 5}, i.e. (a, b) ∈ R if a < b, if (b, a) ∈ R^{1}elements in R^{1} are
A = {1, 2, 3, 4}
B = {1, 3, 5}
(a, b) ∈ R if a < b
(b, a) ∈ R^{  1}
R ^{ 1} will have all (b, a) pairs where b < a, for all b ∈ B, a ∈ A
R ^{ 1} = {(3, 1), (3, 2), (5, 1), (5, 2), (5, 3), (5, 4)}
Let R be a relation on a finite set A having n elements. Then, the number of relations on A is
Number of relations on A =
Stepbystep explanation:
If there are n elements in set A then the total number of ordered pairs in the set A × A = n²
In other words A × A will have n² elements.
We also know that if a set has N elements then the number of subsets of A are 2^{n}
Therefore, for A × A there can as many relations as the number of subsets of A × A
The number of subsets of A × A =
Therefore the number of relations =
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
(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)
⇒ mn^{2} + m^{2}n = nm^{2} + n^{2}m
⇒ mn^{2} + m^{2}n = mn^{2} + m^{2}n
⇒ 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.
A situation in which significant power is distributed among three or more states is known as what?
Definition of Multipolar system: A multipolar system is a system in which power is distributed at least among 3 significant poles concentrating wealth and/or military capabilities and able to block or disrupt major political arrangements threatening their major interests.
Let R be an equivalence relation on Z, the set of integers.
R = {(a, b): a,b ∈ Z and a – b is a multiple of 3} The Equivalence class of [1] is
Correct Answer : d
Explanation : R = (a,b) : 3 divides (ab)
⇒(a−b) is a multiple of 3.
To find equivalence class 1, put b=1
So, (a−0) is a multiple of 3
⇒ a is a multiple of 3
So, In set z of integers, all the multiple
of 3 will come in equivalence
class {1}
Hence, equivalence class {1} = {3x+1}
{5,2,1,4,7}
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