Show that the relation R in the set A of all the books in a library of a school given by R = {(x, y): x and -Maths 9th

Show that the relation R in the set A of all the books in a library of a school given by R = {(x, y): x and -Maths 9th
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Answer :

Given: A = {All books in a library of a school} R = {(x, y): x and y have the same number of pages} Reflexivity: (x, x) ∈ R ⇒ R is reflexive on A Symmetric: Since books x and y have the same number of pages, so (x, y) ∈ R. Since books y and x have the same number of pages, so (y, x) ∈ R. ⇒ R is symmetric on A. Transitivity: Books x, y, z have the same number of pages ⇒ (x, y) ∈ R and (y, z) ∈ R ⇒ (x, z) ∈ R ⇒ R is transitive on A. Hence, R is an equivalence relation.
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Related questions

Let R be a relation on the set N, defined by {(x, y) : 2x – y = 10} then R is -Maths 9th

Description : Let R be a relation on the set N, defined by {(x, y) : 2x – y = 10} then R is -Maths 9th

Last Answer : (a) ReflexiveGiven, {(\(x\), y) : 2\(x\) – y = 10} Reflexive, \(x\) R \(x\) = 2\(x\) – \(x\) = 10 ⇒ \(x\) = 10 ⇒ y = 10 ∴ Point (10, 10) ∈ N ⇒ R is reflexive.

The relation R in the set {1, 2, 3} given by R = {(1, 2), (2, 1)} is: -Maths 9th

Description : The relation R in the set {1, 2, 3} given by R = {(1, 2), (2, 1)} is: -Maths 9th

Last Answer : (c) Symmetric onlyAs (1, 1), (2, 2) ∉ R so R is not reflexive. A relation a R b is said to symmetric if (a, b) ∈ R ⇒ (b, a) ∈ R. Here (1, 2) ∈ R and (2, 1) ∈ R so it is symmetric. Also, as (1, 2) ∈ R, but (2, 3), (1, 3) ∉ R, so R is not transitive.

Given the relation R = {(1, 2), (2, 3)} on the set A = {1, 2, 3}, -Maths 9th

Description : Given the relation R = {(1, 2), (2, 3)} on the set A = {1, 2, 3}, -Maths 9th

Last Answer : (d) 7A = {1, 2, 3}. R = {(1, 2), (2, 3)} To make R an equivalence relation, it should be: (i) Reflexive: So three more ordered pairs (1, 1), (2, 2), (3, 3) should be added to R to make it ... 2, 1), (3, 2), (1, 3), (3, 1)} is an equivalence relation. So minimum 7 ordered pairs are to be added.

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Description : Let R be a relation on the set of integers given by a = 2^k .b for some integer k. Then R is -Maths 9th

Last Answer : (c) equivalence relationGiven, a R b = a = 2k .b for some integer. Reflexive: a R a ⇒ a = 20.a for k = 0 (an integer). True Symmetric: a R b ⇒ a = 2k b ⇒ b = 2-k . a ⇒ b R a as k, -k are both ... = 2k1 + k2 c, k1 + k2 is an integer. ∴ a R b, b R c ⇒ a R c True ∴ R is an equivalence relation.

Let 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)}. Then R is -Maths 9th

Description : Let 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)}. Then R is -Maths 9th

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Description : Let R be a relation from A = {1, 2, 3, 4, 5, 6} to B = {1, 3, 5} which is defined as “x is less than y”. -Maths 9th

Last Answer : R = {a, b : a < b, a ∈ A, b ∈ B}, where A = {1, 2, 3, 4, 5, 6} and B = {1, 3, 5}. ∴ R = {(1, 3), (1, 5), (2, 3), (2, 5), (3, 5), (4, 5)} Domain of R = {1, 2, 3, 4} Range of R = {3, 5} Codomain of R = {1, 3, 5}.

If R is a relation in N × N defined by (a, b) R (c, d) if and only if ad = bc, show that R is an equivalence relation. -Maths 9th

Description : If R is a relation in N × N defined by (a, b) R (c, d) if and only if ad = bc, show that R is an equivalence relation. -Maths 9th

Last Answer : (i) R is reflexive. For all (a, b) ∈ N N we have (a, b) R (a, b) because ab = ba ⇒ R is reflexive. (ii) R is symmetric. Suppose (a, b) R (c, d) Then (a, b) R (c, d) ⇒ ... ) R (e, f) ⇒ R is transitive. Since R is reflexive, symmetric and transitive, therefore, R is an equivalence relation on N N.

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Description : Choose the correct option. The relation R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)} on a set A = {1, 2, 3} is -Maths 9th

Last Answer : R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)} on set A = {1, 2, 3}. (i) Since (1, 1), (2, 2), (3, 3) ∈ R ⇒ R is reflexive. (ii) (1, 2) ∈ R but (2, 1) ∉ R ⇒ R is not symmetric. (iii) (1, 2) ∈ R, (2, 3) ∈ R and (1, 3) ∈R ⇒ R is transitive. ∴ Option (a) is the right answer.

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Description : Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as R = {(a, b): b = a + 1} is reflexive, symmetric or transitive. -Maths 9th

Last Answer : Reflexive: R = {(a, b) : b = a +1} = {(a, a + l) : a, a + 1∈{l, 2, 3, 4, 5, 6}} = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)} ⇒ R is not reflexive since (a, a) ∉R for all a. Symmetric: R is not symmetric as (a ... as (a, b) ∈ R and (b, c) ∈ R but (a, c) ∉ R e.g., (1, 2) ∈ R (2, 3) ∈ R but (1, 3) ∉R

Let R = {(3, 3), (6, 6), (9, 9), (12, 12), (6, 12), (3, 12), (3, 6)} be a relation on set A = {3, 6, 9, 12}. The relation is -Maths 9th

Description : Let R = {(3, 3), (6, 6), (9, 9), (12, 12), (6, 12), (3, 12), (3, 6)} be a relation on set A = {3, 6, 9, 12}. The relation is -Maths 9th

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Description : Let R be a relation defined on the set A of all triangles such that R = {(T1, T2) : T1 is similar to T2}. Then R is -Maths 9th

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Description : On a set N of all natural numbers is defined the relation R by a R b iff the GCD of a and b is 2, then R is -Maths 9th

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Description : On the set R of all real numbers, a relation R is defined by R = {(a, b) : 1 + ab > 0}. Then R is -Maths 9th

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Description : If R is a relation defined on the set of natural numbers N such that (a, b) R (c, d) if and only if a + d = b + c, then R is -Maths 9th

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Description : Let A = {1, 2, 3, 4}, B = {5, 6, 7, 8}. Then R = {(1, 5), (1, 7), (2, 6)} is a relation from set A to B defined as : -Maths 9th

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If R is a relation on a finite set A having n elements, then the number of relations on A is -Maths 9th

Description : If R is a relation on a finite set A having n elements, then the number of relations on A is -Maths 9th

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Show that the relation ‘≅’ congruence on the set of all triangles in Euclidean plane geometry is an equivalence relation. -Maths 9th

Description : Show that the relation ‘≅’ congruence on the set of all triangles in Euclidean plane geometry is an equivalence relation. -Maths 9th

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The temperature of a liquid can be measured in Kelvin units as x K or in Fahrenheit units as y°F. The relation between the two system of measurement of temperature is given by the linear equation y = 9/5 ( x - 273 ) + 32 -Maths 9th

Description : The temperature of a liquid can be measured in Kelvin units as x K or in Fahrenheit units as y°F. The relation between the two system of measurement of temperature is given by the linear equation y = 9/5 ( x - 273 ) + 32 -Maths 9th

Last Answer : Solution :-

While discussing the properties of a parallelogram teacher asked about the relation between two angles x and y of a parallelogram as shown ... -Maths 9th

Description : While discussing the properties of a parallelogram teacher asked about the relation between two angles x and y of a parallelogram as shown ... -Maths 9th

Last Answer : (a) Yes , x < y is correct (b) Ð ADB =Ð DBC = y (alternate int. angles) since BC < CD (angle opp. to smaller side is smaller) there for, x < y (c) Truth value

Find the relation between x and y if points (2, 1), (x, y) and (7, 5) are collinear. -Maths 9th

Description : Find the relation between x and y if points (2, 1), (x, y) and (7, 5) are collinear. -Maths 9th

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Let R be a relation defined as a Rb if | a – b | > 0, then the relation is -Maths 9th

Description : Let R be a relation defined as a Rb if | a – b | > 0, then the relation is -Maths 9th

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For any two real number a b and , we defined aRb if and only if sin^2a + cos^2b = 1. The relation R is -Maths 9th

Description : For any two real number a b and , we defined aRb if and only if sin^2a + cos^2b = 1. The relation R is -Maths 9th

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Let A = {(2, 5, 11)}, B = {3, 6, 10} and R be a relation from A to B defined by R = {(a, b) : a and b are co-prime}. Then R is -Maths 9th

Description : Let A = {(2, 5, 11)}, B = {3, 6, 10} and R be a relation from A to B defined by R = {(a, b) : a and b are co-prime}. Then R is -Maths 9th

Last Answer : (c) {(2, 3), (5, 3), (5, 6), (11, 3), (11, 6), (11, 10)}

Let A = {1, 2, 3} and R = {(1, 2), (1, 1), (2, 3)} be a relation on A. -Maths 9th

Description : Let A = {1, 2, 3} and R = {(1, 2), (1, 1), (2, 3)} be a relation on A. -Maths 9th

Last Answer : (a) 1For the relation R to become transitive: (1, 2) ∈R and (2, 3) ∈R should imply (1, 3) ∈R ∴ Minimum one ordered pair (1, 3) should be added to R.

Let N be the set of natural numbers. Describe the following relation in words giving its domain -Maths 9th

Description : Let N be the set of natural numbers. Describe the following relation in words giving its domain -Maths 9th

Last Answer : The given relation stated in words is R = {(x, y) : x is the fourth power of y; x ∈ N, y ∈ {1, 2, 3, 4}}.

Which of the following is an equivalence relation defined on set A = {1, 2, 3} -Maths 9th

Description : Which of the following is an equivalence relation defined on set A = {1, 2, 3} -Maths 9th

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The relation “is parallel to” on a set S of all straight lines in a plane is : -Maths 9th

Description : The relation “is parallel to” on a set S of all straight lines in a plane is : -Maths 9th

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The relation ‘is less than’ on a set of natural numbers is -Maths 9th

Description : The relation ‘is less than’ on a set of natural numbers is -Maths 9th

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If both x – 2 and x -(1/2) are factors of px2+ 5x+r, then show that p = r. -Maths 9th

Description : If both x – 2 and x -(1/2) are factors of px2+ 5x+r, then show that p = r. -Maths 9th

Last Answer : Show that p = r.

If both x – 2 and x -(1/2) are factors of px2+ 5x+r, then show that p = r. -Maths 9th

Description : If both x – 2 and x -(1/2) are factors of px2+ 5x+r, then show that p = r. -Maths 9th

Last Answer : Show that p = r.

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Description : If both (x -2) and (x - 1/2) are factors of px2 + 5x + r, show that p = r. -Maths 9th

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If a relation `R` on the set `N` of natural numbers is defined as `(x,y)hArrx^(2)-4xy+3y^(2)=0,Aax,yepsilonN`. Then the relation `R` is

Description : If a relation `R` on the set `N` of natural numbers is defined as `(x,y)hArrx^(2)-4xy+3y^(2)=0,Aax,yepsilonN`. Then the relation `R` is

Last Answer : If a relation `R` on the set `N` of natural numbers is defined as `(x,y)hArrx^(2) ... symmetric B. reflexive C. transitive D. an equivalence relation.

Consider the following relations R = {(x, y) | x, y are real numbers and x = wy for some rational number w}; -Maths 9th

Description : Consider the following relations R = {(x, y) | x, y are real numbers and x = wy for some rational number w}; -Maths 9th

Last Answer : (c) S is an equivalence relation but R is not an equivalence relationR = {(x, y) | x, y ∈ R, x = wy, w is a rational number} Reflexive: x R x ⇒ x = wx ⇒ w = 1, (a rational number) Hence R is reflexive. Symmetric ... \(rac{r}{s}\) ⇒ \(rac{m}{n}\) S \(rac{r}{s}\) (True)∴ S is an equivalence relation.

If the roots of the equation x^2 + x + 1 = 0 are in the ratio of m : n, then which one of the following relation holds ? -Maths 9th

Description : If the roots of the equation x^2 + x + 1 = 0 are in the ratio of m : n, then which one of the following relation holds ? -Maths 9th

Last Answer : answer:

If x+y=10 and x=z then show that z+y=10 by using appropriate eculids axioms? -Maths 9th

Description : If x+y=10 and x=z then show that z+y=10 by using appropriate eculids axioms? -Maths 9th

Last Answer : This answer was deleted by our moderators...

If x+y =10 and x=z then show that z+y =10 -Maths 9th

Description : If x+y =10 and x=z then show that z+y =10 -Maths 9th

Last Answer : It is given that x+y =10 Also x= z Therefore, x+y =10 Z+y =10 ( x = z)

If x+y=10 and x=z then show that z+y=10 by using appropriate eculids axioms? -Maths 9th

Description : If x+y=10 and x=z then show that z+y=10 by using appropriate eculids axioms? -Maths 9th

Last Answer : This answer was deleted by our moderators...

If x+y =10 and x=z then show that z+y =10 -Maths 9th

Description : If x+y =10 and x=z then show that z+y =10 -Maths 9th

Last Answer : It is given that x+y =10 Also x= z Therefore, x+y =10 Z+y =10 ( x = z)

if xylogxy/x+y, yzlogyz/y+z and zxlogzx/z+x are mutually equal, then show that x^x= y^y=z^z -Maths 9th

Description : if xylogxy/x+y, yzlogyz/y+z and zxlogzx/z+x are mutually equal, then show that x^x= y^y=z^z -Maths 9th

Last Answer : NEED ANSWER

if xylogxy/x+y, yzlogyz/y+z and zxlogzx/z+x are mutually equal, then show that x^x= y^y=z^z -Maths 9th

Description : if xylogxy/x+y, yzlogyz/y+z and zxlogzx/z+x are mutually equal, then show that x^x= y^y=z^z -Maths 9th

Last Answer : This answer was deleted by our moderators...

Using factor theorem,show that (x-y) is a factor of x(y(square) - z(square)) + y(z(square) - x(square)) + z(x(square) - y(square) ) -Maths 9th

Description : Using factor theorem,show that (x-y) is a factor of x(y(square) - z(square)) + y(z(square) - x(square)) + z(x(square) - y(square) ) -Maths 9th

Last Answer : Solution :-

If the points (2, 1) and (1, – 2) are equidistant from the point (x, y), show that x + 3y = 0. -Maths 9th

Description : If the points (2, 1) and (1, – 2) are equidistant from the point (x, y), show that x + 3y = 0. -Maths 9th

Last Answer : (a) The distance d between any two points say P(x1, y1) and Q(x2, y2) is given by:d = \(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)⇒ d2 = (x2 - x1)2 + (y2 - y1)2 ⇒ d = \(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)( ... distance of a point P(x1, y1) form the origin= \(\sqrt{(x_2-0)^2+(y_2-0)^2}\) = \(\sqrt{x^2_1+y^2_1}\)

If the points (x, 1), (1, 2) and (0, y + 1) are collinear show that -Maths 9th

Description : If the points (x, 1), (1, 2) and (0, y + 1) are collinear show that -Maths 9th

Last Answer : Two lines are parallel if their slopes are equal∴ \(rac{0-(-8)}{3-(-5)}\) = \(rac{a-3}{4-6}\) ⇒ \(rac{8}{8}\) = \(rac{a-3}{-2}\) ⇒ a – 3 = –2 ⇒ a = 1.

Let X be the set of all graduates in India. Elements x and y in X are said to be related, if they are graduates of the same university. -Maths 9th

Description : Let X be the set of all graduates in India. Elements x and y in X are said to be related, if they are graduates of the same university. -Maths 9th

Last Answer : R = {(x, y)}: x and y are graduates of same university, x, y ∈ {All graduates of India}. R is reflexive as (x, x) ∈ R, since x and x are graduates from the same university. R ... graduates from the same university ⇒ (x, z) ∈ R. R being reflexive, symmetric and transitive is an equivalence relation.

3. ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus. -Maths 9th

Description : 3. ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus. -Maths 9th

Last Answer : Solution: Given in the question, ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Construction, Join AC and BD. To Prove, PQRS is a rhombus. Proof: In ΔABC P and Q ... (ii), (iii), (iv) and (v), PQ = QR = SR = PS So, PQRS is a rhombus. Hence Proved

2. ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle. -Maths 9th

Description : 2. ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle. -Maths 9th

Last Answer : Solution: Given in the question, ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. To Prove, PQRS is a rectangle. Construction, Join AC and BD. Proof: In ΔDRS and ... , In PQRS, RS = PQ and RQ = SP from (i) and (ii) ∠Q = 90° , PQRS is a rectangle.

ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig 8.29). AC is a diagonal. Show that: (i) SR || AC and SR = 1/2 AC (ii) PQ = SR (iii) PQRS is a parallelogram. -Maths 9th

Description : ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig 8.29). AC is a diagonal. Show that: (i) SR || AC and SR = 1/2 AC (ii) PQ = SR (iii) PQRS is a parallelogram. -Maths 9th

Last Answer : . Solution: (i) In ΔDAC, R is the mid point of DC and S is the mid point of DA. Thus by mid point theorem, SR || AC and SR = ½ AC (ii) In ΔBAC, P is the mid point of AB and Q is the mid point of BC. ... ----- from question (ii) ⇒ SR || PQ - from (i) and (ii) also, PQ = SR , PQRS is a parallelogram.

ABCD is a parallelogram in which P and Q are the mid-points of opposite sides AB and CD (Fig. 8.48). If AQ intersects DP at S and BQ intersects CP at R, show that -Maths 9th

Description : ABCD is a parallelogram in which P and Q are the mid-points of opposite sides AB and CD (Fig. 8.48). If AQ intersects DP at S and BQ intersects CP at R, show that -Maths 9th

Last Answer : Solution :-

ABCD is a rectangle and p q r s are the mid points of the side AB BC CD AND DA respectively. Show that the quadrilateral PQRS is a rhombus -Maths 9th

Description : ABCD is a rectangle and p q r s are the mid points of the side AB BC CD AND DA respectively. Show that the quadrilateral PQRS is a rhombus -Maths 9th

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Let P(–3, 2), Q(–5, –5), R(2, –3) and S(4, 4) be four points in a plane. Then show that PQRS is a rhombus. Is it a square ? -Maths 9th

Description : Let P(–3, 2), Q(–5, –5), R(2, –3) and S(4, 4) be four points in a plane. Then show that PQRS is a rhombus. Is it a square ? -Maths 9th

Last Answer : Let P(1, -1), Q \(\big(rac{-1}{2},rac{1}{2}\big)\) and R(1,2) be the vertices of the ΔPQR.Then, PQ = \(\sqrt{\big(rac{-1}{2}-1\big)^2+\big(rac{1}{2}+1\big)^2}\) = \(\sqrt{rac{9}{4}+rac{9}{4}} ... {3\sqrt2}{2}\)PR = \(\sqrt{(1-1)^2+(2+1)^2}\) = \(\sqrt9\) = 3∵ PQ = QR, the triangle PQR is isosceles.