Category: Mathematics

In this article, you will learn about simplifying algebraic fractions, reducing the fraction to its lowest term, performing arithmetical operations on algebraic fractions. Get the solved example questions on algebraic fractions to understand the concept better. Each and Every Problem is explained with Step by Step Solutions so that you can learn the Procedure on how to solve related problems easily.

Solved Examples on Algebraic Fractions

Example 1.

Reduce the algebraic fractions to their lowest terms?

(i) (3x² – 6y²) / (6x – 12y)

(ii) (5x² – 5y²) / (25x² + 50xy + 25y²)

(iii) (3ab – 3a²) / (3a² – 6ab + 3b²)

Solution:

(i) (6x² – 6y²) / (12x – 12y)

Factorizing the numerator and denominator separately and cancel the common factors we get,

= (6 (x² – y²)) / ((12 (x – y))

= (x² – y²) / (2 (x – y))

= ((x – y) (x + y)) / (2 (x – y))

= (x + y) / 2

(ii) (5x² – 5y²) / (25x² + 50xy + 25y²)

Factorizing the numerator and denominator separately and cancel the common factors we get,

= (5 (x² – y²)) / (25 (x² + 2xy + y²))

= (x² – y²) / (5 (x² + 2xy + y²))

= [(x + y) (x – y)] / [5 (x² + xy + xy + y²)]

= [(x + y) (x – y)] / [5 (x( x + y) + y (x + y)]

= [(x + y) (x – y)] / [5 ((x + y) (x + y)]

= (x – y) / [5 (x + y)]

(iii) (3ab – 3a²) / (3a² – 6ab + 3b²)

Factorizing the numerator and denominator separately and cancel the common factors we get,

= [3a (b – a)] / [(3a² – 3ab – 3ab + 3b²)]

= [3a (b – a)] / [(3a(a – b) – 3b(a – b))]

= [3a (b – a)] / [(3a – 3b) (a – b)]

= [-3a (a – b)] / [(3a – 3b) (a – b)]

= -3a / 3(a – b)

= -a / (a – b)

Example 2.

Simplify the algebraic fractions?

(i) [1/x + 1/y] / [1/x² – 1/y²]

(ii) [(u + v) / 2u – 2v) + (v – u) / (2v + 2u) + 2v² / (u² – v²)] [1/v – 1/u]

(iii) [(a³ – ab² + b³) / (a – b)³ – b / (a – b)] [(a² – 2ab + 2b²) / (a² – ab + b²) – b/a]

Solution:

(i) [1/x + 1/y] / [1/x² – 1/y²]

Factorize the numerator

1/x + 1/y = (y + x) / (xy)

Factorize the denominator

1/x² – 1/y² = (y² – x²) / (x²y²)

= (y + x) (y – x) / x²y²

Simplification of the given expression after factorizing the numerator and the denominator:

[(y + x) / (xy)] / [(y + x) (y – x) / x²y²]

= [(y + x) * x²y²] / [(y + x) (y – x) * xy]

= xy / (y – x)

(ii) [(u + v) / 2u – 2v) + (v – u) / (2v + 2u) + 2v² / (u² – v²)] * [1/v – 1/u]

Factorize the denominators

2(u – v), 2 (u + v), (u +v) (u – v)

L.C.M of first expression is 2 (u + v) (u – v), second expression is uv

= [((u + v) * (u + v) / 2(u – v) (u + v)) + ((v – u) * (v – u) / 2 (u + v) (u – v)) + 2v² * 2 / 2 (u + v) (u – v)] * [u – v / vu]

= [(u + v)² / 2(u – v) (u + v) + (v – u)² / 2(u – v) (u + v) + 4v² / 2(u – v) (u + v)] * [u – v / vu]

= [((u + v)² + (v – u)² + 4v²) / 2(u – v) (u + v)] * [u – v / vu]

= [(u² + v² + 2uv + v² + u² – 2uv + 4v²) / 2(u – v) (u + v)] * [u – v / vu]

= [(2u² + 6v²) / 2(u – v) (u + v)] * [u – v / vu]

= [2u³ -2u²v + 6v²u + 6v³] / [2(u – v) (u + v)vu]

= 2u² (u – v) + 6v² (u + v) / [2(u – v) (u + v)vu]

= 2[u² (u – v) + 3v² (u + v)] / [2(u – v) (u + v)vu]

= [u² (u – v) + 3v² (u + v)] / [(u – v) (u + v)vu]

(iii) [(a³ – ab² + b³) / (a – b)³ – b / (a – b)] [(a² – 2ab + 2b²) / (a² – ab + b²) – b/a]

Factorize the denominators

(a – b)³, (a – b) and a² – ab + b², a

L.C.M of (a – b)³, (a – b) is (a – b)³, L.C.M of a² – ab + b², a is a (a² – ab + b²)

Express all fractions in terms of the lowest common denominator.

= [(a³ – ab² + b³) / (a – b)³ – (b (a – b)²) / (a – b)³] [(a (a² – 2ab + 2b²)) / a (a² – ab + b²) – b (a² – ab + b²) / a (a² – ab + b²)]

= [(a³ – ab² + b³) / (a – b)³ – (b (a² – 2ab + b²) / (a – b)³] * [(a³ – 2a²b + 2ab²)) / a (a² – ab + b²) – (a²b – ab² + b³) / a (a² – ab + b²)]

= [(a³ – ab² + b³) / (a – b)³ – (a²b – 2ab² + b³) / (a – b)³] * [(a³ – 2a²b + 2ab²) / a (a² – ab + b²) – (a²b – ab² + b³) / a (a² – ab + b²)]

= [(a³ – ab² + b³ – a²b + 2ab² – b³) / (a – b)³] * [(a³ – 2a²b + 2ab² – a²b + ab² – b³) / a (a² – ab + b²)]

= [(a³ + ab² – a²b) / (a – b)³] * [(a³ – 3a²b + 3ab² – b³) / a (a² – ab + b²)]

= [(a³ + ab² – a²b) / (a – b)³] * [(a – b)³ / a (a² – ab + b²)]

= [(a³ + ab² – a²b) (a – b)³] / [(a – b)³ a (a² – ab + b²)]

= [a (a² – ab + b²)] / [a (a² – ab + b²)]

= 1

Example 3.

Simplify the sum and difference of algebraic fractions?

(i) (2x – 3y) / x + (4x² – 5y²) / xy

(ii) x / ac – x / bc + x / ab

Solution:

(i) (2x – 3y) / x + (4x² – 5y²) / xy

L.C.M of denominators is xy.

Express all fractions in terms of the lowest common denominator.

= y(2x – 3y) / xy + (4x² – 5y²) / xy

= (2xy – 3y² + 4x² – 5y²) / xy

= (4x² + 2xy – 8y²) / xy

= 2(2x² + xy – 4y²) / xy

(ii) x / ac – x / bc + x / ab

L.C.M of all denominators is abc

Express all fractions in terms of the lowest common denominator.

= bx / abc – ax / abc + cx / abc

= (bx – ax + cx) / abc

= x(b – a + c) / abc

Example 4.

Simplify the product and quotient of algebraic fractions

(i) (3x² – 3y²) / (12x – 12y)

(ii) (x – y) : (1/x + 1/y)

Solution:

(i) (3x² – 3y²) / (12x – 12y)

Factorize numerators and denominators

= 3(x² – y²) / 12 (x – y)

= (x + y) (x – y) / 4 (x – y)

= (x + y) / 4

(ii) (x – y) : (1/x + 1/y)

= (x – y) : (y + x) / xy)

= xy (x – y) / (x + y)

Learn about how to perform sum and difference of two or more algebraic fractions on this page. We are giving a detailed step by step procedure that helps to calculate the addition, subtraction of algebraic fractions easily. Have a look at some example questions and answers for a better understanding of the concept.

How to Add and Subtract Algebraic Fractions?

You may feel that performing addition or subtraction of algebraic fractions is a bit difficult. To help you out in solving those questions, we are providing the step by step procedure in the below sections of this page. Follow these steps while solving the questions.

• If the denominator of fractions is the same, then just add or subtract the numerators and keep the denominator as it is.
• If the denominator of the algebraic fractions is different, then find the lowest common multiple of those denominators.
• Express all fractions in terms of the lowest common denominator.
• Perform the required operation among the numerators to obtain the result.

Solved Example Questions on Sum & Difference of Algebraic Fractions

Example 1.

Find the sum of a / (a – b) + b / (a² – b²)?

Solution:

We can observe that the denominators of the fractions are different. Those are (a-b) and (a² – b²).

The factors of denominators are (a – b), and (a + b) (a – b).

L.C.M of (a-b), (a² – b²) is (a – b) (a + b)

To make the two fractions having common denominator both the numerator and denominator of these are to be multiplied by (a * (a + b)) / ((a + b) (a – b)) in case of a / (a – b), (b * 1) / ((a – b) (a + b)) in case of b / (a² – b²).

Therefore, a / (a – b) + b / (a² – b²)

= (a * (a + b)) / ((a + b) (a – b)) + b / ((a + b) (a – b))

= (a (a + b) + b) / ((a + b) (a – b))

= (a² + ab + b) / (a² – b²).

Example 2:

Find the difference of (x² + 5x + 6) / (7x + 7y) – (y² – 8y + 16) / (x² – xy)?

Solution:

We can observe that the denominators of the fractions are different. Those are (7x + 7y), (x² – xy).

The factors of denominators are 7 (x + y), x (x – y).

Least common multiple of denominators (7x + 7y), (x² – xy) is 7x (x + y) (x – y)

To make the two fractions having common denominator both the numerator and denominator of these are to be multiplied by [(x² + 5x + 6) * x (x – y)] / [7x (x + y) (x – y)] in case of (x² + 5x + 6) / (7x + 7y), [(y² – 8y + 16) * 7 (x + y)] / [7x (x + y) (x – y)] in case of (y² – 8y + 16) / (x² – xy).

Therefore, (x² + 5x + 6) / (7x + 7y) – (y² – 8y + 16) / (x² – xy)

= [(x² + 5x + 6) * x (x – y)] / [7x (x + y) (x – y)] – [(y² – 8y + 16) * 7 (x + y)] / [7x (x + y) (x – y)]

= [x (x – y) (x² + 3x + 2x + 6)] / [7x (x + y) (x – y)] – [7 (x + y) (y² – 4y – 4y + 16)] / [7x (x + y) (x – y)]

= [x (x – y) (x (x + 3) + 2 (x + 3))] / [7x (x + y) (x – y)] – [7 (x + y) (y (y – 4) – 4 (y – 4))] / [7x (x + y) (x – y)]

= [x (x – y) (x + 3) (x + 2)] / [7x (x + y) (x – y)] – [7 (x + y) (y – 4) (y – 4)] / [7x (x + y) (x – y)]

= [x (x – y) (x + 3) (x + 2) – 7 (x + y) (y – 4) (y – 4)] / [7x (x + y) (x – y)]

= [x (x – y) (x + 3) (x + 2) – 7 (x + y) (y – 4)²] / [7x (x + y) (x – y)].

Example 3.

Simplify the algebraic fractions 1 / (m – n) – 1 / (m + n) + 2n / (m² – n²)?

Solution:

We can say that all the denominators are different, those are (m – n), (m + n), (m² – n²)

The factors of denominators are (m – n), (m + n), (m + n) (m – n)

L.C.M of denominators is (m + n) (m – n)

To make the fractions having common denominator both the numerator and denominator of these are to be multiplied by (m + n) / [(m + n) (m – n)] in case of 1 / (m – n), (m – n) / [(m + n) (m – n)] in case of 1 / (m + n), 2n / [(m + n) (m – n)] in case of 2n / (m² – n²).

Therefore, 1 / (m – n) – 1 / (m + n) + 2n / (m² – n²)

= (m + n) / [(m + n) (m – n)] – (m – n) / [(m + n) (m – n)] + 2n / [(m + n) (m – n)]

= [m + n – (m – n) + 2n] / [(m + n) (m – n)]

= [m + n – m + n + 2n] / [(m + n) (m – n)]

= 4n / [(m + n) (m – n)].

On this page, you will learn completely about the rule of separation of division of algebraic fractions. Get to see the solved example questions in the coming sections of this article. Solved Examples on Separation of Division of Algebraic Fractions will make you familiar with the concept in a better way and you can solve related problems easily.

Rule of Separation of Division

• (x + y) / z = x/z + y/z
• (x – y) / z = x/z – y/z
• z / (x + y) ≠ z/x + z/y

From the above three expressions, we can observe that the denominator of the fractions should be the same to perform addition or subtraction operator between them. The rule of separation of division says that to calculate sum or difference between two or more fractions, you need to make a common denominator for them and add or subtract numerators.

Examples on Separation of Division of Algebraic Fractions

Example 1.

Find the difference of fractions by taking the common denominator: x / bc – y / ab?

Solution:

We observe the two denominators are bc and ab and their L.C.M. is abc. So, abc is the least quantity which is divisible by ab and bc. To subtract those fractions, you must make a common denominator i.e abc. To make denominator as abc for x / bc multiply it with a, and multiply y / ab with c.

Therefore, we can write

(x * a) / abc – (y * c) / abc

= (ax + cy) / abc

Example 2.

Find the sum of fractions by taking the common denominator: a / xy + b / xz + c / yz.

Solution:

There are three denominators xy, xz, and yz, and their L.C.M. is xyz. To make the fractions with the common denominator, the numerator and denominator of these are to be multiplied by xyz ÷ xy = z in case of a/xy, xyz ÷ yz = x in case of c/yz, xyz ÷ xz = y in case of b/xz.

Therefore, we can write

a / xy + b / xz + c / yz

= (a.z) / xyz + (b.y) / xyz + (c.x) / xyz

= (az + by + cx) / xyz

Example 3.

Solve fractions p/qr + q/pr – r/pq by taking the common denominator.

Solution:

We can observe that three fractions have denominators as qr, pr, and pq their L.C.M is pqr. To make a common denominator for three fractions, their numerators should be multiplied by pqr ÷ qr = p for p/qr, pqr ÷ pr = q for q/pr, and pqr ÷ pq = r for r/pq.

Therefore, we can write

p/qr + q/pr – r/pq

= (p.p) / pqr + (q.q) / pqr – (r.r) / pqr

= (p² + q² – r²) / pqr.

If you are searching everywhere for assistance regarding the concept functions then take the help of our Math Practice Test on Functions. Solve as many questions as possible from the Functions Practice Test and get various question models. We have provided step by step solution for several examples of Functions so that you can understand them better. Become familiar with the entire concept of Functions by using the Practice Test on Functions.

• Relations and Mapping
• Ordered Pair
• Cartesian Product of Two Sets
• Relation in Math
• Domain and Range of a Relation
• Practice Test on Math Relation
• Functions or Mapping
• Domain Co-domain and Range of Function

1. Consider A = {4, 5, 6} and B = {3, 5, 7, 8}. Find if R = {(4, 3) (5, 7) (6, 8)} is a mapping from A to B and give reasons to support your answer?

Solution:

Since R = {(4, 3) (5, 7) (6, 8)}

No two ordered pairs have the same first components, R is Mapping from A to B.

2. Which of the following relations are functions?

(a) R₁ = {(4, 7) (5, 7) (6, 7) (7, 7)}

(b) R₂ = {(1, 3) (1, 4) (1, 5) (1, 6)}

(c) R₃ = {(a,b) (b, c) (c, d) (d, e)}

Solution:

R₃ is a function since each element in one set is associated with a unique element in the other set. Thus, it is a function.

3. Given f(x) = 2x+3 find

(i) f(1) (ii) f(-2) (iii) f(3)

Solution:

Given f(x) = 2x+3

(i) f(1) = 2*1+3 = 2+3 = 5

(ii) f(-2) = 2*(-2)+3 = -4+3 = -1

(iii) f(3) = 2*3+3 = 6+3 = 9

4. If f(x) = 4x – 3, x∈R and f(x) = 15 find the value of x?

Solution:

Given f(x) = 4x – 3

f(x) = 15

Equating both we can get the value of x as such

4x-3 = 15

4x = 15+3

4x = 18

x = 18/4

= 9/2

5. If A = {a, b, c, d} B = {x, y, z} Is R = {(a, x) (a, y) (a, z) (b, x) (b, y) (b, z) (c, x) (c, y) (d, z)} a function from A to B. Give reasons to support your answer?

Solution:

No, Relation R is not a function since the first components are repeated.

6. Which set of ordered pairs is not a function?

(a) (3, 2) (2, 6) (3, 5) (b) (4, 2) (7, 6) (2, 5)

(c) (1, 3) (3, 5) (4, 9) (d) (1, 2) (5, 2) (6, 5)

Solution:

(a) (3, 2) (2, 6) (3, 5)

The above option isn’t a function since the set of ordered pairs given has the repeated first component.

7. Recalling function notation we want x to be y and f(x) = 2x+3/2, find f(y)?

Solution:

f(x) = 2x+3/2

since we want function notation x to be y replace x with y in the given function

f(y) = (2y+3)/2

8. The Revenue of a Company is given by R(x) = X²+800X and Cost C(x) = 400X+150. Find the Profit gained by the Company R(x)- C(x)?

Solution:

R(x) = X²+800X

C(x) = 400X+150

Profit = R(x)- C(x)

= X²+800X – (400X+150)

= X²+800X – 400X – 150

= X²+ 400X – 150

9. The function S(r) =4πr² gives the surface area of a sphere having radius r. What is the surface area of a sphere for  radius 2?

Solution:

Given S(r) =4πr²

Substitute 2 in place of the radius in the given function to find surface area of sphere with radius 2.

S(2) = 4.π.2²

= 4π.4

= 16π

Surface Area of Sphere for Radius 2 is 16π.

10. Does the following arrow diagram represent a function?

Solution:

Every element in the first set is associated with only element in the other set. Thus, the relation F qualifies to be a function.

You are familiar with the terms Domain and Range of a Function let’s dive deep into the article to know what a function is. You need to be familiar with relations to better understand the concept of functions. A relation is a rule that relates an element from one set to the other set. The function is a special kind of relation.

Suppose a Relation F from Set A to B. A Relation F is said to be a function if each element in Set A is associated with only one element in Set B. Understand the difference between Relation and Domain by checking out the following sections.

     Fig 1 Arrow Diagram for Relation R

The above diagram Fig 1 is an example of relation because it relates elements from one set to the other set.

   Fig 2 Arrow Diagram for Function F

Fig 2 is an Example of Function since each element in set A is associated with only one element in Set B. It is not related to more than one element thus it is a function.

• Relations and Mapping
• Ordered Pair
• Cartesian Product of Two Sets
• Relation in Math
• Domain and Range of a Relation
• Practice Test on Math Relation
• Functions or Mapping
• Math Practice Test on Functions

Domain and Range of a Function

Do remember that the domain might not be the same in the left arrow diagram. This is because some of the elements in the Set may not have images in the other set whereas, in the case of functions, the domain will always be the first set.  However, Range and Codomain are similar to the way they are defined in Relations.

Consider function f represented from A to B i.e. f: A → B (f be a function from A to B), then

● Set A is the domain of the function ‘f’

● Set B is the co-domain of the function ‘f’

● The set of all f-images of all the elements of A is known as the range of “f’ and is denoted by f(A).

The set of all Possible values considered as inputs of the function is called the Domain of the function. Set of all the outputs of a function are known as Range of a Function.

Solved Examples on Domain Codomain and Range of a Function

1.  Does the arrow diagrams mentioned below represents mapping? Give reasons supporting your answer?

Solution:

The above diagram depicted is Mapping since each element in Set A is associated with only one element in Set B.

2.  Does the arrow diagrams mentioned below represent mapping? Give reasons supporting your answer?

Solution:

In the above arrow diagram element 2 in Set X is associated with B, C in Set Y. In fact, few elements in Set X are not related. Thus, the arrow diagram is not a Mapping.

3. Find out whether R is a mapping from A to B or not?

(i) Let A = {1, 2, 3} and B= {4, 5, 6, 7} and R = {(1, 4) (2, 5) (3, 6)}

(ii) Let A = {4, 5, 6} and B= {7, 8} and R = {(4, 7); (5, 8); (6, 7); (6, 8)}

Solution:

(i) Let A = {1, 2, 3} and B= {4, 5, 6, 7} and R = {(1, 4) (2, 5) (3, 6)}

Since, R ={(1, 4) (2, 5) (3, 6)} then Domain (R) = {1, 2, 3} = A

No two ordered pairs have the same first components therefore, R is a Mapping From A to B.

(ii) Let A = {4, 5, 6} and B= {7, 8} and R = {(4, 7); (5, 8); (6, 7); (6, 8)}

Since R = {(4, 7); (5, 8); (6, 7); (6, 8)}

Ordered Pairs (6, 7), (6, 8) has the same first components R is not a Mapping from A to B.

4. Let A = {0, 1, 2} and B = {0, 1, 3, 6, 7, 9}

Consider a rule f (x) = 2x² + 1, x∈A, then

Solution:

Using f (x) = 2x² + 1, x ∈ A we have

f(0)= 2*0+1 = 1

f(1) = 2*1+1 = 3

f(2) =2*4+1= 9

Every element in Set A has a unique image in Set B. Thus, f is a Mapping from A to B.

In general, a Function is a kind of relation in which each element in the domain is paired with only one element in the range. To help you understand the concept of Functions or Mapping we have provided a detailed explanation along with solved examples. Mapping Diagram consists of two columns in which the first column denotes the domain and the second column denotes the range.

A mapping diagram usually represents the relationship between input and output values. A mapping diagram is called a function if each input value is paired with only one output value.

• Relations and Mapping
• Ordered Pair
• Cartesian Product of Two Sets
• Relation in Math
• Domain and Range of a Relation
• Practice Test on Math Relation
• Domain Co-domain and Range of Function
• Math Practice Test on Functions

Introduction to Mapping or Function

Let us assume there are two sets A and B and the relation between Set A to Set B is called the function or Mapping.

Every element of Set A is associated with a unique element of Set B. Function f from A to B is represented by f : A → B. Relation will have a set of ordered pairs. In the Ordered Pairs, the second element is called the image of the first element and the first element is the preimage of the second element.

If f: A → B and x ∈ A, then f(x) ∈ B where f(x) is called the image of x under f & x is called the pre-image of f(x) under ‘f’.

For f to be Mapping from A to B

Different Elements of A can have the Same Image in B. Thus, the following figure represents Mapping.

No element of A should have more than one image. The below figure doesn’t represent mapping since the element in Set A is having two images i.e. I, III.

Every element of A must have an Image in Set B. The adjacent figure doesn’t represent mapping since 1, 2 are not associated with the elements in Set B.

Thus, we can infer that every mapping is a relation but not every relation is mapping.

Function as a Special Kind of Relation

Suppose A and B are two non-empty sets then rule f associates each element of A with a unique element in B is known as function or mapping from A to B.

We can denote f as a mapping from A to B in f: A → B and read as f is a function from A to B.

If f: A → B and x ∈ A and y ∈ B then y is called the image of element x under f and function f is given by f(x).

Thus can we write as y = f(x)

where x is the pre-image of y.

Therefore, For a function from A to B.

● Set A and Set B should be non-empty.
● Every element of Set A should have an image in Set B.
● No Element in Set A should have more than one image in Set B.
● Two or more elements of Set A can have the same image in Set B.
● f: x → y means that under the function of ‘f’ from A to B, an element x in Set A has image y in Set B.
● It is necessary that every f image is in Set B but there may be some elements in Set B that are not f images of any elements of Set A.

How to Identify Functions from Mapping Diagrams?

A Mapping Diagram represents a function if each input value is paired with only one output value.

Example:

The following figure represents a function since each element in Set A is paired with only one element in Set B.

Practice Test on Math Relations helps students to get a grip on the concept of Relations. Get to know about various questions like Representation of a Relation, finding Domain and Range, Cartesian Product of Two Sets, etc. Learn topics of Relations easily and solve the questions in Math Relation Worksheet whenever you want. Identify the knowledge gap and concentrate on the areas you are lagging and improvise on them.

1. Let A = {3, 4, 5, 6} and B = {x, y, z}. Let R be a relation from A into B defined by R = {(3, x), (4, y), (5, z), (6, x), (6, y)} find the domain and range of R.

Solution:

Given Relation R = {(3, x), (4, y), (5, z), (6, x), (6, y)}

The domain is the first component of the ordered pairs in the Relation R whereas Range is the second component of the ordered pairs. Repetition is not allowed.

Domain = {3, 4, 5, 6}

Range = {x, y, z}

• Relations and Mapping
• Ordered Pair
• Cartesian Product of Two Sets
• Relation in Math
• Domain and Range of a Relation
• Functions or Mapping
• Domain Co-domain and Range of Function
• Math Practice Test on Functions

2. Find the Domain and Range Value from the given tabular form

Solution:

Domain = {-18, -16, -10, -8, 3, 7}

Range = {11}

3. Determine the domain and range of the given functions:
{7, -9), (1, -5), (4, 3), (8, 5), (8, -14)}

Solution:

Given Relation = {7, -9), (1, -5), (4, 3), (8, 5), (8, -14)}

The domain is the first component of the ordered pairs in the Relation R whereas Range is the second component of the ordered pairs. Repetition is not allowed.

Domain = {7, 1, 4, 8} Range R = {-9, -5, 3, 5, -14}

4. What is the domain of this set of ordered pairs?

{ (3, 7), (4, -3), (1, 5), (-10, 6) }

Solution:

The domain is the first component in the ordered pairs.

Domain = { 3, 4, 1, -10}

5. If Set A = { 3, 4, 5}, Set B = {x, y, z} find the Cartesian Product of Two Sets?

Solution:

Set A = { 3, 4, 5}, Set B = {x, y, z}

AxB = {(3, x), (4, x), (5, x), (3, y), (4, y), (5, y), (3, z), (4, z), (5, z)}

6. Which of the following statement is true

a) All the relations are functions.
b) In every relation, each input value has exactly one output value.
c) A relation is defined as a set of input and output values that are related in some way.
d) All the above statements are true.

Solution:

c) A relation is defined as a set of input and output values that are related in some way.

7. The following Arrow Diagram shows a Relation from Set A to Set B. Find the Domain and Range?

Solution:

The domain is the first component of the ordered pairs in the Relation R whereas Range is the second component of the ordered pairs. Repetition is not allowed.

Domain = { -2, 2, 4, 5, 6}

Range = { 4, 16, 25, 36}

8.  The below figure shows a relation between Set x and Set y. Write the same in Roster Form, Set Builder Form, and find the domain and Range?

Solution:

In the Set Builder Form R = {(x, y): x is the square of y, x ∈ X, y ∈ Y}

In Roster Form R = {(2, 1)(4, 2)}

Domain = {2, 4}

Range = {1, 2}

9. What can you say about the ordered pairs (x, y) and (y, x)?

Solution:

Ordered Pairs (x, y)  ≠ (y, x).

In case of Ordered Pairs Order Matters.

10. If A × B = {(a, 2); (a, 6); (a, 3); (b, 3); (b, 6); (b, 2)}, find B × A.

Solution:

Given AxB = {(a, 2); (a, 6); (a, 3); (b, 3); (b, 6); (b, 2)}

BxA = {(2, a); (6, a); (3, a); (3, b); (6, b); (2, b)}

A Relation is an Important Concept in Set Theory. Sets, Relations, and Functions are all interlinked. Relations define the Operations Performed on Sets. You can know the connection between given sets using Relations. Get to know about various types of Relations, Representation of Relations in Different Forms, etc. in detail.

• Relations and Mapping
• Ordered Pair
• Cartesian Product of Two Sets
• Domain and Range of a Relation
• Practice Test on Math Relation
• Functions or Mapping
• Domain Co-domain and Range of Function
• Math Practice Test on Functions

Relation Definition

A Relation in math defines the relationship between two different sets of information. If there are two sets then the relation between them is built if there is a connection between elements of two or more non-empty sets.

Types of Relations

There are 8 major types of Relations. We have listed the definitions of each relation along with a Solved Example in the coming modules.

• Empty Relation
• Universal Relation
• Identity Relation
• Inverse Relation
• Reflexive Relation
• Symmetric Relation
• Transitive Relation
• Equivalence Relation

Empty Relation:

In Empty Relation, there will be no relation between any elements of the set. It is also known as Empty Relation and is denoted by R = φ ⊂ A × A

Universal Relation:

In the case of a Universal Relation, every element of the set is associated with each other. Let us assume a set A = { a, b, c}. Universal Relation is given by R = {x, y} where |x – y| ≥ 0. For Universal Relation R = AxA.

Identity Relation:

Every element of a set is related to itself. Consider the Set A = {a, b, c} then Identity Relation is given by I = {a, a}, {b, b}, {c, c}. Identity Relation is given by I = {(a, a), a ∈ A}.

Inverse Relation:

If a set has elements that are inverse pairs of another set then they are said to be in Inverse Relation. If Set A = {(u, v), (w,x)} then inverse relation R^-1 = {(v, u), (x, w)}. For an inverse relation R^-1 = {(b, a): (a, b) ∈ R}.

Reflexive Relation:

Every element mapped to itself is called a Reflexive Relation. Consider a Set A = {3, 4} then reflexive relation is given by R = {(3,3), (4, 4) (3, 4), (4, 3)}. In general, Reflexive Relation is denoted by (a, a) ∈ R.

Symmetric Relation:

If a= b is true, then b = a is also true in the case of a Symmetric Relation. You can say that a Relation is Symmetric if (a, b) ∈ R is true then (b, a) ∈ R is true. Symmetric Relation is denoted by

aRb ⇒ bRa, ∀ a, b ∈ A

Transitive Relation:

A Relation is said to be transitive if (x, y) ∈ R, (y, z) ∈ R, then (x, z) ∈ R. Transitive Relation is denoted by aRb and bRc ⇒ aRc ∀ a, b, c ∈ A

Equivalence Relation:

If a Relation is Transitive, Symmetric and Reflexive at the Same Time is Known as Equivalence Relation.

Representation of Relation

The relationship between Set A to Set B can be expressed in different forms and they are as under

• Roster Form
• Set Builder Form
• Arrow Diagram

Roster Form: In this form, the relation from one set to the other set is expressed using the ordered pairs. Consider two sets A and B then the first component in each ordered pair is from Set A whereas the second component is from Set B. Remember the Relation dealing with is (>, <, etc.)

Example

If A ={x, y, z} B = {4, 5, 6}

then R = {(x, 4), ((y,5), (z,6)}

Thus, R ⊆ A × B.

Set Builder Form: In the Set Builder Form relation between Set A and Set B is given by R = {(a, b): a ∈ A, b ∈ B, a…b} where blank space denotes the rule that associates a and b.

Example

Let A = { 2, 3, 4, 5, 6} B = { 6, 7, 8, 9} then R = {(2,6), (3, 7), (4,8), (5, 9)}. Here R is in the Set Builder Form in which R = {a, b} : a ∈ A, b ∈ B, a is 4 less than b.

Arrow Diagram: 

• Draw two circles denoting Set A and Set B.
• Note down the corresponding elements of the Set A in Circle A and elements of the Set B in the Circle B.
• Draw Arrows from A to B that satisfy the relation and indicate the ordered pairs too.

Example:

If A = {1, 2, 3} and B ={5, 7, 9} then Relation R from Set A to Set B is denoted by the arrow diagram as such

Here R = {(1,5), (1,7), (2, 7), (3, 9)}

Relation from Set A to Set B is shown by drawing arrows from the 1st components to the 2nd components of all ordered pairs that belong to R.

Let us consider A and B to be two non-empty sets and the Cartesian Product is given by AxB set of all ordered pairs (a, b) where a ∈ A and b ∈ B.

AxB = {(a,b) | a ∈ A and b ∈ B}. Cartesian Product is also known as Cross Product.

Consider Set A = { 3, 4, 5} B = {x, y} then AxB is given by

A

3 4 5

B

x

y

AxB = {(3,x), (4,x), (5, x), (3, y), (4, y), (5, y)}

In the same way, we can find the value of BxA

BxA = {(x,3), (x, 4), (x, 5), (y, 3), (y, 4), (y, 5)}

Thus from the example, we can say that AxB and BxA don’t have the same ordered pairs. Therefore, AxB ≠ BxA.

If A = B then AxB is called the Cartesian Square of Set A and is represented as A².

A² = {(a,b) a ∈ A and b ∈ A}

• Relations and Mapping
• Ordered Pair
• Relation in Math
• Domain and Range of a Relation
• Practice Test on Math Relation
• Functions or Mapping
• Domain Co-domain and Range of Function
• Math Practice Test on Functions

Solved Examples

1. If A = {3, 4, 5} B = {1, 2} find the value of AxB, BxA, A², B²?

Solution:

Given A = {3, 4, 5} B = {1, 2}

AxB = {3, 4, 5}x{1, 2}

= {(3,1), (3, 2), (4, 1), (4, 2), (5, 1), (5, 2)}

BxA = {1, 2}x{3, 4, 5}

= {(1, 3), (1,4), (1,5), (2, 3), (2, 4), (2,5)}

A² = {3, 4, 5}x{3, 4, 5}

= {(3, 3), (3, 4), (3, 5), (4, 3), (4, 4), (4, 5), (5, 3), (5, 4), (5, 5)}

B² = {1, 2}x{ 1, 2}

= {(1,1), (1,2), (2, 1), (2,2)}

2. If A = {x, y,z} then B = {y, z} find the Cartesian Product AxB?

Solution:

A = {x, y,z}

B = {y, z}

AxB = {x, y,z}x{y, z}

= {(x,y), (x,z), (y, y), (y, z), (z, y), (z, z)}

3. If A = { 4, 5, 6} B = {7, 8} find the Cartesian Product of AxB?

Solution:

A = { 4, 5, 6}

B = {7, 8}

AxB = {4, 5, 6}x{7, 8}

AxB = {(4,7), (4, 8), (5,7), (5, 8), (6, 7), (6, 8)}

Grouping of data plays an important role while dealing with a large amount of data. This information can also be displayed using a bar graph or pictograph. You can the grouping of data definition, solved examples in the below sections of this article. Furthermore, you will be familiar with the steps to draw a frequency distribution table for grouped data in the coming modules.

What is Meant by Grouping of Data?

When the number of observations is very large, we may divide the data into several groups, by using the grouping of data concept. The data formed by arranging the individual observations of a variable into groups, so that the frequency distribution table of these groups is a convenient way of summarizing the data. The advantages of grouping data are, it improves the accuracy/ efficiency of estimation, helps to focus on the important subpopulations, and ignores irrelevant ones.

Steps to Draw Frequency Distribution Table for Grouped Data

• From the given data, divide the data into some groups.
• Arrange the given observations in ascending order.
• Get the frequency of each observation.
• Write the frequency, group name in the frequency distribution table.

Get more useful information regarding the data handling such as the types of data, definition, and terms used in the data handling.

Solved Example Questions & Answers

Example 1.

The mass of 40 students in a class is given below. The measurement of the weight will be in kgs.

55, 70, 57, 73, 55, 59, 64, 72, 60, 48, 58, 54, 69, 51, 63, 78, 75, 64, 65, 57, 71, 78, 76, 62, 49, 66, 62, 76, 61, 63, 63, 76, 52, 76, 71, 61, 53, 56, 67, 71

State the frequency distribution table for the grouped data?

Solution:

Given that,

The weight of 40 students in a class are 55, 70, 57, 73, 55, 59, 64, 72, 60, 48, 58, 54, 69, 51, 63, 78, 75, 64, 65, 57, 71, 78, 76, 62, 49, 66, 62, 76, 61, 63, 63, 76, 52, 76, 71, 61, 53, 56, 67, 71

The ascending order of the students weight is 48, 49, 51, 52, 53, 54, 55, 55, 56, 57, 57, 58, 59, 60, 61, 61, 62, 62, 63, 63, 63, 64, 64, 65, 66, 67, 69, 70, 71, 71, 71, 72, 73, 75, 76, 76, 76, 76, 78, 78

The range = 78 – 48 = 30

The intervals should separate the scale into equal parts. We could choose intervals of 5. We then begin the scale with 45 and end with 79.

Frequency Distribution Table for Grouped Data is

Mass in Kg	Frequency
45 – 49	2
50 – 54	4
55 – 59	7
60 – 64	10
65 – 69	4
70 – 74	6
75 – 79	7
Total	40

Example 2.

The marks obtained by 40 students of Class VII in an examination are given below:
16, 17, 18, 3, 7, 23, 18, 13, 10, 21, 7, 1, 13, 21, 13, 15, 19, 24, 16, 2, 23, 5, 12, 18, 8, 12, 6, 8, 16, 5, 3, 5, 0, 7, 9, 12, 20, 10, 2, 23

State the frequency distribution table for the grouped data?

Solution:

Given data,

The marks scored by 40 students is 16, 17, 18, 3, 7, 23, 18, 13, 10, 21, 7, 1, 13, 21, 13, 15, 19, 24, 16, 2, 23, 5, 12, 18, 8, 12, 6, 8, 16, 5, 3, 5, 0, 7, 9, 12, 20, 10, 2, 23

The ascending order of the marks is 0, 1, 2, 2, 3, 3, 5, 5, 5, 6, 7, 7, 7, 8, 8, 9, 10, 10, 12, 12, 12, 13, 13, 13, 15, 16, 16, 16, 17, 18, 18, 18, 19, 20, 21, 21, 23, 23, 23, 24

Range is = 24 – 0 = 24

The intervals should separate the scale into equal parts. We could choose intervals of 5. We then begin the scale with 0 and end with 24.

Frequency Distribution Table for Grouped Data is

Marks Scored	Number of Students (Frequency)
0 – 4	6
5 – 9	10
10 – 14	8
15 – 19	9
20 – 24	7
Total	40

Example 3.

The height (in cms) of 35 persons are given below:

140, 125, 128, 126, 130, 134, 134, 146, 125, 140, 140, 127, 125, 137, 132, 134, 126, 128, 128, 129, 150, 151, 131, 138, 128, 129, 126, 127, 129, 130, 135, 132, 134, 140, 152

State the frequency distribution table for the grouped data.

Solution:

Given that,

The height of 35 persons are 140, 125, 128, 126, 130, 134, 134, 146, 125, 140, 140, 127, 125, 137, 132, 134, 126, 128, 128, 129, 150, 151, 131, 138, 128, 129, 126, 127, 129, 130, 135, 132, 134, 140, 152

Ascending order of the 35 persons height is 125, 125, 125, 126, 126, 126, 127, 127, 128, 128, 128, 128, 129, 129, 129, 130, 130, 131, 132, 132, 134, 134, 134, 134, 135, 137, 138, 140, 140, 140, 140, 146, 150, 151, 152

The range is 152 – 125 = 27

The intervals should separate the scale into equal parts. We could choose intervals of 4. We then begin the scale with 125 and end with 152.

Frequency Distribution Table for Grouped Data is

Height (in Cms)	Number of Persons (Frequency)
125 – 128	12
129 – 132	8
133 – 136	5
137 – 140	6
141 – 144	0
145 – 148	1
149 – 152	3
Total	35

Frequently Asked Questions on Grouping of Data

1. What is grouped and ungrouped data?

Grouped data is the data given in intervals whereas ungrouped data is the data without a frequency distribution.

2. What is a group of data called?

The frequency table is also called the grouped data. Grouped data is used in data analysis.

3. How can we convert ungrouped data to grouped data?

The first step of the conversion is to determine how many classes you have and find the range of data. And then divide the number of classes into groups.

Earlier we have learned to write a set in different forms, studied operations on sets and Venn diagrams,. In Coordinate System you will learn about Ordered Pair. Usually, Ordered Pair is used to locate a point in the Coordinate System. In Order Pair, we have a pair of integers in which the first one is called abscissa and the second one is called the ordinate.

Ordered Pair (3, 4) is not equal to the Ordered Pair (4, 3). Thus, Order in a Pair is important and Order Pair contains two elements written in a fixed order.

• Relations and Mapping
• Cartesian Product of Two Sets
• Relation in Math
• Domain and Range of a Relation
• Practice Test on Math Relation
• Functions or Mapping
• Domain Co-domain and Range of Function
• Math Practice Test on Functions

Ordered Pairs Definition

Pair of elements that occur in a particular order and enclosed within brackets is called a set of ordered pairs.

Let us consider a, b are two elements an Ordered Pair is represented as (a, b) where a is called the first component and b is called the second component. If Position of Components is Changed then Ordered Pair is Changed i.e it becomes (b, a) then (a, b) ≠ (b, a).

Equality of Ordered Pairs

Two Ordered Pairs are said to be equal if the corresponding first components are equal alongside the corresponding second components are equal. Consider two ordered pairs (u, v) and (x,y). The two ordered pairs are equal only if u = x, v = y i.e. (u, v) = (x, y).

Solved Examples

1.  Find the values of (2x – 2, y – 1) = (x + 3, 4)?

Solution:

As per the equality of ordered pairs, the two ordered pairs will be equal only if the corresponding first components and second components are equal.

As per the above statement, we can write the equation as

2x-2 = x+3 , y-1 = 4

Solve the expressions and find the values of x, y easily

2x-x = 3+2

x = 5

y-1= 4

y = 4+1

y = 5

Therefore, the values of x, y are 5, 5.

2. If (4a, 4) = (3a+2, b – 1) find the values of a, b?

Solution:

As per the equality of ordered pairs, the two ordered pairs will be equal only if the corresponding first components and second components are equal.

As per the above statement, we can write the equation as

4a = 3a+2

4a-3a = 2

a = 2

4 = b-1

4+1 = b

4 = b-1

b =5

Therefore, the values of a, b are 2, 5.

3. What is the value of x, y if Ordered Pairs (x, y) and (3, 5) are equal?

Solution:

As per the equality of ordered pairs, the two ordered pairs will be equal only if the corresponding first components and second components are equal.

(x,y) = (3, 5)

The values of x, y are 3, 5.