11th Maths Chapter Set Theory

Class XI Maths Chapter Set Sets

1.1 Sets and their Representations

1.2 The Empty Set

1.3 Finite and Infinite Sets

1.4 Equal Sets

1.5 Subsets

1.6 Power Set

1.7 Universal Set

1.8 Venn Diagrams

1.9 Operations on Sets

1.10 Complement of a Set

The theory of sets was developed by German mathematician Georg Cantor (1845-1918). He first encountered sets while working on “problems on trigonometric series”.

A set is a well define collection of object.

Greek symbol ∈ (epsilon) is used to denote the phrase ‘belongs to’.

If ‘a’ is an element of a set A, we write a ∈ A.

If ‘b’ is not an element of a set A, we write b ∉ A and read “b does not belong to A”.

There are two methods we used to represents a set

a. Roaster or Tabular form: We use commas to separate elements of sets within braces { } without repainting elements. The order of listing elements has no relevance.

For example, the set of all even positive integers less than 5 is described in roster form as {2, 4,}.

b. Set builder form: We use this form if all the elements of a set possess a single common property which is not possessed by any element outside the set.

Example: V = {x : x is a vowel in English alphabet}

V is read as “the set of all x such that x is a vowel of the English alphabet”. In this description the braces stand for “the set of all”, the colon stands for “such that”.

The set of all natural numbers which divide 42 is written as   A= {x : x is a natural number which divides 42}

Q. Write the set {x: x is a positive integer and x² < 40} in the roster form.

Solution:  x² < 40 here,  x = 1, 2, 3, 4, 5, 6.(Whose square is less than 40 )

x  = {1, 2, 3, 4, 5, 6)

Q. Write the set A = {1, 4, 9, 16, 25…} in set-builder form

Ans: 1, 4, 9, 16, 25… are square of 1, 2, 3, 4, 5… respectively. Or Square of natural number

A = {x: x is the square of a natural number}

                               Or,

A = { x : x = n²  , Where n ∈ N}

Q. Write the set {1/2 ,  2/3, ¾, 4/5, 5/6, 6/7 }  in the set-builder form

Solution We see that each member in the given set has the denominator one  more than  the numerator

Also, the numerators begin from 1 and do not exceed 6.

{x : x = n/n+1 where is a natural number and 1£ n£6}

Finite and Infinite set

Let us see new set: B = { x : x is a student presently studying in both Classes X and XI }

We observe that a student cannot study simultaneously in both Classes X and XI. Thus, the set B contains no element at all. Such type of set is called Empty Set.

A set which does not contain any element is called the empty set or the null set or the void set.  The empty set is denoted by the symbol φ or { }

Let us see another new set 

A = {men living presently in your town}

here we do not know the numbers of element of this set

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

This set has a definite number of elements

Such type of sets ate called Infinite and Finite respectively.

Hence, A set which is empty or consists of a definite number of elements is called finite otherwise, the set is called infinite.

Note : All infinite sets cannot be described in the roster form. For example, the set of real numbers cannot be described in this form, because the elements of this set do not follow any particular pattern.

Q. State which of the following sets are finite or infinite:

(i) {x : x ∈ N and (x – 1) (x –2) = 0}

(ii) {x : x ∈ N and x2 = 4}

(iii) {x : x ∈ N and 2x –1 = 0}

(iv) {x : x ∈ N and x is prime}

(v) {x : x ∈ N and x is odd}

Solution (i) Given set = {1, 2}. Hence, it is finite.

(ii) Given set = {2}. Hence, it is finite.

(iii) Given set = φ. Hence, it is finite.

(iv) The given set is the set of all prime numbers and since set of prime numbers is infinite. Hence the given set is infinite

(v) Since there are infinite numbers of odd numbers, hence, the given set is infinite.

Equal Sets

Two set A and B are said to be equal if they have exactly the same elements and we

Q. write A = B. Otherwise, the sets are said to be unequal

Q. Find the pairs of equal sets, if any, give reasons:

A = {0}, B = {x : x > 15 and x < 5},

C = {x : x – 5 = 0 }, D = {x: x2 = 25},

E = {x : x is an integral positive root of the equation x2 – 2x –15 = 0}.

Solution:  Since 0 ∈ A and 0 does not belong to any of the sets B, C, D and E, it follows that, A ≠ B, A ≠ C, A ≠ D, A ≠ E.

Since B = φ but none of the other sets are empty. Therefore B ≠ C, B ≠ D and B ≠ E. Also C = {5} but –5 ∈ D, hence C ≠ D.

Since E = {5}, C = E. Further, D = {–5, 5} and E = {5}, we find that, D ≠ E. 

Thus, the only pair of equal sets is C and E.

Subsets

Consider the sets : X = set of all students in your school, Y = set of all students in your

class.

We note that every element of Y is also an element of X; we say that Y is a subset  of X. The fact that Y is subset of X is expressed in symbols as Y ⊂ X. The symbol ⊂ stands for ‘is a subset of’ or ‘is contained in’.

Definition 4 A set A is said to be a subset of a set B if every element of A is also an element of B

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