Math, asked by sivaramakrishnab81, 2 months ago

Two sets having identical elements are known as
a.
null sets
b. equal sets
C.
disjoint sets
d. None of the above​

Answers

Answered by tennetiraj86
5

Option b

Step-by-step explanation:

Given:-

Two sets having identical elements

To find:-

What type of these sets ?

Solution:-

Two sets having identical elements then the two sets are called Equal sets .

If A and B are having identical elements then both A and B are called equal sets and it is denoted by A=B.

That means All elements in A are in B and all elements in B area also in A.

Answer:-

Two sets having identical elements then the two sets are called Equal sets .

Options wise explanation:-

Null sets :-

A set having no element in it is called a null or void or empty set.

Ex:-

A is a set of whole numbers less than 0

Dis joint sets:-

If two sets having no elements in common then they are called Dis joint sets.

If A and B are dis joint sets then AnB = { }

n(AnB) = 0.

Answered by preronakakoti262
0

Solution The required numbers are 1, 2, 3, 4, 5, 6. So, the given set in the roster form

is {1, 2, 3, 4, 5, 6}.

Example 3 Write the set A = {1, 4, 9, 16, 25, . . . }in set-builder form.

Solution We may write the set A as

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

Alternatively, we can write

A = {x : x = n2

, where n ∈ N}

Example 4 Write the set

123456 { }

234567

,,,,, in the set-builder form.

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

the denominator. Also, the numerator begin from 1 and do not exceed 6. Hence, in the

set-builder form the given set is

where is a natural number and 1 6

1

n

x:x , n n

n

⎧ ⎫ ⎨ ⎬ = ≤≤ ⎩ ⎭ +

Example 5 Match each of the set on the left described in the roster form with the

same set on the right described in the set-builder form :

(i) {P, R, I, N, C, A, L} (a) { x : x is a positive integer and is a divisor of 18}

(ii) { 0 } (b) { x : x is an integer and x2 – 9 = 0}

(iii) {1, 2, 3, 6, 9, 18} (c) {x : x is an integer and x + 1= 1}

(iv) {3, –3} (d) {x : x is a letter of the word PRINCIPAL}

Solution Since in (d), there are 9 letters in the word PRINCIPAL and two letters P and I

are repeated, so (i) matches (d). Similarly, (ii) matches (c) as x + 1 = 1 implies

x = 0. Also, 1, 2 ,3, 6, 9, 18 are all divisors of 18 and so (iii) matches (a). Finally, x2 – 9 = 0

implies x = 3, –3 and so (iv) matches (b).

EXERCISE 1.1

1. Which of the following are sets ? Justify your answer.

(i) The collection of all the months of a year beginning with the letter J.

(ii) The collection of ten most talented writers of India.

(iii) A team of eleven best-cricket batsmen of the world.

(iv) The collection of all boys in your class.

(v) The collection of all natural numbers less than 100.

(vi) A collection of novels written by the writer Munshi Prem Chand.

hope \: this \: will \: help \: you

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