Proving relationship between sets method I

using `A-B = AcapB'`

Replace `-` sign using above formula and than use Commutative , Associative distributive and De Morgan’s laws.

`color{blue}(ul"Some Properties of the Operation of Union") :`

(i) `A ∪ B = B ∪ A` (Commutative law)

(ii) `( A ∪ B ) ∪ C = A ∪ ( B ∪ C)` (Associative law )

(iii) ` A ∪ φ = A` (Law of identity element, `φ` is the identity of `∪`)

(iv) `A ∪ A = A` (Idempotent law)

(v) `U ∪ A = U` (Law of `U`)

`color{green}(=>" Idempotent Law" ):`

`color{blue}(ul"De Morgan’s laws")`

For any two finite sets `A` and `B;`

`(i)\ \ (A U B)' = A' ∩ B'` (which is a De Morgan's law of union).

`(ii) \ \ (A ∩ B)' = A' U B'` (which is a De Morgan's law of intersection).

Some more important Relationships

1. `A-B = A∩B'`

2. `A-(B∩C) = (A-B) ∪ (A-C)`

3. `A-(B∪C) = (A-B) ∩ (A-C)`

4.`A∩(B-C) = (A∩B) - (A∩C)`

using `A-B = AcapB'`

Replace `-` sign using above formula and than use Commutative , Associative distributive and De Morgan’s laws.

`color{blue}(ul"Some Properties of the Operation of Union") :`

(i) `A ∪ B = B ∪ A` (Commutative law)

(ii) `( A ∪ B ) ∪ C = A ∪ ( B ∪ C)` (Associative law )

(iii) ` A ∪ φ = A` (Law of identity element, `φ` is the identity of `∪`)

(iv) `A ∪ A = A` (Idempotent law)

(v) `U ∪ A = U` (Law of `U`)

`color{green}(=>" Idempotent Law" ):`

`color{blue}(ul"De Morgan’s laws")`

For any two finite sets `A` and `B;`

`(i)\ \ (A U B)' = A' ∩ B'` (which is a De Morgan's law of union).

`(ii) \ \ (A ∩ B)' = A' U B'` (which is a De Morgan's law of intersection).

Some more important Relationships

1. `A-B = A∩B'`

2. `A-(B∩C) = (A-B) ∪ (A-C)`

3. `A-(B∪C) = (A-B) ∩ (A-C)`

4.`A∩(B-C) = (A∩B) - (A∩C)`

Q 1774601556

For all sets `A` and `B, A- (A cap B) = A - B`.

NCERT Exemplar

NCERT Exemplar

`LHS = A- (A cap B)= A cap (A cap B)' quad [∵ A - B = A cap B']`

` = A cap (A' cup B') quad [∵ (A cap B)'= A' cup B']`

` = (A cap A') cup (A cap B') = phi cup (A cap B')`

` = A cap B' quad [ ∵ phi cup A = A]`

` = A - B = RHS`

Q 1744601553

For all sets `A` and `B, A cup (B - A) = A cup B`.

NCERT Exemplar

NCERT Exemplar

`LHS = A cup (B - A)= A cup (B cap A') quad [∵ A - B = A cap B']`

`= (A cup B) cap (A cup A') = (A cup B) cap U quad [∵ A cup A' = U]`

`= A cup B = RHS quad [ ∵ A cap U = A]`

Proving relationship between sets method 2

eg. to Prove De Morgan’s laws.

`(AcupB)' = A'cap B'`

Let `x in (Acup B)'`

Prove `x in A'cap B' => (AcupB)' subseteq A'capB' ...........................(1)`

Now take `y in A'cap B'`

Prove `Y in (Acup B)'`

`=> A'cap B' subseteq (A cup B)' ...............................................(2)`

from (I) & (2) `(A cupB)' = A' cap B'`

eg. to Prove De Morgan’s laws.

`(AcupB)' = A'cap B'`

Let `x in (Acup B)'`

Prove `x in A'cap B' => (AcupB)' subseteq A'capB' ...........................(1)`

Now take `y in A'cap B'`

Prove `Y in (Acup B)'`

`=> A'cap B' subseteq (A cup B)' ...............................................(2)`

from (I) & (2) `(A cupB)' = A' cap B'`

Q 1764701655

If `A, B` and `C` be sets. Then, show that `A cap (B cup C) = (A cap B) cup (A cap C)`.

NCERT Exemplar

NCERT Exemplar

Let ` x in A cap (B cup C)`

` => x in A` and `x in (B cup C)`

` => x in A` and `(x in B` or `x in C)`

` => (x in A` and `x in B )` or `(x in A` and `x in C )`

` => x in A cap B` or `x in A cap C`

` x in (A cap B) cup (A cap C)`

` => A cap (B cup C) subset (A cap B) cup (A cap C)` .........(1)

Again, let ` y in (A cap B) cup (A cap C)`

` => y in (A cap B)` or `y in (A cap C)`

` => (y in A` and `y in B)` or `(y in A` and `y in C)`

` => y in A` and (`y in B` or `y in C`)

` => y in A` and `y in B cup C`

` => y in A cap (B cup C)`

` => (A cap B) cup (A cap C) subset A cap (B cup C)` ............(2)

From Eqs.(1) and( 2),

` A cap (B cup C) = (A cap B) cup (A cap C)`

If A and B are finite sets, then

` n ( A ∪ B ) = n ( A ) + n ( B ) – n ( A ∩ B )`

- If A ∩ B = φ, then

` n ( A ∪ B ) = n ( A ) + n ( B )`

` n ( A ∪ B) = n ( A – B) + n ( A ∩ B ) + n ( B – A )`

If A, B and C are finite sets, then

`n ( A ∪ B ∪ C ) = n ( A ) + n ( B ) + n ( C ) – n ( A ∩ B ) – n ( B ∩ C) – n ( A ∩ C ) + n ( A ∩ B ∩ C )`

`n [ A ∩ ( B ∪ C ) ] = n ( A ∩ B ) + n ( A ∩ C ) – n [ ( A ∩ B ) ∩ (A ∩ C)]`

` n ( A ∪ B ) = n ( A ) + n ( B ) – n ( A ∩ B )`

- If A ∩ B = φ, then

` n ( A ∪ B ) = n ( A ) + n ( B )`

` n ( A ∪ B) = n ( A – B) + n ( A ∩ B ) + n ( B – A )`

If A, B and C are finite sets, then

`n ( A ∪ B ∪ C ) = n ( A ) + n ( B ) + n ( C ) – n ( A ∩ B ) – n ( B ∩ C) – n ( A ∩ C ) + n ( A ∩ B ∩ C )`

`n [ A ∩ ( B ∪ C ) ] = n ( A ∩ B ) + n ( A ∩ C ) – n [ ( A ∩ B ) ∩ (A ∩ C)]`

Q 2315767660

Out of `32` persons, `30` invest in National Savings

Certificates and `17` invest in shares. What is the

number of persons who invest in both?

NDA Paper 1 2010

Certificates and `17` invest in shares. What is the

number of persons who invest in both?

NDA Paper 1 2010

(A)

`13`

(B)

`15`

(C)

`17`

(D)

`19`

`n(N cup S)= 32, n(N) = 30, n(S) = 17`

We know that,

`n(N cup S)= n(N) + n(S) - n(N cap S)`

`=> 32 = 30 + 17 - n (N cap S)`

`=> n(N cap S) = 47 - 32 = 15`

Correct Answer is `=>` (B) `15`

Q 1762067835

In a survey of `25` students, it was found that `15` had taken

Mathematics, `12` had taken Physics and `11` had taken

Chemistry, `5` had taken Mathematics and Chemistry, `9` had

taken Mathematics and Physics, `4` had taken Physics and

Chemistry and `3` had taken all the three subjects.

The number of students who had taken only two subjects

is

NDA Paper 1 2014

Mathematics, `12` had taken Physics and `11` had taken

Chemistry, `5` had taken Mathematics and Chemistry, `9` had

taken Mathematics and Physics, `4` had taken Physics and

Chemistry and `3` had taken all the three subjects.

The number of students who had taken only two subjects

is

NDA Paper 1 2014

(A)

`7`

(B)

`8`

(C)

`9`

(D)

`10`

Given that,

Total number of surveyed students `= 25`

Number of students, who had taken all three subjects `= 3`

Number of students, who had taken Physics and Chemistry

` = 4`

Number of students, who had taken Mathematics and

Physics `= 9`

Number of students who had taken Mathematics and

Chemistry `= 5`

Number of students, who had taken Chemistry `= 11`

Number of students, who had taken Physics `= 12`

Number of students, who had taken Mathematics `= 15`

The number of students who had taken only two

subjects `= 6 + 2 + 1 = 9`

Correct Answer is `=>` (C) `9`