1. De Morganâ€™s Laws for Set Difference 2. De Morganâ€™s Laws for Complementation

**De Morganâ€™s Laws**

Augustus
De Morgan (1806 â€“ 1871) was a British mathematician. He was born on 27^{th}
June 1806 in Madurai, Tamilnadu, India. His father was posted in India by the East
India Company. When he was seven months old, his family moved back to England. De
Morgan was educated at Trinity College, Cambridge, London. He formulated laws for
set difference and complementation. These are called De Morganâ€™s laws.

** **

These
laws relate the set operations union, intersection and set difference.

Let
us consider three sets A, B and C as *A* = {âˆ’5, âˆ’2,1, 3}, *B* =
{âˆ’3,âˆ’2,
0, 3, 5}
and C = {âˆ’2,âˆ’1,
0, 4,5}.

Now, *B *âˆª*C *=* *{-3,-2,-1,
0, 3, 4,5}

A âˆ’ (B âˆªC) = { âˆ’5,1 } ... (1)

Then,

A âˆ’ B = {-5, 1}
and A âˆ’ C = { âˆ’5, 1, 3}

(Aâˆ’B)âˆª(Aâˆ’C) = { âˆ’5,1, 3 } ... (2)

(Aâˆ’B)âˆ©(Aâˆ’C) =
{ âˆ’5,1 } ... (3)

From (1) and (2), we see that

Aâˆ’(B âˆªC) â‰ (Aâˆ’B)âˆª(Aâˆ’C)

But note that from (1) and (3), we see that

Aâˆ’(B âˆªC) =
(Aâˆ’B)âˆ©(Aâˆ’C)

**Thinking Corner **

(Aâˆ’B) âˆª(Aâˆ’C) âˆª(A âˆ© B)=____

Now, B âˆ©C = {
âˆ’2, 0, 5 }

Aâˆ’(B âˆ©C) = { âˆ’5,1, 3 } ... (4)

From (3)
and (4) we see that

* A*âˆ’(*B *âˆ©*C*)* *â‰ *
*(*A*âˆ’*B*)âˆ©(*A*âˆ’*C*)

But
note that from (2) and (4), we get *A* âˆ’ (*B* âˆ©
*C* ) = (*A* âˆ’ *B* ) âˆª (*A* âˆ’*C*)

**De Morganâ€™s laws for set difference : **For any three sets** ***A*,**
***B*** **and** ***C*

(i) *A *âˆ’* *(*B *âˆª* C *)* *=* *(*A *âˆ’* B *)* *âˆ©* *(*A *âˆ’*C*)* *(ii)* A *âˆ’* *(*B *âˆ©* C *)* *=* *(*A *âˆ’* B *)* *âˆª* *(*A *âˆ’*C*)

**Example 1.23**

Verify
*A* âˆ’ (*B* âˆª
*C* ) = (*A* âˆ’ *B* ) âˆ©
(*A* âˆ’*C*) using Venn diagrams.

*Solution*

From
(1) and (2), we get *A* âˆ’ (*B* âˆª*C* ) = (*A* âˆ’
*B* ) âˆ© (*A* âˆ’*C*) . Hence it is verified.

**Example 1.24**

If
*P* = {*x* : *x*âˆˆ
**W** and 0 < *x* < 10}, *Q*
= {*x* : *x* = 2*n*+1, *n*âˆˆ
**W** and *n*<5} and *R* =
{ 2, 3, 5, 7,11, 13}, then verify *P* âˆ’ (*Q* âˆ©
*R* ) = (*P* âˆ’*Q* ) âˆª
(*P* âˆ’ *R*)

** Solution **The roster form of sets

*P
*=*
*{1,2,3,4,5,6,7,8,9} , Q = {1,3,5,7,9} and R = {2, 3, 5, 7,11, 13}

First, we find (Q âˆ© R) = { 3,5,7 }

Then, P âˆ’ (Q âˆ© R) = { 1,2,4,6,8,9 } ... (1)

Next, Pâ€“Q = { 2,4,6,8 } and

Pâ€“R = { 1,4,6,8,9 }

and so, (P âˆ’ Q) âˆª (P âˆ’ R) =
{ 1,2, 4,6, 8,9} ... (2)

Hence
from (1) and (2), it is verified that *P* âˆ’ (*Q* âˆ©
*R* ) = (*P* âˆ’*Q* ) âˆª
(*P* âˆ’ *R*).

**Finding the elements of
set Q**

Given, *x = 2n* + 1

*n *=* *0* *â†’* x *=* *2(0)* *+1* *=* *0* *+1* *=* *1

*n *=* *1* *â†’* x *=* *2(1)* *+1* *=* *2* *+1* *=* *3

*n *=* *2* *â†’* x *=* *2(2)* *+1* *=* *4* *+1* *=* *5

*n *=* *3* *â†’* x *=* *2(3)* *+1* *=* *6* *+1* *=* *7

*n *=* *4* *â†’* x *=* *2(4)* *+1* *=* *8* *+1* *=* *9

Therefore, *x* takes values such as 1, 3, 5, 7 and 9.

** **

These
laws relate the set operations on union, intersection and complementation.

**Thinking Corner : **Check whether Aâˆ’B =Aâˆ©Bâ€²

Let
us consider universal set *U*={0,1,2,3,4,5,6}, *A*={1,3,5} and *B*={0,3,4,5}.

Now, A âˆª B = {0,1,
3, 4,5}

Then, (A âˆª B)â€² = {2,6}
.....(1)

Next, *A*â€² =
{0, 2, 4, 6} and Bâ€² = {1,2,6}

Then, Aâ€² âˆ© Bâ€² = {2,6} .....(2)

From (1) and (2), we get (A âˆª B)â€² = Aâ€² âˆ© Bâ€²

Also, A âˆ© B = {3, 5},

**Thinking Corner** (Aâˆ’ B) âˆª (B âˆ’ Aâ€² ) = ____

(A âˆ© B)â€² = {0,1,2,
4, 6} .....(3)

Aâ€² = { 0,2,4,6 } and Bâ€² = { 1,2,6 }

Aâ€² âˆª Bâ€² = { 0,1,2,
4,6 } .....(4)

From
(3) and (4), we get (*A* âˆ© *B*)^{â€²}
=
*A*â€²
âˆª *B*â€²

**De Morganâ€™s laws for complementation : **Let â€˜*U*â€™ be the universal
set containing finite sets A and B. Then (i) (*A* âˆª *B*)â€² = *A*â€² âˆ© *B*â€² (ii) (*A* âˆ© *B*)â€² = *A*â€² âˆª *B*â€²

**Example 1.25**

Verify
(*A* âˆª *B*)â€²
=
*A*â€² âˆ©
*B*â€²
using Venn diagrams.

*Solution*

From
(1) and (2), it is verified that (*A* âˆª *B*)â€²
=
*A*â€² âˆ©
*B*â€²

**Example 1.26**

If
*U* = {*x* : *x* âˆˆ
â„¤, âˆ’ 2 â‰¤ *x* â‰¤
10},
*A *=* *{*x *:* x *=*
*2* p *+*
*1,* p *âˆˆ* *â„¤,* *âˆ’* *1* *â‰¤* p *â‰¤*
*4},* B *=*
*{*x
*:* x *=*
*3*q *+*
*1,* q *âˆˆ* *â„¤,* *âˆ’* *1* *â‰¤* q *<*
*4}, verify De Morganâ€™s laws for complementation.

*Solution*

Given
*U* = {âˆ’2, âˆ’1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
,

A= {âˆ’1, 1, 3, 5, 7, 9} and *B* = {âˆ’2,
1, 4, 7,10}

**Thinking Corner** A âˆ© (A âˆª B)â€² =____

Law (i) (*A* âˆª *B*)â€² = *A*â€² âˆ© *B*â€²

Now, A âˆª B = {-2,-1,1,
3, 4,5,7,9,10}

(A âˆª B)â€² = {0,2,6, 8} ..... (1)

Then, *A*â€² =
{-2, 0, 2, 4, 6, 8,10} and Bâ€² = {-1, 0,2, 3,5,6, 8,9}

* A*â€² âˆ© *B*â€² = {0,
2, 6, 8} ..... (2)

From (1) and (2), it is verified that (AâˆªB)â€² = Aâ€² âˆ© B â€²

Law (ii) (A âˆ© B)â€² =Aâ€² âˆªBâ€²

Now, A âˆ© B = {1,7}

(A âˆ© B)â€² = {-2,-1,
0,2, 3, 4,5,6, 8,9,10} ..... (3)

Then, Aâ€² âˆª Bâ€² = {-2,-1,
0,2, 3, 4,5,6, 8,9,10} .....
(4)

From (3) and (4), it is verified that (Aâˆ©B)â€² = Aâ€² âˆª Bâ€²

**Thinking Corner **(*A* âˆª *B*)â€² âˆª (*A*â€² âˆ© *B*) =___

Tags : Definition, Example Solved Problems | Set Language | Maths , 9th Maths : UNIT 1 : Set Language

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9th Maths : UNIT 1 : Set Language : De Morganâ€™s Laws | Definition, Example Solved Problems | Set Language | Maths

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