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# De MorganŌĆÖs Laws

1. De MorganŌĆÖs Laws for Set Difference 2. De MorganŌĆÖs Laws for Complementation

De MorganŌĆÖs Laws

Augustus De Morgan (1806 ŌĆō 1871) was a British mathematician. He was born on 27th 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 MorganŌĆÖs laws.

## 1. De MorganŌĆÖs Laws for Set Difference

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

Let us consider three sets A, B and C as A = {ŌłÆ5, ŌłÆ2,1, 3}, B = {ŌłÆ3,ŌłÆ2, 0, 3, 5} and C = {ŌłÆ2,ŌłÆ1, 0, 4,5}.

Now, B Ōł¬C = {-3,-2,-1, 0, 3, 4,5}

A ŌłÆ (B Ōł¬C) = { ŌłÆ5,1 }  ... (1)

Then,

A ŌłÆ B = {-5, 1} and A ŌłÆ C = { ŌłÆ5, 1, 3}

(AŌłÆB)Ōł¬(AŌłÆC) = { ŌłÆ5,1, 3 } ... (2)

(AŌłÆB)Ōł®(AŌłÆC) = { ŌłÆ5,1 } ... (3)

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

AŌłÆ(B Ōł¬C) ŌēĀ (AŌłÆB)Ōł¬(AŌłÆC)

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

AŌłÆ(B Ōł¬C) = (AŌłÆB)Ōł®(AŌłÆC)

Thinking Corner

(AŌłÆB) Ōł¬(AŌłÆC) Ōł¬(A Ōł® B)=____

Now,  B Ōł®C = { ŌłÆ2, 0, 5 }

AŌłÆ(B Ōł®C) = { ŌłÆ5,1, 3 } ... (4)

From (3) and (4) we see that

AŌłÆ(B Ōł®C) ŌēĀ (AŌłÆB)Ōł®(AŌłÆC)

But note that from (2) and (4), we get A ŌłÆ (B Ōł® C ) = (A ŌłÆ B ) Ōł¬ (A ŌłÆC)

De MorganŌĆÖs laws for set difference : For any three sets A, B and C

(i) A ŌłÆ (B Ōł¬ C ) = (A ŌłÆ B ) Ōł® (A ŌłÆC) (ii) A ŌłÆ (B Ōł® C ) = (A ŌłÆ B ) Ōł¬ (A ŌłÆC)

Example 1.23

Verify A ŌłÆ (B Ōł¬ C ) = (A ŌłÆ B ) Ōł® (A ŌłÆC) using Venn diagrams.

Solution From (1) and (2), we get A ŌłÆ (B Ōł¬C ) = (A ŌłÆ B ) Ōł® (A ŌłÆC) . Hence it is verified.

Example 1.24

If P = {x : xŌłł W and 0 < x < 10}, Q = {x : x = 2n+1, nŌłł W and n<5} and R = { 2, 3, 5, 7,11, 13}, then verify P ŌłÆ (Q Ōł® R ) = (P ŌłÆQ ) Ōł¬ (P ŌłÆ R)

Solution The roster form of sets P, Q and R are

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 Ōł® R) = {  3,5,7 }

Then, P ŌłÆ (Q Ōł® R) = { 1,2,4,6,8,9 } ... (1)

Next, PŌĆōQ = { 2,4,6,8 } and

PŌĆōR = { 1,4,6,8,9 }

and so, (P ŌłÆ Q) Ōł¬ (P ŌłÆ R) = { 1,2, 4,6, 8,9} ... (2)

Hence from (1) and (2), it is verified that P ŌłÆ (Q Ōł® R ) = (P ŌłÆQ ) Ōł¬ (P ŌłÆ R).

Finding the elements of set Q

Given, x = 2n + 1

n = 0 ŌåÆ x = 2(0) +1 = 0 +1 = 1

n = 1 ŌåÆ x = 2(1) +1 = 2 +1 = 3

n = 2 ŌåÆ x = 2(2) +1 = 4 +1 = 5

n = 3 ŌåÆ x = 2(3) +1 = 6 +1 = 7

n = 4 ŌåÆ x = 2(4) +1 = 8 +1 = 9

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

## 2. De MorganŌĆÖs Laws for Complementation

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

Thinking Corner : Check whether AŌłÆB =AŌł®BŌĆ▓

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 Ōł¬ B = {0,1, 3, 4,5}

Then, (A Ōł¬ B)ŌĆ▓ = {2,6} .....(1)

Next, AŌĆ▓ = {0, 2, 4, 6} and BŌĆ▓ = {1,2,6}

Then, AŌĆ▓ Ōł® BŌĆ▓ = {2,6} .....(2)

From (1) and (2), we get (A Ōł¬ B)ŌĆ▓ = AŌĆ▓ Ōł® BŌĆ▓

Also, A Ōł® B = {3, 5},

Thinking Corner (AŌłÆ B) Ōł¬ (B ŌłÆ AŌĆ▓ ) = ____

(A Ōł® B)ŌĆ▓ = {0,1,2, 4, 6} .....(3)

AŌĆ▓ = { 0,2,4,6 } and BŌĆ▓ = { 1,2,6 }

AŌĆ▓ Ōł¬ BŌĆ▓ = { 0,1,2, 4,6 } .....(4)

From (3) and (4), we get (A Ōł® B)ŌĆ▓ = AŌĆ▓ Ōł¬ BŌĆ▓

De MorganŌĆÖs laws for complementation : Let ŌĆśUŌĆÖ be the universal set containing finite sets A and B. Then (i) (A Ōł¬ B)ŌĆ▓ = AŌĆ▓ Ōł® BŌĆ▓ (ii) (A Ōł® B)ŌĆ▓ = AŌĆ▓ Ōł¬ BŌĆ▓

Example 1.25

Verify (A Ōł¬ B)ŌĆ▓ = AŌĆ▓ Ōł® BŌĆ▓ using Venn diagrams.

Solution From (1) and (2), it is verified that (A Ōł¬ B)ŌĆ▓ = AŌĆ▓ Ōł® BŌĆ▓

Example 1.26

If U = {x : x Ōłł Ōäż, ŌłÆ 2 Ōēż x Ōēż 10}, A = {x : x = 2 p + 1, p Ōłł Ōäż, ŌłÆ 1 Ōēż p Ōēż 4}, B = {x : x = 3q + 1, q Ōłł Ōäż, ŌłÆ 1 Ōēż q < 4}, verify De MorganŌĆÖs laws for complementation.

Solution

Given U = {ŌłÆ2, ŌłÆ1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} ,

A= {ŌłÆ1, 1, 3, 5, 7, 9} and B = {ŌłÆ2, 1, 4, 7,10}

Thinking Corner  A Ōł® (A Ōł¬ B)ŌĆ▓ =____

Law (i) (A Ōł¬ B)ŌĆ▓ = AŌĆ▓ Ōł® BŌĆ▓

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

(A Ōł¬ B)ŌĆ▓ = {0,2,6, 8} ..... (1)

Then, AŌĆ▓ = {-2, 0, 2, 4, 6, 8,10} and BŌĆ▓ = {-1, 0,2, 3,5,6, 8,9}

AŌĆ▓ Ōł® BŌĆ▓ = {0, 2, 6, 8} ..... (2)

From (1) and (2), it is verified that (AŌł¬B)ŌĆ▓ =  AŌĆ▓ Ōł® B ŌĆ▓

Law (ii) (A Ōł® B)ŌĆ▓ =AŌĆ▓ Ōł¬BŌĆ▓

Now, A Ōł® B = {1,7}

(A Ōł® B)ŌĆ▓ = {-2,-1, 0,2, 3, 4,5,6, 8,9,10}     ..... (3)

Then, AŌĆ▓ Ōł¬ BŌĆ▓ = {-2,-1, 0,2, 3, 4,5,6, 8,9,10}      ..... (4)

From (3) and (4), it is verified that (AŌł®B)ŌĆ▓ = AŌĆ▓ Ōł¬ BŌĆ▓

Thinking Corner (A Ōł¬ B)ŌĆ▓ Ōł¬ (AŌĆ▓ Ōł® B) =___

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9th Maths : UNIT 1 : Set Language : De MorganŌĆÖs Laws | Definition, Example Solved Problems | Set Language | Maths