We first take up the properties of set operations on union and intersection.

**Properties of Set Operations**

It
is an interesting investigation to find out if operations among sets (like union,
intersection, etc) follow mathematical properties such as Commutativity, Associativity,
etc., We have seen numbers having many of these properties; whether sets also possess
these, is to be explored.

We
first take up the properties of set operations on union and intersection.

In
set language, commutative situations can be seen when we perform operations. For example, we can look into the Union (and Intersection)
of sets to find out if the operation is commutative.

**Note**

For any set *A*,

*â€¢ A*** **âˆª** ***A*** **=** ***A*** **and** ***A*** **âˆ© ** ***A*** **=** ***A*** **[Idempotent Laws].

â€¢ A âˆª *Ï•* = A and A âˆ© U = A [Identity Laws].

Let *A*
=
{2,
3, 8,10}
and *B* = {1, 3,10, 13}
be two sets.

Then, *A*
âˆª *B* =
{1,
2, 3, 8,10, 13} and

*B
*âˆª* A *=*
*{1,2,
3, 8,10,13}

From the
above, we see that *A* âˆª
*B* = *B* âˆª
*A*.

This is called
Commutative property of union of sets.

Now, *A*
âˆ©
*B* = {3, 10} and *B* âˆ©
*A* = {3, 10}. Then, we see that *A* âˆ©
*B* = *B* âˆ© *A* .

This
is called Commutative property of intersection of sets.

**Commutative property:**** **For any two sets** ***A*** **and** ***B*

(i) *A* âˆª *B =B* âˆª *A*

(ii) *A* âˆ© *B =B* âˆ© *A*

**Example 1.19**

If A = {*b, e, f ,g*} and B
= {*c, e, g,h*}, then verify the commutative
property of (i) union of sets (ii) intersection
of sets.

*Solution*

Given, A = { *b, e, f ,g *}
and *B* = { *c, e, g,h* }

(i) *A* âˆª *B* = { *b, c,e, f ,g , h* } ... (1)

* B* âˆª *A* = { *b, c, e, f, g, h* } ... (2)

From
(1) and (2) we have *A* âˆª
*B* = *B* âˆª
*A*

It
is verified that union of sets is commutative.

(ii) Aâˆ©B = { *e, g* } ... (3)

Bâˆ©A= {* e, g *} ... (4)

From
(3) and (4) we get, *A* âˆ© *B* =
*B* âˆ© *A*

It
is verified that intersection of sets is commutative.

**Thinking Corner**

Given, *P *= { *l*, *n* , *p *} and *P*
âˆª *Q = *{ *j *,*l *,*m *,* n*,*o *,* p*}. If* **P** *and* **Q *are disjoint sets, then what will be* *Q and *P* âˆ©*Q* ?

**Note**

Recall that subtraction on numbers is not commutative. Is set difference
commutative? We expect that the set difference is not commutative as well. For instance,
consider *A* = { *a , b, c* }, B = { *b, c, d* }. Aâˆ’B = { *a* }, Bâˆ’A= { *d* } ; we see that
A âˆ’ B â‰ B âˆ’ A.

Now, we perform
operations on union and intersection for three sets.

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

Now,
B âˆª C = {-3, 0,1,2, 3, 4}

Aâˆª(B âˆªC) = {âˆ’1, 0,1,2} âˆª{âˆ’3, 0,1,2, 3, 4}

= {-3,-1, 0,1,2, 3, 4} ... (1)

Then,
A âˆª B = {-3,-1, 0,1,2, 3}

(AâˆªB)âˆªC = {âˆ’3,âˆ’1, 0,1,2, 3} âˆª{0,1, 3, 4}

= {-3,-1, 0,1,2, 3, 4} ... (2)

From
(1) and (2), *A* âˆª
(*B* âˆª*C* ) = (*A* âˆª *B* ) âˆª*C* .

This
is associative property of union among sets*A*, *B*, and *C*.

Now,
B âˆ©C = {0, 3}

Aâˆ©(B âˆ©C) = {âˆ’1, 0,1,2} âˆ©{0, 3}

= {0} ... (3)

Then,
A âˆ© B = {0,2}

(Aâˆ©B)âˆ©C = {0,2} âˆ© {0,1, 3, 4}

= {0} ... (4)

From
(3) and (4), *A* âˆ© (*B* âˆ©*C* ) = (*A* âˆ©
*B* ) âˆ©*C* .

This
is associative property of intersection among sets*A*, *B* and *C*.

Associative property: For any three sets A, B and C

(i) *A *âˆª* *(*B *âˆª*C *)* *=* *(*A *âˆª* B *)* *âˆª*C*

(ii) *A* âˆ© (*B* âˆ©*C* ) = (*A* âˆ© *B* ) âˆ©*C*

**Example 1.20**

If then verify *A* âˆ© (*B* âˆ©*C* ) = (*A* âˆ©
*B* ) âˆ©*C* .

*Solution*

From
(1) and (2), it is verified that

(*A*âˆ©*B*)âˆ©*C* = *A*âˆ©(*B*
âˆ©*C*)

**Note**

The set difference in general is not associative

that is, (*A*â€“*B*)â€“*C *â‰ *A*â€“(*B*â€“*C*).

But, if the sets *A*, *B* and *C* are mutually disjoint
then the set difference is associative

that is, (*A*â€“*B*)â€“*C*=*A*â€“(*B*â€“*C*).

** **

In
lower classes, we have studied distributive property of multiplication over
addition on numbers. That is, *a* Ã— (*b* +*c* ) = (*a* Ã—*b* ) + (*a* Ã—*c*). In the same way we can define
distributive properties on sets.

**Distributive property**:** **For any three sets** ***A*,** ***B*** **and** ***C*

(i) *A *âˆ©* *(*B *âˆª*C *)* *=* *(*A *âˆ©* B *)* *âˆª* *(*A *âˆ©*C*)* *[Intersection over union]

(ii) *A *âˆª* *(*B *âˆ©*C *)* *=* *(*A *âˆª* B *)* *âˆ©* *(*A *âˆª*C*)* *[Union over intersection]

**Example 1.21**

If
*A* = {0, 2, 4, 6, 8},
*B* = {*x* : *x* is a prime number and *x* < 11}
and C= {*x* : *x* âˆˆ
*N* and 5 â‰¤ *x* <
9 } then verify *A* âˆª
(*B* âˆ©*C* ) = (*A* âˆª *B* ) âˆ©
(*A* âˆª*C*) .

*Solution*

Given A = {
0,2,4,6,8 } , B = { 2,3,5,7 } and C = { 5,6,7,8 }

First, we find B âˆ©C = {5,7 }, A âˆª(B âˆ©C) = {0,2,4,5, 6,7,8} ... (1)

Next, A âˆª B = { 0,2,3,4,5,6,7,8
} , AâˆªC= { 0,2,4,5,6,7,8 }

Then, (A âˆª B) âˆ© (A âˆªC) = { 0,2,4,5,6,7,8 } ... (2)

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

**Example 1.22**

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

*Solution*

From
(1) and (2), *A *âˆ©*
*(*B *âˆª*C *)* *=* *(*A *âˆ©* B *)* *âˆª* *(*A *âˆ©*C*)* *is verified.

Tags : Set Language | Maths , 9th Maths : UNIT 1 : Set Language

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9th Maths : UNIT 1 : Set Language : Properties of Set Operations | Set Language | Maths

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