Question

7. For A = {x|x is a prime factor of 42}, B {x|5 < x < 12,x E N} ! N and
C= {1,4,5,6} verify A+B(b+c)=(A+B)+C.

Answer 1

Question :

For A = {x|x is a prime factor of 42}, B ={x|5 < x ≤ 12, x ∈ N} and C = {1,4,5,6} verify A U (B U C) = (A U B) U C.

Union of two sets :
The union of the sets A and B is the set of all the element that belongs to either A or B or both. It is denoted by A U B(“A union B”).

Set builder form or rule method :
In this method the elements are described by a common property of the members.
We can represent a set of by stating a property which its elements satisfy. Thus,the set of natural numbers less than 10 is,
A = {x | x is a natural number,x <10}
Or
A = {x : x is a natural number and x <10}
•We can read this as “ A is the set of all elements x such that x is a natural number and x is less than 10.”
• ∈ - belongs to

SOLUTION:

A = {x|x is a prime factor of 42}
A = {2,3,7}

B ={x|5 < x ≤ 12, x ∈ N}
B = {6,7,8,9,10,11,12}

C = {1,4,5,6}

L.H.S -:
A U (B U C)

(B U C) = {6,7,8,9,10,11,12} U {1,4,5,6}
= {1,4,5,6,7,8,9,10,11,12}

A U (B U C) = {2,3,7} U {1,4,5,6,7,8,9,10,11,12}

A U (B U C) = {1,2,3,4,5,6,7,8,9,10,11,12}

R.H.S -:
(A U B) U C

(A U B) = {2,3,7} U {6,7,8,9,10,11,12}
= {2,3,6,7,8,9,10,11,12}

(A U B) U C = {2,3,6,7,8,9,10,11,12} U {1,4,5,6}

(A U B) U C = {1,2,3,5,6,7,8,9,10,11,12}

L.H S = R.H.S

HOPE THIS WILL HELP YOU….

Answer 2

Hi ,

factors of 42 = { 1 , 2 , 3 , 6, 7 , 14, 21 , 42 }

prime factors of 42 = { 2 , 3 , 7 }

A = { x/x is a prime factor of 42 }

A = { 2 , 3 , 7 }

B = { x/ 5 < x ≤ 12 , x € N }

B = { 6 , 7 , 8 , 9 , 10 , 11,12 }

C = { 1 , 4 , 5 , 6 }

Now ,

A U ( B U C )

= A U ( { 6 ,7 , 8 ,9 , 10,11,12 } U { 1 , 4 , 5 , 6 } )

= A U { 1 , 4 , 5, 6, 7, 8 , 9 , 10 , 11,12}

= { 2,3,7 } U { 1,4,5,6,7,8,9,10,11,12 }

= { 1,2,3,4,5,6,7,8,9,10,11,12 } ----( 1 )

( A U B ) U C

= ( { 2 ,3 , 7 } U { 6,7,8,9,10,11,12 } ) U C

= { 2 ,3,6,7,8,9,10,11,12 } U C

= { 2,3,6,7,8,9,10,11,12 } U { 1 , 4 , 5 ,6 }

= { 1 , 2 , 3 , 4 ,5 , 6 , 7 , 8 , 9 , 10 , 11,12 } ---( 2 )

from ( 1 ) and ( 2 ) , we conclude that

LHS = RHS

associative proporty .

I hope this helps you.

: )