Question: If \(A = \left\\{ {3,5,7,9,11} \right\\},B = \left\\{ {7,9,11,13} \right\\},C = \left\\{ {11,13,15} ...

Question

If $A = \left\\{ {3,5,7,9,11} \right\\},B = \left\\{ {7,9,11,13} \right\\},C = \left\\{ {11,13,15} \right\\},D = \left\\{ {15,17} \right\\};$
Find
i. $A \cap B$
ii. $B \cap C$
iii. $A \cap C \cap D$
iv. $A \cap C$
v. $B \cap D$
vi. $A \cap \left( {B \cup C} \right)$
vii. $A \cap D$
viii. $A \cap \left( {B \cup D} \right)$
ix. $\left( {A \cap B} \right) \cap \left( {B \cap C} \right)$
x. $\left( {A \cup D} \right) \cap \left( {B \cup C} \right)$

Answer 1

Solution

To solve the above part we must have the information of what is intersection of the sets and what is the union of the sets, intersection of the set is the collection of the common elements in that set hence for intersection we need to find the common elements between the sets and union is the collection of all the elements of the set that means if we find the union of two sets then in that case the union will be the collection of all elements in both the sets.

Complete step-by-step answer:
To determine one by one
i. $A \cap B$
In the above we need to find the intersection that means we need to find the common elements in the set
$A = \left\\{ {3,5,7,9,11} \right\\}$
And
$B = \left\\{ {7,9,11,13} \right\\}$
If we have a look for the common elements that comes out to be $7,9,11$
Hence, the collection of the common elements between two sets is known as the intersection
$A \cap B = \left\\{ {3,5,7,9,11} \right\\} \cap \left\\{ {7,9,11,13} \right\\} = \left\\{ {7,9,11} \right\\}$
Hence, the above represent the required result

ii. $B \cap C$
We have provided that $B = \left\\{ {7,9,11,13} \right\\}$ and $C = \left\\{ {11,13,15} \right\\}$
In the above we need to find the intersection that means we need to find the common elements in the set
$B = \left\\{ {7,9,11,13} \right\\}$
And
$C = \left\\{ {11,13,15} \right\\}$
If we have a look for the common elements that comes out to be $11,13$
Hence, the collection of the common elements between two sets is known as the intersection
$B \cap C = \left\\{ {7,9,11,13} \right\\} \cap \left\\{ {11,13,15} \right\\} = \left\\{ {11,13} \right\\}$
Hence, the above represent the required result
iii. $A \cap C \cap D$
We have the information
$A = \left\\{ {3,5,7,9,11} \right\\},C = \left\\{ {11,13,15} \right\\},D = \left\\{ {15,17} \right\\}$
Now, to make the above part to be simple let us cut it into the respective terms
$A \cap C \cap D = \left( {A \cap C} \right) \cap D$
Now, let us first simplify the above part which is present in bracket that is
$A \cap C = \left\\{ {3,5,7,9,11} \right\\} \cap \left\\{ {11,13,15} \right\\}$
When we look to find the common element, we get
$A \cap C = \left\\{ {3,5,7,9,11} \right\\} \cap \left\\{ {11,13,15} \right\\} = \left\\{ {11} \right\\}$
Now the intersection with D
$A \cap C \cap D = \left\\{ {11} \right\\} \cap \\{ 15,17\\}$
No common elements hence the required result is
$A \cap C \cap D = \left\\{ {11} \right\\} \cap \\{ 15,17\\} = \phi$
The above is the required solution
iv. $A \cap C$
From the above part

$A \cap C = \left\\{ {3,5,7,9,11} \right\\} \cap \left\\{ {11,13,15} \right\\}$
When we look to find the common element, we get
$A \cap C = \left\\{ {3,5,7,9,11} \right\\} \cap \left\\{ {11,13,15} \right\\} = \left\\{ {11} \right\\}$
So, the above represent the required result.

v. $B \cap D$
In the above we need to find the intersection that means we need to find the common elements in the set
$B = \left\\{ {7,9,11,13} \right\\}$
And
$D = \left\\{ {15,17} \right\\}$
If we have a look for the common elements that comes out to be no one
Hence, the collection of the common elements between two sets is known as the intersection
$A \cap B = \left\\{ {15,17} \right\\} \cap \left\\{ {7,9,11,13} \right\\} = \phi$
Hence, the above represent the required result
vi. $A \cap \left( {B \cup C} \right)$
To solve such type of question let us first simplify the bracket part
$A = \left\\{ {3,5,7,9,11} \right\\},B = \left\\{ {7,9,11,13} \right\\},C = \left\\{ {11,13,15} \right\\}$
Hence
$\left( {B \cup C} \right) = \left\\{ {7,9,11,13} \right\\} \cup \left\\{ {11,13,15} \right\\}$
Union represent the collection of all the element which is present in both set
$\left( {B \cup C} \right) = \left\\{ {7,9,11,13} \right\\} \cup \left\\{ {11,13,15} \right\\} = \left\\{ {7,9,11,13,15} \right\\}$
And now we need to find the intersection with A, we get
$A \cap \left( {B \cup C} \right) = \left\\{ {3,5,7,9,11} \right\\} \cap \left\\{ {7,9,11,13,15} \right\\} = \left\\{ {7,9,11} \right\\}$
The above represent the required result
vii. $A \cap D$
In the above we need to find the intersection that means we need to find the common elements in the set
$A = \left\\{ {3,5,7,9,11} \right\\}$
And
$D = \left\\{ {15,17} \right\\}$
If we have a look for the common elements that comes out to be $7,9,11$
Hence, the collection of the common elements between two sets is known as the intersection
$A \cap D = \left\\{ {3,5,7,9,11} \right\\} \cap \left\\{ {15,17} \right\\} = \phi$
Hence, the above represent the required result
viii. $A \cap \left( {B \cup D} \right)$
Now, let is first find out the bracket part
To solve such type of question let us first simplify the bracket part
$A = \left\\{ {3,5,7,9,11} \right\\},B = \left\\{ {7,9,11,13} \right\\},D = \left\\{ {15,17} \right\\}$
Hence
$\left( {B \cup D} \right) = \left\\{ {7,9,11,13} \right\\} \cup \left\\{ {15,17} \right\\}$
Union represent the collection of all the element which is present in both set
$\left( {B \cup D} \right) = \left\\{ {7,9,11,13} \right\\} \cup \left\\{ {15,17} \right\\} = \left\\{ {7,9,11,13,15,17} \right\\}$
And now we need to find the intersection with A, we get
$A \cap \left( {B \cup D} \right) = \left\\{ {3,5,7,9,11} \right\\} \cap \left\\{ {7,9,11,13,15,17} \right\\} = \left\\{ {7,9,11} \right\\}$
The above represent the required result

ix. $\left( {A \cap B} \right) \cap \left( {B \cap C} \right)$
$A \cap B$
In the above we need to find the intersection that means we need to find the common elements in the set
$A = \left\\{ {3,5,7,9,11} \right\\}$
And
$B = \left\\{ {7,9,11,13} \right\\}$
If we have a look for the common elements that comes out to be $7,9,11$
Hence, the collection of the common elements between two sets is known as the intersection
$A \cap B = \left\\{ {3,5,7,9,11} \right\\} \cap \left\\{ {7,9,11,13} \right\\} = \left\\{ {7,9,11} \right\\}$
Hence, the above represent the required result

$B \cap C$
We have provided that $B = \left\\{ {7,9,11,13} \right\\}$ and $C = \left\\{ {11,13,15} \right\\}$
In the above we need to find the intersection that means we need to find the common elements in the set
$B = \left\\{ {7,9,11,13} \right\\}$
And
$C = \left\\{ {11,13,15} \right\\}$
If we have a look for the common elements that comes out to be $11,13$
Hence, the collection of the common elements between two sets is known as the intersection
$B \cap C = \left\\{ {7,9,11,13} \right\\} \cap \left\\{ {11,13,15} \right\\} = \left\\{ {11,13} \right\\}$
Now, we need to find the common elements between
$A \cap B = \left\\{ {3,5,7,9,11} \right\\} \cap \left\\{ {7,9,11,13} \right\\} = \left\\{ {7,9,11} \right\\}$
$B \cap C = \left\\{ {7,9,11,13} \right\\} \cap \left\\{ {11,13,15} \right\\} = \left\\{ {11,13} \right\\}$
$\left( {A \cap B} \right) \cap \left( {B \cap C} \right) = \left\\{ {7,9,11} \right\\} \cap \left\\{ {11,13} \right\\} = \left\\{ {11} \right\\}$

x. $\left( {A \cup D} \right) \cap \left( {B \cup C} \right)$
Now simplifying the bracket parts we get
$A = \left\\{ {3,5,7,9,11} \right\\},D = \left\\{ {15,17} \right\\};$
We get
$A \cup D = \left\\{ {3,5,7,9,11} \right\\} \cup \left\\{ {15,17} \right\\} = \left\\{ {3,5,7,9,11,15,17} \right\\}$
$B = \left\\{ {7,9,11,13} \right\\},C = \left\\{ {11,13,15} \right\\}$
$B \cup C = \left\\{ {7,9,11,13} \right\\} \cup \left\\{ {11,13,15} \right\\} = \left\\{ {7,9,11,13,15} \right\\}$
Now
$\left( {A \cup D} \right) \cap \left( {B \cup C} \right) = \left\\{ {3,5,7,9,11,15,17} \right\\} \cap \left\\{ {7,9,11,13,15} \right\\} = \left\\{ {7,9,11,15} \right\\}$
The above represent the required result.

Note: Trick to solve the tricky question we need to first simplify the bracket and after the simplification of the bracket we can move further to the required solution it is extremely helpful incase where both union and intersection is present.