Answer :
Certainly! Let's analyze the problem step by step.
### Step (a): Determine if sets [tex]\( P \)[/tex] and [tex]\( Q \)[/tex] are disjoint
- Set [tex]\( P \)[/tex]: [tex]\( \{2, 4, 6, 8, 10\} \)[/tex]
- These are the even numbers between 2 and 10.
- Set [tex]\( Q \)[/tex]: [tex]\( \{2, 3, 5\} \)[/tex]
- These are the prime numbers up to 6.
For two sets to be disjoint, they must have no elements in common. Let's find the intersection of sets [tex]\( P \)[/tex] and [tex]\( Q \)[/tex]:
- The common elements between [tex]\( P \)[/tex] and [tex]\( Q \)[/tex] are just {2} which means the intersection is {2}
Since {2} is not an empty set, sets [tex]\( P \)[/tex] and [tex]\( Q \)[/tex] are not disjoint. Therefore, there is at least one element common to both sets. The answer is: No, [tex]\( P \)[/tex] and [tex]\( Q \)[/tex] are not disjoint sets.
### Step (b): Proper subsets of set [tex]\( Q \)[/tex]
The set Q is [tex]\( \{2, 3, 5\} \)[/tex].
A proper subset of a set is any subset of that set which is not equal to the set itself. Proper subsets do not include the set itself.
Let's list all the proper subsets:
1. [tex]\( \emptyset \)[/tex] (The empty set)
2. [tex]\( \{2\} \)[/tex]
3. [tex]\( \{3\} \)[/tex]
4. [tex]\( \{5\} \)[/tex]
5. [tex]\( \{2, 3\} \)[/tex]
6. [tex]\( \{2, 5\} \)[/tex]
7. [tex]\( \{3, 5\} \)[/tex]
Therefore, all the proper subsets of set [tex]\( Q \)[/tex] are:
[tex]\[ \emptyset, \{2\}, \{3\}, \{5\}, \{2, 3\}, \{2, 5\}, \{3, 5\} \][/tex]
These are all the proper subsets of the set [tex]\( Q \)[/tex].
In summary:
- (a) [tex]\( P \)[/tex] and [tex]\( Q \)[/tex] are not disjoint sets because they share a common element, 2.
- (b) The proper subsets of [tex]\( Q \)[/tex] are [tex]\( \emptyset \)[/tex], [tex]\( \{2\} \)[/tex], [tex]\( \{3\} \)[/tex], [tex]\( \{5\} \)[/tex], [tex]\( \{2, 3\} \)[/tex], [tex]\( \{2, 5\} \)[/tex], and [tex]\( \{3, 5\} \)[/tex].
### Step (a): Determine if sets [tex]\( P \)[/tex] and [tex]\( Q \)[/tex] are disjoint
- Set [tex]\( P \)[/tex]: [tex]\( \{2, 4, 6, 8, 10\} \)[/tex]
- These are the even numbers between 2 and 10.
- Set [tex]\( Q \)[/tex]: [tex]\( \{2, 3, 5\} \)[/tex]
- These are the prime numbers up to 6.
For two sets to be disjoint, they must have no elements in common. Let's find the intersection of sets [tex]\( P \)[/tex] and [tex]\( Q \)[/tex]:
- The common elements between [tex]\( P \)[/tex] and [tex]\( Q \)[/tex] are just {2} which means the intersection is {2}
Since {2} is not an empty set, sets [tex]\( P \)[/tex] and [tex]\( Q \)[/tex] are not disjoint. Therefore, there is at least one element common to both sets. The answer is: No, [tex]\( P \)[/tex] and [tex]\( Q \)[/tex] are not disjoint sets.
### Step (b): Proper subsets of set [tex]\( Q \)[/tex]
The set Q is [tex]\( \{2, 3, 5\} \)[/tex].
A proper subset of a set is any subset of that set which is not equal to the set itself. Proper subsets do not include the set itself.
Let's list all the proper subsets:
1. [tex]\( \emptyset \)[/tex] (The empty set)
2. [tex]\( \{2\} \)[/tex]
3. [tex]\( \{3\} \)[/tex]
4. [tex]\( \{5\} \)[/tex]
5. [tex]\( \{2, 3\} \)[/tex]
6. [tex]\( \{2, 5\} \)[/tex]
7. [tex]\( \{3, 5\} \)[/tex]
Therefore, all the proper subsets of set [tex]\( Q \)[/tex] are:
[tex]\[ \emptyset, \{2\}, \{3\}, \{5\}, \{2, 3\}, \{2, 5\}, \{3, 5\} \][/tex]
These are all the proper subsets of the set [tex]\( Q \)[/tex].
In summary:
- (a) [tex]\( P \)[/tex] and [tex]\( Q \)[/tex] are not disjoint sets because they share a common element, 2.
- (b) The proper subsets of [tex]\( Q \)[/tex] are [tex]\( \emptyset \)[/tex], [tex]\( \{2\} \)[/tex], [tex]\( \{3\} \)[/tex], [tex]\( \{5\} \)[/tex], [tex]\( \{2, 3\} \)[/tex], [tex]\( \{2, 5\} \)[/tex], and [tex]\( \{3, 5\} \)[/tex].
