Sure! Let's break down each part of the problem step by step.
### Part (a)
We are given the Cartesian product [tex]\( P \times Q = \{(4, 5), (4, 6), (4, 7), (5, 5), (5, 6), (5, 7)\} \)[/tex].
#### Step 1: Find P
We can notice that in the ordered pairs of [tex]\( P \times Q \)[/tex], the first component is either 4 or 5. Therefore, set [tex]\( P \)[/tex] must be:
[tex]\[ P = \{4, 5\} \][/tex]
#### Step 2: Find Q
Similarly, from the ordered pairs, the second component is 5, 6, or 7. Therefore, set [tex]\( Q \)[/tex] must be:
[tex]\[ Q = \{5, 6, 7\} \][/tex]
#### Step 3: Calculate [tex]\( n(P \times Q) \)[/tex]
The number of elements in the Cartesian product [tex]\( P \times Q \)[/tex] can be calculated as follows:
[tex]\[ n(P \times Q) = \text{Number of elements in } P \times \text{Number of elements in } Q \][/tex]
Given that:
[tex]\[ n(P) = 2 \][/tex]
and
[tex]\[ n(Q) = 3 \][/tex]
Therefore:
[tex]\[ n(P \times Q) = 2 \times 3 = 6 \][/tex]
#### Step 4: Verification
We have:
[tex]\[ n(P \times Q) = 6 \][/tex]
[tex]\[ n(P) \times n(Q) = 2 \times 3 = 6 \][/tex]
Both calculations confirm that:
[tex]\[ n(P \times Q) = n(P) \times n(Q) \][/tex]
### Part (b)
We are given the Cartesian product [tex]\( R \times S = \{(1, 2), (1, 3), (1, 4), (2, 2), (2, 3), (2, 4)\} \)[/tex].
#### Step 1: Find R
From the ordered pairs of [tex]\( R \times S \)[/tex], the first component is either 1 or 2. Therefore, set [tex]\( R \)[/tex] must be:
[tex]\[ R = \{1, 2\} \][/tex]
#### Step 2: Find S
Similarly, from the ordered pairs, the second component is 2, 3, or 4. Therefore, set [tex]\( S \)[/tex] must be:
[tex]\[ S = \{2, 3, 4\} \][/tex]
#### Step 3: Calculate [tex]\( n(R \times S) \)[/tex]
The number of elements in the Cartesian product [tex]\( R \times S \)[/tex] can be calculated as follows:
[tex]\[ n(R \times S) = \text{Number of elements in } R \times \text{Number of elements in } S \][/tex]
Given that:
[tex]\[ n(R) = 2 \][/tex]
and
[tex]\[ n(S) = 3 \][/tex]
Therefore:
[tex]\[ n(R \times S) = 2 \times 3 = 6 \][/tex]
#### Step 4: Verification
We have:
[tex]\[ n(R \times S) = 6 \][/tex]
[tex]\[ n(R) \times n(S) = 2 \times 3 = 6 \][/tex]
Both calculations confirm that:
[tex]\[ n(R \times S) = n(R) \times n(S) \][/tex]
So, we have found the sets [tex]\( P, Q, R, \)[/tex] and [tex]\( S \)[/tex], and we have verified that the number of elements in the Cartesian products [tex]\( P \times Q \)[/tex] and [tex]\( R \times S \)[/tex] is equal to the product of the number of elements in the respective sets.
