
Let A and B be two sets such that \[n[P(A)] = 8\] and \[n[P(B)] = 4\]. If $(a,2),(b,3),(c,3)$ are in $A \times B$. Find (i) A and B, where a, b, c are distinct and (ii) $A \times B$.
Answer
568.2k+ views
Hint: We will first use the formula of number of elements in the power set and using that, we will calculate the number of elements in A and B. Now, we already have some elements of $A \times B$. Using these elements, we can easily find A and B and thus $A \times B$ as well.
Complete step-by-step answer:
We know that if a set (Say) X has $x$ elements, then \[n[P(X)] = {2^x}\].
Since, we already have \[n[P(A)] = 8\].
We can write it as:
\[ \Rightarrow n[P(A)] = {2^3}\]
Hence, A has 3 elements.
Similarly, we are given that \[n[P(B)] = 4\].
We can write it as:
\[ \Rightarrow n[P(B)] = {2^2}\]
Hence, B has 2 elements.
Now, we are given some elements of $A \times B$ which are $(a,2),(b,3),(c,3)$.
We know that $(x,y) \in X \times Y$ iff $x \in X,y \in Y$.
Therefore, by looking at $(a,2),(b,3),(c,3)$ as elements of $A \times B$.
We see that $a,b,c \in A$ and $2,3 \in B$.
Since A has 3 elements and a, b and c are given to be distinct. Therefore, $A = \{ a,b,c\} $.
B also has 2 elements. Therefore, $B = \{ 2,3\} $
Now, we know that $(x,y) \in X \times Y$ iff $x \in X,y \in Y$.
Therefore, we have $A = \{ a,b,c\} $ and $B = \{ 2,3\} $.
So, $A \times B = \{ (a,2),(a,3),(b,2),(b,3),(c,2),(c,3)\} $.
Note: The students might wonder how did we use the condition of a, b and c being distinct. But you must remember that if we have elements written in a set form, one element cannot be repeated again and again. It appears only once in the seat. So, if a, b and c or any two of them, if, would have been the same, we would not have been able to get the 3 elements as our set A has.
The students must remember the formula of number of elements in the power set because that plays a major role here. If we would not have known about the number of elements of a set, it would be impossible for us to say what the sets will be.
Must remember: $A \times B = \{ (a,b):a \in A,b \in B\} $
This is the definition of the cross product of two non-empty sets A and B.
Complete step-by-step answer:
We know that if a set (Say) X has $x$ elements, then \[n[P(X)] = {2^x}\].
Since, we already have \[n[P(A)] = 8\].
We can write it as:
\[ \Rightarrow n[P(A)] = {2^3}\]
Hence, A has 3 elements.
Similarly, we are given that \[n[P(B)] = 4\].
We can write it as:
\[ \Rightarrow n[P(B)] = {2^2}\]
Hence, B has 2 elements.
Now, we are given some elements of $A \times B$ which are $(a,2),(b,3),(c,3)$.
We know that $(x,y) \in X \times Y$ iff $x \in X,y \in Y$.
Therefore, by looking at $(a,2),(b,3),(c,3)$ as elements of $A \times B$.
We see that $a,b,c \in A$ and $2,3 \in B$.
Since A has 3 elements and a, b and c are given to be distinct. Therefore, $A = \{ a,b,c\} $.
B also has 2 elements. Therefore, $B = \{ 2,3\} $
Now, we know that $(x,y) \in X \times Y$ iff $x \in X,y \in Y$.
Therefore, we have $A = \{ a,b,c\} $ and $B = \{ 2,3\} $.
So, $A \times B = \{ (a,2),(a,3),(b,2),(b,3),(c,2),(c,3)\} $.
Note: The students might wonder how did we use the condition of a, b and c being distinct. But you must remember that if we have elements written in a set form, one element cannot be repeated again and again. It appears only once in the seat. So, if a, b and c or any two of them, if, would have been the same, we would not have been able to get the 3 elements as our set A has.
The students must remember the formula of number of elements in the power set because that plays a major role here. If we would not have known about the number of elements of a set, it would be impossible for us to say what the sets will be.
Must remember: $A \times B = \{ (a,b):a \in A,b \in B\} $
This is the definition of the cross product of two non-empty sets A and B.
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