Question
Question: 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 ...
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×B. Find (i) A and B, where a, b, c are distinct and (ii) A×B.
Solution
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×B. Using these elements, we can easily find A and B and thus A×B as well.
Complete step-by-step answer:
We know that if a set (Say) X has x elements, then n[P(X)]=2x.
Since, we already have n[P(A)]=8.
We can write it as:
⇒n[P(A)]=23
Hence, A has 3 elements.
Similarly, we are given that n[P(B)]=4.
We can write it as:
⇒n[P(B)]=22
Hence, B has 2 elements.
Now, we are given some elements of A×B which are (a,2),(b,3),(c,3).
We know that (x,y)∈X×Y iff x∈X,y∈Y.
Therefore, by looking at (a,2),(b,3),(c,3) as elements of A×B.
We see that a,b,c∈A and 2,3∈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)∈X×Y iff x∈X,y∈Y.
Therefore, we have A=a,b,c and B=2,3.
So, A×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×B=(a,b):a∈A,b∈B
This is the definition of the cross product of two non-empty sets A and B.