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Question

Question: If we are given the sets as A = {1, 2, 3, 4, 5}, B = {6, 8, 10, 12 }, C = {2, 3, 5, 7, 11} and D = {...

If we are given the sets as A = {1, 2, 3, 4, 5}, B = {6, 8, 10, 12 }, C = {2, 3, 5, 7, 11} and D = {1, 4, 9, 16}. Find (A \cap C) \cap (B \cup D).

Explanation

Solution

We will apply here the definition of intersection which means to select those terms from the sets which are in common. This fact of commonness is used between the sets which are considered under intersection.

Complete step-by-step solution:
Here we will consider the expression (A \cap C) \cap (B \cup D). After looking at the expression we clearly comes to know that the sets which are taken under the intersection operation are A and C where A = {1, 2, 3, 4, 5}, B = {6, 8, 10, 12 }, C = {2, 3, 5, 7, 11} and D = {1, 4, 9, 16}. Now we will start one by one.
We will first take the expression (A \cap C). Now we will consider the sets A = {1, 2, 3, 4, 5} and C = {2, 3, 5, 7, 11}. Since, we are taking an intersection operation here. So, by the definition of intersection which means that the collection of those elements between the sets which are in common. Therefore, the common elements between A and C are 2, 3 and 5. We can write it as,
(A \cap C) = {1, 2, 3, 4, 5} \cap {2, 3, 5, 7, 11}
Therefore, we have (A \cap B) = {2, 3, 5}.
Now we will find the Union of the sets for B and D for the expression (B \cup D). For that we have sets B = {6, 8, 10, 12} and D = {1, 4, 9, 16}. And we take all the elements present in both the sets which are 1,4,6,8,9,10,12 and 16. And we can write it as,
(B \cup D) = {6, 8, 10, 12} \cup {1, 4, 9, 16 }
Therefore, we have (B \cup D) = {1, 4, 6, 8, 9, 10, 12, 16}.
Thus we have that (A \cap C) \cap (B \cup D) = {2, 3, 5} \cap {1, 4, 6, 8, 9, 10, 12, 16}. This results in an empty set as there are no common elements between these sets.
Hence, (A \cap C) \cap (B \cup D) = ϕ\phi .

Note: While applying intersection between the sets if no element is coming as a common element then it does not mean that the answer is not correct. This is because there can be an empty set as we have in this case. An empty set can be represented as a slashed o as we have written in this answer. Also, we can write an empty set as \left\\{ {} \right\\}.