Question

Show that \[A \cap B = A \cap C\] need not imply \[B = C\] .

Answer 1

Hint:
Let the elements of set A, B and C in such a way that it satisfies the above condition and from there we can observe whether \[B = C\] is always true or not. And hence from there we can observe our answer.

Complete step by step solution:
The given condition is \[A \cap B = A \cap C\]
For example,
Let set \[A = \{ 3,6,9\} \] and set \[B = \{ 2,4,6\} \] and randomly set \[C = \{ 2,4,6,8\} \]
Now, calculating \[A \cap B\] ,we get,
\[ \Rightarrow A \cap B = \{ 6\} \]
Now, calculating \[A \cap C\] , we get,
\[ \Rightarrow A \cap C = \{ 6\} \]
Hence, \[A \cap B = A \cap C\]
But we can see that, \[B \ne C\]

Hence, \[A \cap B = A \cap C\] need not imply \[B = C\] is shown.

Note:
In mathematics, the intersection of two sets A and B, denoted by \[A \cap B\] , is the set containing all elements of A that also belong to B.
The above question can also proceed with Venn diagram’s method and hence the above given condition is not always true can be stated.
If there are no elements in at least one of the sets we are trying to find the intersection of, then the two sets have no elements in common. In other words, the intersection of any set with the empty set will give us the empty set. Take the examples properly in such a way that correct analysis can be provided.