Let A and B be sets, if \[A \cap X = B \cap X = \phi \] and \[A \cup X = B \cup X\] for some set X, show that \[A = B\].

Hint:
As using the above given information in the question we can say that first calculate the value of set A and similarly calculate the value of set B and from there observe whether they are having the same answers or not if yes then the above stated can be shown and if know the above given will be wrong.

Complete step by step solution:
As given that \[A \cap X = B \cap X = \phi \] and \[A \cup X = B \cup X\] for some set X.
We can write A as,
\[A = A \cap (A \cup X)\]
On using \[(A \cup X = B \cup X)\] we can say that,
\[ \Rightarrow A = A \cap (B \cup X)\;\;{\text{ }}\;{\text{ }}\;\;\]
Now, apply distributive law for above equation,
\[ \Rightarrow A\]\[ = (A \cap B) \cup (A \cap X)\;\]
As we know that \[A \cap X = B \cap X = \phi \]
\[ \Rightarrow A\]\[ = (A \cap B) \cup \phi \;\;{\text{ }}\;{\text{ }}\;\]
As \[A \cup \phi = A\], we get,
\[ \Rightarrow A\]\[ = A \cap B\;\]…..(1)
We can write B as,
\[ \Rightarrow \]\[B = B \cap (B \cup X)\]
On using \[(A \cup X = B \cup X)\] we can say that,
\[ \Rightarrow B\]\[ = B \cap (A \cup X)\;\;{\text{ }}\;{\text{ }}\;{\text{ }}\]
Now, apply distributive law for above equation,
\[ \Rightarrow B\]\[ = (B \cap A) \cup (B \cap X)\]
As we know that \[A \cap X = B \cap X = \phi \]
\[ \Rightarrow B\]\[ = (B \cap A) \cup \phi \]
As \[A \cup \phi = A\] , we get,
\[ \Rightarrow B\]\[ = B \cap A\;\]…..(2)
As we know that \[B \cap A = A \cap B\;\;{\text{ }}\]
So from 1 and 2 we have
\[A = B\]

Hence, if \[A \cap X = B \cap X = \phi \] and \[A \cup X = B \cup X\] for some set X, then \[A = B\].

Note:
The union of two sets is a new set that contains all of the elements that are in at least one of the two sets.
The intersection of two sets is a new set that contains all of the elements that are in both sets.
To “distribute” means to divide something or give a share or part of something. According to the distributive property, multiplying the sum of two or more addends by a number will give the same result as multiplying each addend individually by the number and then adding the products together. Calculate properly using proper operations and laws and compare the equation exactly as given.