Question

For any two sets a and b prove that a minus b intersection b minus equals to 5

Answer 1

We have to prove that,

$\textsf{For any two sets A and B , prove that,}\\ \begin{center}\Large (A-B)\cap(B-A)=\O\end{center}$

⇒  $(A-B)$  contains the elements that are in  $A$  but not in  $B.$  

⇒  $(B-A)$  contains the elements that are in  $B$  but not in  $A.$  

Since both  $(A-B)$  and  $(B-A)$  contain no common elements, the intersection of both will be an empty set.

A proof by contradiction is given below.

Suppose there exists an element  $x \in [(A-B)\cap (B-A)]$

This means that the element  $x$  is included in both  $(A-B)$  and  $(B-A).$  

Hence,

$\begin{aligned}&x\in (A-B)\ \wedge\ x\in (B-A)\\ \\ \Longrightarrow\ \ &(x\in A\ \wedge\ x\notin B)\ \wedge\ (x\in B\ \wedge\ x\notin A)\\ \\ \Longrightarrow\ \ &x\in A\ \wedge\ x\notin B\ \wedge\ x\in B\ \wedge\ x\notin A\\ \\ \Longrightarrow\ \ &x\in A\ \wedge\ x\notin A\ \wedge\ x\in B\ \wedge\ x\notin B\end{aligned}$

Here a contradiction occurs that  $x$  is included in both  $A$  and  $B$  but not included in both  $A$  and  $B$ !!!

Hence Proved!

An example is given below.

Let  $A=\{1,\ 2,\ 3,\ 4\}\ \wedge\ B=\{3,\ 4,\ 5,\ 6\}.$

$(A-B) = \{1,\ 2,\ 3,\ 4\}-\{3,\ 4,\ 5,\ 6\}=\{1,\ 2\}\\ \\ (B-A)=\{3,\ 4,\ 5,\ 6\}-\{1,\ 2,\ 3,\ 4\}=\{5,\ 6\}\\ \\ \\ \\ (A-B)\cap(B-A)=\{1,\ 2\}\cap \{5,\ 6\}=\O$

Answer 2

Answer:

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