Question

Show that\(A \oplus B = \left( {A - B} \right) \cup \left( {B - A} \right)\).

Short answer

Step by Step solution

Thus, here we proved that \(A \oplus B = \left( {A - B} \right) \cup \left( {B - A} \right)\).

Step-by-step solution

Definition

We know that the definition of symmetric difference,

\(A \oplus B = \left( {A \cup B} \right) - \left( {A \cap B} \right)\).

We need to prove that,

\(\left( {A - B} \right) \cup \left( {B - A} \right) = \left( {A \cup B} \right) - \left( {A \cap B} \right)\).

concept

Let\(x \in \left( {A \cup B} \right) - \left( {A \cap B} \right)\).

\( \Rightarrow x \in \left( {A \cup B} \right)\)and\(x \notin A \cap B\).

\( \Rightarrow x \in A\)or \(x \in B\) but x is not in both sets.

solution

Case:-1

Suppose \(x \in A\) and \(x \notin B\)

\( \Rightarrow x \in A - B\)

Case:-2

Suppose \(x \notin A\)but\(x \in B\).

\( \Rightarrow x \in B - A\)

So,

\( \Rightarrow x \in \left( {A - B} \right) \cup \left( {B - A} \right)\)

\( \Rightarrow \left( {A \cup B} \right) - \left( {A \cap B} \right) \subseteq \left( {A - B} \right) \cup \left( {B - A} \right)\)_____________(i)

Converse part

Let \(x \in \left( {A - B} \right) \cup \left( {B - A} \right)\)

Either ,\(x \in A - B\) or \(x \in B - A\).

Case:-1

\(x \in A - B\)then \(x \in A\)but \(x \notin B\).

So, \(x \in \left( {A \cup B} \right)\) but \(x \notin \left( {A \cap B} \right)\).

\( \Rightarrow x \in \left( {A \cup B} \right) - \left( {A \cap B} \right)\)

Case:-2

\(x \in B - A\)then \(x \in B\)but\(x \notin A\).

So, \(x \in \left( {B \cup A} \right)\)but\(x \notin A\left( {B \cap A} \right)\).

So, \(x \in \left( {A \cup B} \right) - \left( {A \cap B} \right)\)

\( = \left( {A - B} \right) \cup \left( {B - A} \right) \subseteq \left( {A \cup B} \right) - \left( {A \cap B} \right)\)____________(ii)

From (i) and (ii)

\( \Rightarrow \left( {A \cup B} \right) - \left( {A \cap B} \right) = \left( {A - B} \right) \cup \left( {B - A} \right)\)

Hence, \(A \oplus B = \left( {A - B} \right) \cup \left( {B - A} \right)\).