Question:medium

Show that for any sets A and B, \(A = (A ∩ B) ∪ (A – B)\) and \(A ∪ (B – A) = (A ∪ B)\)

Updated On: Jan 21, 2026
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Solution and Explanation

We are required to show the following identities for any two sets \( A \) and \( B \):

\[ \text{(i) } A = (A \cap B) \cup (A - B) \]

\[ \text{(ii) } A \cup (B - A) = A \cup B. \]


Proof of (i):

Let \( x \in A \).

Then either \( x \in B \) or \( x \notin B \).

Case 1: If \( x \in B \), then

\[ x \in A \cap B. \]

Case 2: If \( x \notin B \), then

\[ x \in A - B. \]

Hence, in both cases,

\[ x \in (A \cap B) \cup (A - B). \]

Therefore,

\[ A \subseteq (A \cap B) \cup (A - B). \]

Conversely, let \[ x \in (A \cap B) \cup (A - B). \]

Then either \[ x \in A \cap B \quad \text{or} \quad x \in A - B. \]

In both cases, we have \[ x \in A. \]

Therefore,

\[ (A \cap B) \cup (A - B) \subseteq A. \]

Hence,

\[ A = (A \cap B) \cup (A - B). \]


Proof of (ii):

Consider the left-hand side:

\[ A \cup (B - A). \]

Let \( x \in A \cup (B - A) \).

Then either \[ x \in A \quad \text{or} \quad x \in B - A. \]

If \( x \in A \), then \[ x \in A \cup B. \]

If \( x \in B - A \), then

\[ x \in B \quad \text{and} \quad x \notin A, \]

which implies

\[ x \in A \cup B. \]

Hence,

\[ A \cup (B - A) \subseteq A \cup B. \]

Conversely, let \[ x \in A \cup B. \]

Then either \[ x \in A \quad \text{or} \quad x \in B. \]

If \( x \in A \), then \[ x \in A \cup (B - A). \]

If \( x \in B \) and \( x \notin A \), then

\[ x \in B - A, \]

which implies

\[ x \in A \cup (B - A). \]

Therefore,

\[ A \cup B \subseteq A \cup (B - A). \]

Hence,

\[ A \cup (B - A) = A \cup B. \]

Thus, both the identities are proved.

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