Answer
See explanation
Work Step by Step
We are asked to prove the following identity using an **element argument** (i.e., reasoning about individual elements):
\[
A \times (B \cup C) = (A \times B) \cup (A \times C)
\]
Assume all sets are subsets of some universal set \(U\). Let's prove this identity by showing both sides are subsets of each other.
---
### 🔹 Part 1: Show \(A \times (B \cup C) \subseteq (A \times B) \cup (A \times C)\)
Let \((a, x) \in A \times (B \cup C)\).
Then:
- \(a \in A\),
- \(x \in B \cup C\), so \(x \in B\) **or** \(x \in C\)
So either:
- \((a, x) \in A \times B\), or
- \((a, x) \in A \times C\)
Thus:
\[
(a, x) \in (A \times B) \cup (A \times C)
\]
✅ Therefore:
\[
A \times (B \cup C) \subseteq (A \times B) \cup (A \times C)
\]
---
### 🔹 Part 2: Show \((A \times B) \cup (A \times C) \subseteq A \times (B \cup C)\)
Let \((a, x) \in (A \times B) \cup (A \times C)\).
Then either:
- \((a, x) \in A \times B\), so \(a \in A\) and \(x \in B\),
⟹ \(x \in B \cup C\)
- or \((a, x) \in A \times C\), so \(a \in A\) and \(x \in C\),
⟹ \(x \in B \cup C\)
In either case:
- \(a \in A\),
- \(x \in B \cup C\)
⟹ \((a, x) \in A \times (B \cup C)\)
✅ Therefore:
\[
(A \times B) \cup (A \times C) \subseteq A \times (B \cup C)
\]
---
### ✅ Final Conclusion:
Since both inclusions hold, we conclude:
\[
\boxed{A \times (B \cup C) = (A \times B) \cup (A \times C)}
\]
This completes the proof.
