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Sagot :
Sure! Let's go through the step-by-step proof for the given properties using the sets [tex]\( A \)[/tex], [tex]\( B \)[/tex], and [tex]\( C \)[/tex]:
1. Proving [tex]\( A \cap (B \cup C) = (A \cap B) \cup (A \cap C) \)[/tex]
Let's consider the following sets:
- [tex]\( A = \{a, b, c\} \)[/tex]
- [tex]\( B = \{b, d, e\} \)[/tex]
- [tex]\( C = \{a, e, f\} \)[/tex]
Step 1: Calculate [tex]\( B \cup C \)[/tex]
[tex]\[ B \cup C = \{b, d, e\} \cup \{a, e, f\} = \{a, b, d, e, f\} \][/tex]
Step 2: Calculate [tex]\( A \cap (B \cup C) \)[/tex]
[tex]\[ A \cap (B \cup C) = \{a, b, c\} \cap \{a, b, d, e, f\} = \{a, b\} \][/tex]
Let this result be [tex]\( \text{LHS}_1 \)[/tex].
Step 3: Calculate [tex]\( A \cap B \)[/tex]
[tex]\[ A \cap B = \{a, b, c\} \cap \{b, d, e\} = \{b\} \][/tex]
Step 4: Calculate [tex]\( A \cap C \)[/tex]
[tex]\[ A \cap C = \{a, b, c\} \cap \{a, e, f\} = \{a\} \][/tex]
Step 5: Calculate [tex]\( (A \cap B) \cup (A \cap C) \)[/tex]
[tex]\[ (A \cap B) \cup (A \cap C) = \{b\} \cup \{a\} = \{a, b\} \][/tex]
Let this result be [tex]\( \text{RHS}_1 \)[/tex].
Since [tex]\( \text{LHS}_1 = \text{RHS}_1 = \{a, b\} \)[/tex], we have proven that:
[tex]\[ A \cap (B \cup C) = (A \cap B) \cup (A \cap C) \][/tex]
2. Proving [tex]\( A \cup (B \cap C) = (A \cup B) \cap (A \cup C) \)[/tex]
Step 1: Calculate [tex]\( B \cap C \)[/tex]
[tex]\[ B \cap C = \{b, d, e\} \cap \{a, e, f\} = \{e\} \][/tex]
Step 2: Calculate [tex]\( A \cup (B \cap C) \)[/tex]
[tex]\[ A \cup (B \cap C) = \{a, b, c\} \cup \{e\} = \{a, b, c, e\} \][/tex]
Let this result be [tex]\( \text{LHS}_2 \)[/tex].
Step 3: Calculate [tex]\( A \cup B \)[/tex]
[tex]\[ A \cup B = \{a, b, c\} \cup \{b, d, e\} = \{a, b, c, d, e\} \][/tex]
Step 4: Calculate [tex]\( A \cup C \)[/tex]
[tex]\[ A \cup C = \{a, b, c\} \cup \{a, e, f\} = \{a, b, c, e, f\} \][/tex]
Step 5: Calculate [tex]\( (A \cup B) \cap (A \cup C) \)[/tex]
[tex]\[ (A \cup B) \cap (A \cup C) = \{a, b, c, d, e\} \cap \{a, b, c, e, f\} = \{a, b, c, e\} \][/tex]
Let this result be [tex]\( \text{RHS}_2 \)[/tex].
Since [tex]\( \text{LHS}_2 = \text{RHS}_2 = \{a, b, c, e\} \)[/tex], we have proven that:
[tex]\[ A \cup (B \cap C) = (A \cup B) \cap (A \cup C) \][/tex]
Thus, the properties have been proven as follows:
1. [tex]\( A \cap (B \cup C) = (A \cap B) \cup (A \cap C) \)[/tex]
2. [tex]\( A \cup (B \cap C) = (A \cup B) \cap (A \cup C) \)[/tex]
1. Proving [tex]\( A \cap (B \cup C) = (A \cap B) \cup (A \cap C) \)[/tex]
Let's consider the following sets:
- [tex]\( A = \{a, b, c\} \)[/tex]
- [tex]\( B = \{b, d, e\} \)[/tex]
- [tex]\( C = \{a, e, f\} \)[/tex]
Step 1: Calculate [tex]\( B \cup C \)[/tex]
[tex]\[ B \cup C = \{b, d, e\} \cup \{a, e, f\} = \{a, b, d, e, f\} \][/tex]
Step 2: Calculate [tex]\( A \cap (B \cup C) \)[/tex]
[tex]\[ A \cap (B \cup C) = \{a, b, c\} \cap \{a, b, d, e, f\} = \{a, b\} \][/tex]
Let this result be [tex]\( \text{LHS}_1 \)[/tex].
Step 3: Calculate [tex]\( A \cap B \)[/tex]
[tex]\[ A \cap B = \{a, b, c\} \cap \{b, d, e\} = \{b\} \][/tex]
Step 4: Calculate [tex]\( A \cap C \)[/tex]
[tex]\[ A \cap C = \{a, b, c\} \cap \{a, e, f\} = \{a\} \][/tex]
Step 5: Calculate [tex]\( (A \cap B) \cup (A \cap C) \)[/tex]
[tex]\[ (A \cap B) \cup (A \cap C) = \{b\} \cup \{a\} = \{a, b\} \][/tex]
Let this result be [tex]\( \text{RHS}_1 \)[/tex].
Since [tex]\( \text{LHS}_1 = \text{RHS}_1 = \{a, b\} \)[/tex], we have proven that:
[tex]\[ A \cap (B \cup C) = (A \cap B) \cup (A \cap C) \][/tex]
2. Proving [tex]\( A \cup (B \cap C) = (A \cup B) \cap (A \cup C) \)[/tex]
Step 1: Calculate [tex]\( B \cap C \)[/tex]
[tex]\[ B \cap C = \{b, d, e\} \cap \{a, e, f\} = \{e\} \][/tex]
Step 2: Calculate [tex]\( A \cup (B \cap C) \)[/tex]
[tex]\[ A \cup (B \cap C) = \{a, b, c\} \cup \{e\} = \{a, b, c, e\} \][/tex]
Let this result be [tex]\( \text{LHS}_2 \)[/tex].
Step 3: Calculate [tex]\( A \cup B \)[/tex]
[tex]\[ A \cup B = \{a, b, c\} \cup \{b, d, e\} = \{a, b, c, d, e\} \][/tex]
Step 4: Calculate [tex]\( A \cup C \)[/tex]
[tex]\[ A \cup C = \{a, b, c\} \cup \{a, e, f\} = \{a, b, c, e, f\} \][/tex]
Step 5: Calculate [tex]\( (A \cup B) \cap (A \cup C) \)[/tex]
[tex]\[ (A \cup B) \cap (A \cup C) = \{a, b, c, d, e\} \cap \{a, b, c, e, f\} = \{a, b, c, e\} \][/tex]
Let this result be [tex]\( \text{RHS}_2 \)[/tex].
Since [tex]\( \text{LHS}_2 = \text{RHS}_2 = \{a, b, c, e\} \)[/tex], we have proven that:
[tex]\[ A \cup (B \cap C) = (A \cup B) \cap (A \cup C) \][/tex]
Thus, the properties have been proven as follows:
1. [tex]\( A \cap (B \cup C) = (A \cap B) \cup (A \cap C) \)[/tex]
2. [tex]\( A \cup (B \cap C) = (A \cup B) \cap (A \cup C) \)[/tex]
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