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Sagot :
Let's address both parts of the question step-by-step.
### Part i) Verification of [tex]\( A \times (B \cap C) = (A \times B) \cap (A \times C) \)[/tex]
1. Find [tex]\( B \cap C \)[/tex]:
[tex]\[ B = \{1, 2, 3, 4\} \][/tex]
[tex]\[ C = \{5, 6\} \][/tex]
[tex]\[ B \cap C = \{ \} \][/tex]
Since there are no common elements between [tex]\( B \)[/tex] and [tex]\( C \)[/tex], [tex]\( B \cap C \)[/tex] is the empty set, [tex]\( \{ \} \)[/tex].
2. Compute [tex]\( A \times (B \cap C) \)[/tex]:
[tex]\[ A \times \{ \} = \{ (a, b) \mid a \in A, b \in \{ \} \} = \{ \} \][/tex]
So, [tex]\( A \times (B \cap C) = \{ \} \)[/tex].
3. Find [tex]\( A \times B \)[/tex]:
[tex]\[ A = \{1, 2\} \][/tex]
[tex]\[ B = \{1, 2, 3, 4\} \][/tex]
[tex]\[ A \times B = \{ (1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4) \} \][/tex]
4. Find [tex]\( A \times C \)[/tex]:
[tex]\[ A = \{1, 2\} \][/tex]
[tex]\[ C = \{5, 6\} \][/tex]
[tex]\[ A \times C = \{ (1, 5), (1, 6), (2, 5), (2, 6) \} \][/tex]
5. Compute [tex]\( (A \times B) \cap (A \times C) \)[/tex]:
[tex]\[ (A \times B) \cap (A \times C) = \{ \} \][/tex]
Since there are no common ordered pairs between [tex]\( A \times B \)[/tex] and [tex]\( A \times C \)[/tex].
6. Verification:
[tex]\[ A \times (B \cap C) = \{ \} \][/tex]
[tex]\[ (A \times B) \cap (A \times C) = \{ \} \][/tex]
Therefore:
[tex]\[ A \times (B \cap C) = (A \times B) \cap (A \times C) \][/tex]
This verifies the first part.
### Part ii) Verification of [tex]\( A \times (B \cup C) = (A \times B) \cup (A \times C) \)[/tex]
1. Find [tex]\( B \cup C \)[/tex]:
[tex]\[ B = \{1, 2, 3, 4\} \][/tex]
[tex]\[ C = \{5, 6\} \][/tex]
[tex]\[ B \cup C = \{ 1, 2, 3, 4, 5, 6 \} \][/tex]
2. Compute [tex]\( A \times (B \cup C) \)[/tex]:
[tex]\[ A = \{1, 2\} \][/tex]
[tex]\[ B \cup C = \{ 1, 2, 3, 4, 5, 6 \} \][/tex]
[tex]\[ A \times (B \cup C) = \{ (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6) \} \][/tex]
3. Compute [tex]\( (A \times B) \cup (A \times C) \)[/tex]:
From the previous steps, we have:
[tex]\[ A \times B = \{ (1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4) \} \][/tex]
[tex]\[ A \times C = \{ (1, 5), (1, 6), (2, 5), (2, 6) \} \][/tex]
Combining these sets:
[tex]\[ (A \times B) \cup (A \times C) = \{ (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6) \} \][/tex]
4. Verification:
[tex]\[ A \times (B \cup C) = \{ (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6) \} \][/tex]
[tex]\[ (A \times B) \cup (A \times C) = \{ (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6) \} \][/tex]
Therefore:
[tex]\[ A \times (B \cup C) = (A \times B) \cup (A \times C) \][/tex]
This verifies the second part.
In conclusion, both identities have been verified as true.
### Part i) Verification of [tex]\( A \times (B \cap C) = (A \times B) \cap (A \times C) \)[/tex]
1. Find [tex]\( B \cap C \)[/tex]:
[tex]\[ B = \{1, 2, 3, 4\} \][/tex]
[tex]\[ C = \{5, 6\} \][/tex]
[tex]\[ B \cap C = \{ \} \][/tex]
Since there are no common elements between [tex]\( B \)[/tex] and [tex]\( C \)[/tex], [tex]\( B \cap C \)[/tex] is the empty set, [tex]\( \{ \} \)[/tex].
2. Compute [tex]\( A \times (B \cap C) \)[/tex]:
[tex]\[ A \times \{ \} = \{ (a, b) \mid a \in A, b \in \{ \} \} = \{ \} \][/tex]
So, [tex]\( A \times (B \cap C) = \{ \} \)[/tex].
3. Find [tex]\( A \times B \)[/tex]:
[tex]\[ A = \{1, 2\} \][/tex]
[tex]\[ B = \{1, 2, 3, 4\} \][/tex]
[tex]\[ A \times B = \{ (1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4) \} \][/tex]
4. Find [tex]\( A \times C \)[/tex]:
[tex]\[ A = \{1, 2\} \][/tex]
[tex]\[ C = \{5, 6\} \][/tex]
[tex]\[ A \times C = \{ (1, 5), (1, 6), (2, 5), (2, 6) \} \][/tex]
5. Compute [tex]\( (A \times B) \cap (A \times C) \)[/tex]:
[tex]\[ (A \times B) \cap (A \times C) = \{ \} \][/tex]
Since there are no common ordered pairs between [tex]\( A \times B \)[/tex] and [tex]\( A \times C \)[/tex].
6. Verification:
[tex]\[ A \times (B \cap C) = \{ \} \][/tex]
[tex]\[ (A \times B) \cap (A \times C) = \{ \} \][/tex]
Therefore:
[tex]\[ A \times (B \cap C) = (A \times B) \cap (A \times C) \][/tex]
This verifies the first part.
### Part ii) Verification of [tex]\( A \times (B \cup C) = (A \times B) \cup (A \times C) \)[/tex]
1. Find [tex]\( B \cup C \)[/tex]:
[tex]\[ B = \{1, 2, 3, 4\} \][/tex]
[tex]\[ C = \{5, 6\} \][/tex]
[tex]\[ B \cup C = \{ 1, 2, 3, 4, 5, 6 \} \][/tex]
2. Compute [tex]\( A \times (B \cup C) \)[/tex]:
[tex]\[ A = \{1, 2\} \][/tex]
[tex]\[ B \cup C = \{ 1, 2, 3, 4, 5, 6 \} \][/tex]
[tex]\[ A \times (B \cup C) = \{ (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6) \} \][/tex]
3. Compute [tex]\( (A \times B) \cup (A \times C) \)[/tex]:
From the previous steps, we have:
[tex]\[ A \times B = \{ (1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4) \} \][/tex]
[tex]\[ A \times C = \{ (1, 5), (1, 6), (2, 5), (2, 6) \} \][/tex]
Combining these sets:
[tex]\[ (A \times B) \cup (A \times C) = \{ (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6) \} \][/tex]
4. Verification:
[tex]\[ A \times (B \cup C) = \{ (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6) \} \][/tex]
[tex]\[ (A \times B) \cup (A \times C) = \{ (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6) \} \][/tex]
Therefore:
[tex]\[ A \times (B \cup C) = (A \times B) \cup (A \times C) \][/tex]
This verifies the second part.
In conclusion, both identities have been verified as true.
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