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Exercise 1.1 · Q4

Q.By taking suitable sets A, B, CA,\ B,\ C, verify the following results:

(i) A×(B∩C)=(A×B)∩(A×C)A\times(B\cap C)=(A\times B)\cap(A\times C).
(ii) A×(B∪C)=(A×B)∪(A×C)A\times(B\cup C)=(A\times B)\cup(A\times C).
(iii) (A×B)∩(B×A)=(A∩B)×(B∩A)(A\times B)\cap(B\times A)=(A\cap B)\times(B\cap A).
(iv) C−(B−A)=(C∩A)∪(C∩B′)C-(B-A)=(C\cap A)\cup(C\cap B').
(v) (B−A)∩C=(B∩C)−A=B∩(C−A)(B-A)\cap C=(B\cap C)-A=B\cap(C-A).
(vi) (B−A)∪C=(B∪C)−(A−C)(B-A)\cup C=(B\cup C)-(A-C).
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Take A={1,2,3}, B={2,3,4}, C={1,3,5}A=\{1,2,3\},\ B=\{2,3,4\},\ C=\{1,3,5\}, universal set U=A∪B∪C={1,2,3,4,5}U=A\cup B\cup C=\{1,2,3,4,5\}.

Step 1 (i) A×(B∩C)=(A×B)∩(A×C)A\times(B\cap C)=(A\times B)\cap(A\times C). B∩C={3}B\cap C=\{3\}, so LHS ={(1,3),(2,3),(3,3)}=\{(1,3),(2,3),(3,3)\}. Listing A×BA\times B and A×CA\times C and intersecting gives exactly the pairs with second coordinate 33: {(1,3),(2,3),(3,3)}\{(1,3),(2,3),(3,3)\}. Equal. ✓\checkmark

Step 2 (ii) A×(B∪C)=(A×B)∪(A×C)A\times(B\cup C)=(A\times B)\cup(A\times C). B∪C={1,2,3,4,5}B\cup C=\{1,2,3,4,5\}, so LHS has 3×5=153\times5=15 pairs; by the distributive identity this equals the union of A×BA\times B (9 pairs) and A×CA\times C (9 pairs), which overlap on the 3 pairs with second coordinate 33 -- giving 9+9−3=159+9-3=15, matching. ✓\checkmark

Step 3 (iii) (A×B)∩(B×A)=(A∩B)×(B∩A)(A\times B)\cap(B\times A)=(A\cap B)\times(B\cap A). A∩B={2,3}A\cap B=\{2,3\}, so RHS ={2,3}×{2,3}={(2,2),(2,3),(3,2),(3,3)}=\{2,3\}\times\{2,3\}=\{(2,2),(2,3),(3,2),(3,3)\}. Listing A×BA\times B and B×AB\times A and intersecting gives exactly these four pairs. Equal. ✓\checkmark

Step 4 (iv) C−(B−A)=(C∩A)∪(C∩B′)C-(B-A)=(C\cap A)\cup(C\cap B'). B−A={4}B-A=\{4\}, so LHS =C−{4}={1,3,5}=C-\{4\}=\{1,3,5\}. C∩A={1,3}C\cap A=\{1,3\}; B′=U−B={1,5}B'=U-B=\{1,5\}, so C∩B′={1,5}C\cap B'=\{1,5\}; union ={1,3,5}=\{1,3,5\}. Equal. ✓\checkmark …

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