Question

6. For three sets A, B and C, show that (i) AnB= AnC need not imply B=C. (ii) ACB =>C-B C C-A​

Answer 1

$\large\underline{\sf{Solution-i}}$

Let assume that,

↝ A = { 1, 2 }

↝ B = { 1, 3 }

↝ C = { 1, 4 }

Now,

$\red{\rm :\longmapsto\:A \: \cap \: B \: = \: \{1 \}}$

$\red{\rm :\longmapsto\:A \: \cap \: C \: = \: \{1 \}}$

$\bf\implies \:A \: \cap \: B \: = \: A \: \cap \: C$

But

$\rm :\longmapsto\:B \: \ne \: C$

Hence, Justified

$\green{\large\underline{\sf{Solution-ii}}}$

Given that,

$\rm :\longmapsto\:A \: \sub \: B$

Consider,

Let assume that

$\rm :\longmapsto\:x \: \in \: C - B$

$\rm :\implies\:\:x \: \in \: C \: and \: x \: \cancel\in \: B$

$\rm :\implies\:\:x \: \in \: C \: and \: x \: \cancel\in \: A \: \: \: \: \{ \because \: A \: \sub \: B \}$

$\rm :\implies\:\:x \: \in \: C - A$

$\rm :\implies\:\:C - B \: \sub \: C - A$

Hence, Proved

Additional Information -

1. Commutative Law :-

$\red{ \boxed{ \sf{ \:A \: \cup \: B \: = \: B \: \cup \: A}}}$

$\red{ \boxed{ \sf{ \:A \: \cap \: B \: = \: B \: \cap \: A}}}$

2. Associative Law :-

$\red{ \boxed{ \sf{ \:(A \: \cap \: B ) \: \cap\: C = A \: \cap \:( B \: \cap \: C)}}}$

$\red{ \boxed{ \sf{ \:(A \: \cup \: B ) \: \cup\: C = A \: \cup \:( B \: \cup \: C)}}}$

3. Distributive Law

$\red{ \boxed{ \sf{ \:(A \: \cup \: B) \: \cap \: C \: = \: (A \: \cap \: C) \: \cup \: (B \: \cap \: C) \: \: }}}$

$\red{ \boxed{ \sf{ \:(A \: \cap \: B) \: \cup \: C \: = \: (A \: \cup \: C) \: \cap \: (B \: \cup \: C) \: \: }}}$