Question

If A = {a,b,c,d} and B = {b,d,e,f}, then
prove that
(A U B) – (A ∩ B) = (A ∆ B)​

Answer 1

$\large\underline{\sf{Given- }}$

$\red{\rm :\longmapsto\:A = \: \{a,b,c,d \}}$

and

$\red{\rm :\longmapsto\:B \: = \: \{b,d,e,f \}}$

$\large\underline{\sf{To\:prove - }}$

$\red{\rm :\longmapsto\:(A\cup B) - (A\cap B) = A \: \triangle \: B}$

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

Given that,

${\rm :\longmapsto\:A = \: \{a,b,c,d \}}$

and

${\rm :\longmapsto\:B \: = \: \{b,d,e,f \}}$

So, Consider

$\rm :\longmapsto\:A\cup B = \{a,b,c,d,e,f \}$

and

Now, Consider

$\rm :\longmapsto\:A\cap B = \{b,d \}$

So,

$\rm :\longmapsto\:(A\cup B) - (A\cap B) = \{a,c,e,f \}$

Now, we know that

$\rm :\longmapsto\:A \: \triangle \: B = (A - B)\cup (B - A)$

So,

Consider

$\rm :\longmapsto\:A - B = \{a,c \}$

Consider,

$\rm :\longmapsto\:B - A = \{e,f \}$

Thus,

$\rm :\longmapsto\:(A - B)\cup (B - A)$

$\rm \: = \: \: \{a,c,e,f \}$

Therefore,

$\rm :\longmapsto\:A \: \triangle \: B = \: \: \{a,c,e,f \}$

Hence, We concluded that

$\red{\rm :\longmapsto\:(A\cup B) - (A\cap B) = A \: \triangle \: B}$

Additional Information :-

$\boxed{ \rm{ {A}^{'} \cup {B}^{'} = {(A\cap B)}^{'}}}$

$\boxed{ \rm{ {A}^{'} \cap {B}^{'} = {(A\cup B)}^{'}}}$

$\boxed{ \rm{ A - B = A\cap {B}^{'}}}$

$\boxed{ \rm{ A\cap {A}^{'} = \phi \: }}$

$\boxed{ \rm{ {\phi }^{'} = U}}$

$\boxed{ \rm{ {U}^{'} = \phi }}$

$\boxed{ \rm{ A\cup {A}^{'} = U}}$

$\boxed{ \rm{ A\cap {A}^{'} =\phi }}$