Question

A= {1,2},B={1,2,3,4} ,C={5,6} verify A×(B intersection C) = (A×B)intersection (A×C)​

Answer 1

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

Given that,

$\rm :\longmapsto\:A = \{1,2\}$

$\rm :\longmapsto\:B = \{1,2,3,4\}$

$\rm :\longmapsto\:C = \{5,6\}$

Consider,

$\rm :\longmapsto\:B \: \cap \: C$

$\rm \: = \:\{1,2,3,4\} \: \cap \: \{5,6\}$

$\rm \: = \: \phi$

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

Now, Consider

$\rm :\longmapsto\:A \times (B \: \cap \: C)$

$\rm \: = \:\{1,2\} \times \phi$

$\rm \: = \: \phi$

So,

$\rm :\longmapsto\: \boxed{ \bf{ \: A \times (B \: \cap \: C) \: = \: \phi }}- - - (1)$

Now, Consider,

$\rm :\longmapsto\:A \times B$

$\rm \: = \:\{1,2\} \times \{1,2,3,4\}$

$\rm \: = \:\{(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(2,4)\}$

Now, Consider,

$\rm :\longmapsto\:A \times C$

$\rm \: = \:\{1,2\} \times \{5,6\}$

$\rm \: = \:\{(1,5),(1,6),(2,5),(2,6)\}$

So,

$\rm :\longmapsto\: \boxed{ \bf{ \: (A \times B) \: \cap \: (A \times C) \: = \: \phi \: }} - - - (2)$

So, From equation (1) and (2), we concluded that

$\bf :\longmapsto\:A \times (B \: \cap \: C) = (A \times B) \: \cap \: (A \times C)$

Hence, Proved

Additional Information :-

$\boxed{ \bf{ \: A \times B \: \ne \: B \times A}}$

$\boxed{ \bf{ \: n(A \times B) = n(A) \times n(B)}}$

$\boxed{ \bf{ \: A \: \times \: \phi \: \: = \: \: \phi}}$

Answer 2

$\scriptsize \colorbox{lightyellow} {\text{ \bf♕ Brainliest answer }}$

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

Given that,

$\rm :\longmapsto\:A = \{1,2\}$

$\rm :\longmapsto\:B = \{1,2,3,4\}$

$\rm :\longmapsto\:C = \{5,6\}$

Consider,

$\rm :\longmapsto\:B \: \cap \: C$

$\rm \: = \:\{1,2,3,4\} \: \cap \: \{5,6\}$

$\rm \: = \: \phi$

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

Now, Consider

$\rm :\longmapsto\:A \times (B \: \cap \: C)$

$\rm \: = \:\{1,2\} \times \phi={1,2}$

$\rm \: = \: \phi$

So,

$\rm :\longmapsto\: \boxed{ \bf{ \: A \times (B \: \cap \: C) \: = \: \phi}} - - - (1)$

Now, Consider,

$\rm :\longmapsto\:A \times B$

$\rm \: = \:\{1,2\} \times \{1,2,3,4\}={1,2}×{1,2,3,4}$

$\rm \: = \:\{(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(2,4)\}={(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(2,4)}$

Now, Consider,

$\rm :\longmapsto\:A \times C:⟼A×C$

$\rm \: = \:\{1,2\} \times \{5,6\}={1,2}×{5,6}$

$\rm \: = \:\{(1,5),(1,6),(2,5),(2,6)\}={(1,5),(1,6),(2,5),(2,6)}$

So,

$\rm :\longmapsto\: \boxed{ \bf{ \: (A \times B) \: \cap \: (A \times C) \: = \: \phi \: }}$

So, From equation (1) and (2), we concluded that

$\bf :\longmapsto\:A \times (B \: \cap \: C) = (A \times B) \: \cap \: (A \times C)$

Hence, Proved

Additional Information :-

$\boxed{ \bf{ \: A \times B \: \ne \: B \times A}}$

$\boxed{ \bf{ \: n(A \times B) = n(A) \times n(B)}}$

$\boxed{ \bf{ \: A \: \times \: \phi \: \: = \: \: \phi}}$