Question

If $A=\{1,2,3\},B=\{3,4\}$ and $C=\{1,3,5\}$, then $A\times (B-C)=$

• A. $(A\times B)-(A\times C)$
• B. $(A\times B)+(A\times C)$
• C. $(A\times B)-(B\times C)$
• D. $(A\times B)-(C\times A)$

Answer 1

The correct option is A $(A\times B)-(A\times C)$

$(B-C)$ is set of all elements present in $B$ and not in $C$

$\therefore (B-C)=\left\{4\right\}$

$A\times (B-C)=\{1,2,3\}\times \left\{4\right\}=\left\{\right(1,4),(2,4),(3,4\left)\right\}$

Lets find $(A\times B)$ and $(A\times C)$

$(A\times B)=\{1,2,3\}\times \{3,4\}$

$=\left\{\right(1,3),(2,3),(3,3),(1,4),(2,4),(3,4\left)\right\}$

$(A\times C)=\{1,2,3\}\times \{1,3,5\}$

$=\left\{\right(1,1),(1,3),(1,5),(2,1),(2,3),(2,5),(3,1),(3,3),(3,5\left)\right\}$

$(A\times B)-(A\times C)=\left\{\right(1,3),(2,3),(3,3),(1,4),(2,4),(3,4\left)\right\}-\left\{\right(1,1),(1,3),(1,5),(2,1),(2,3),$

$(2,5),(3,1),(3,3),(3,5)\}$

$(A\times B)-(A\times C)=\left\{\right(1,4),(2,4),(3,4\left)\right\}$

$\therefore A\times (B-C)=(A\times B)-(A\times C)$