Question

For $A=\{x|-3\le x<4,x\in R\},B=\{x|x<5,xonN\}$ and C= { 5, 3, 1, 0, 1, 3}, show that $A\cap (B\cap C)=(A\cap B)\cup (A\cap C)$.

Answer 1

First note that the set A contains all the real numbers (not just integers) that are greater than or equal to -3 and less than 4.
On the other hand the set B contains all the positive integers that are less than 5. So,
$A=\{x|-3\le x<4,x\in R\},B=\{x|x<5,x\in N\}$; that is, A consists of all real numbers from -3 upto 4 but 4 is not included.
Also, $B=\{x|x<5,x\in N\}$ $=1,2,3,4$. Now, we find
$B\cup C$$=1,2,3,4$$\cup$ $-5,-3,-1,0,1,3$
$=1,2,3,4,-5,-3,-1,0$; thus
$A\cup (B\cup C)$= A $\cap$$1,2,3,4,-5,-3,-1,0$
$=-3,-1,0,1,2,3$ (1)
Next, to find $(A\cap B)\cup (A\cap C)$, we consider
$A\cap B=\{x|-3\le x<4,x\in R\}\cap$$\{1,2,3,4\}=\{1,2,3\}$
$A\cap C=\{x|-3\le x<4,x\in R\}\cap$ $\{-5,-3,-1,0,1,3\}$
$=\{-3,-1,0,1,2,3\}$
Hence, $(A\cap B)\cup (A\cap C)$ $=1,2,3$ $\cup$ $\{-3,-1,0,1,3\}$
$=\{-3,-1,0,1,2,3\}.$
Now (1) and (2) imply $A\cap (B\cup C)=(A\cap B)\cup (A\cap C)$