Question

If A = 1, 3, 5….. 17 and B = 2, 4, 6…. 18. The universal set is given as N (The set of natural numbers). Then Show that $A'\cup \left( \left( A\cup B \right)\cap B' \right)=N$.

Answer 1

Hint: Now we are given with sets A and Set B. Hence we will first calculate $A\cup B$ and the set $B'$ since we know that $A\cup B=\{x:x\in A,x\in B\}$ and $B'=\{x:x\in U\And x\notin B\}$ . Now we will take intersection of two sets and find $\left( A\cup B \right)\cap B'$ as we know $X\cap Y=\{x:x\in X\And x\in Y\}$ . Now we will find $A'$ and take union of $\left( A\cup B \right)\cap B'$ and $A'$ as we know $A\cup B=\{x:x\in A,x\in B\}$ Hence we will finally get the value of $A'\cup \left( \left( A\cup B \right)\cap B' \right)$ we will show that this is equal to the set of Natural numbers.

Complete step-by-step solution:
Now we are given that A = 1, 3, 5….. 17 and B = 2, 4, 6…. 18
Now let us first calculate $A\cup B$
Now $A\cup B$ is nothing but the set of all elements of A as well as B.
Now $A\cup B=\{x:x\in A,x\in B\}$
Hence we get $A\cup B=\{1,2,3,4......18\}..................\left( 1 \right)$
Now we know that $B'=\{x:x\in U\And x\notin B\}$
The set B’ is the set of all elements which are in U but not in B.
Now $B = 2, 4, 6…. 18$ and $U = 1, 2, 3, 4, …..$
Hence we get $B'=\{1,3,5,....17,19,20,21,22,23.....\}...................\left( 2 \right)$
Now let us calculate $\left( A\cup B \right)\cap B'$
Now $X\cap Y=\{x:x\in X\And x\in Y\}$
Basically in the intersection, we form a set of elements which are present in both sets
Hence from equation (2) and equation (3), we get that the two sets have no common elements. Hence we get $\left( A\cup B \right)\cap B'=\{1,3,5....17\}..................\left( 3 \right)$ .
Now consider $A'=\{x:x\in U,x\notin A\}$
Hence we get $A'=\{2,4,6,.......18,19,20....\}................\left( 4 \right)$
Now again as we know $A\cup B=\{x:x\in A,x\in B\}$
Hence from equation (3) and equation (4), we get
$A'\cup \left( \left( A\cup B \right)\cap B' \right)=\{1,2,3,4,5,6,7.....\}$
Hence we get $A'\cup \left( \left( A\cup B \right)\cap B' \right)=N$ .

Note: Note that we can also solve this using properties of set theory. Now we know that $A\cup \left( B\cap C \right)=\left( A\cap B \right)\cup \left( A\cap C \right)$ and $A\cap \left( B\cup C \right)=\left( A\cap B \right)\cup \left( A\cap C \right)$ . also we know that $A\cap A'=\varnothing $ and$A\cup A'=U$ where U is universal set hence we can solve this by using properties.