Question

Let U = {x|x $\in$ N, x < 10}
A = {a|a is even, a $\in$ U}
B = {b|b is a factor of 6, b $\in$ U}
Verify that n(A)+ n(B) = n(A$\cup$B) + n(A$\cap$B)

Answer 1

U = {1,2,3,4,5,6,7,8,9}
A = { 2,4,6,8}
B = {1,2,3,6}
n(A) = 4
n(B)= 4

$\mathrm{A}\cup \mathrm{B}$ = {1,2,3,4,6,8}
$\therefore$n ($\mathrm{A}\cup \mathrm{B}$) = 6

$\mathrm{A}\cap \mathrm{B}=\{2,6\}$
$\therefore$n ($\mathrm{A}\cap \mathrm{B}$) = 2

Now, we have:
n(A) + n(B) = 4 + 4 = 8
n($\mathrm{A}\cup \mathrm{B}$) + n($\mathrm{A}\cap \mathrm{B}$) = 6 + 2 = 8

$\therefore$n (A) +n (B ) = n ($\mathrm{A}\cup \mathrm{B}$)+n ($\mathrm{A}\cap \mathrm{B}$)

Hence, proved.