Question
Let $$N$$ denote the set of natural numbers and $$A = \left\{ {{n^2}:n\, \in \,N} \right\}$$ and $$B = \left\{ {{n^3}:n\, \in \,N} \right\}.$$ Which one of the following incorrect ?
A.
$$A \cup B = N$$
B.
The complement of $$\left( {A \cup B} \right)$$ is an infinite set
C.
$$\left( {A \cap B} \right)$$ must be a finite set
D.
$$\left( {A \cap B} \right)$$ must be a proper subset of $$\left\{ {{m^6}:m\, \in \,N} \right\}$$
Answer :
$$A \cup B = N$$
Solution :
$$\eqalign{
& {\text{Let }}A = \left\{ {{n^2}:n\, \in \,N} \right\}{\text{ and }}B = \left\{ {{n^3}:n\, \in \,N} \right\} \cr
& A = \left\{ {1,\,4,\,9,\,16,.....} \right\}{\text{ and }}B = \left\{ {1,\,8,\,27,\,64,.....} \right\} \cr
& {\text{Now, }}A \cap B = \left\{ 1 \right\}{\text{ which is a finite set}} \cr
& {\text{Also, }}A \cup B = \left\{ {1,\,4,\,8,\,9,\,27,.....} \right\} \cr
& {\text{So, complement of }}A \cup B\,\,{\text{is infinite set}}{\text{.}} \cr
& {\text{Hence, }}A \cup B \ne N \cr} $$