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Let $N$ denote the setof natural numbers and $A=\left\{\mathrm{n}^{2}: \mathrm{n} \in N\right\}$ and $\mathrm{B}=\left\{\mathrm{n}^{3}: \mathrm{n} \in N\right\} .$ Which one of the following is incorrect?
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Verified Answer
The correct answer is:
$A \cup B=N$
Let $\mathrm{A}=\left\{n^{2}: n \in \mathrm{N}\right\}$ and $\mathrm{B}=\left\{n^{3}: n \in \mathrm{N}\right\}$
$\mathrm{A}=\{1,4,9,16, \ldots . .\}$
and $\mathrm{B}=\{1,8,27,64, \ldots \ldots\}$
Now, $\mathrm{A} \cap \mathrm{B}=\{1\}$ which is a finite set.
Also, $\mathrm{A} \cup \mathrm{B}=\{1,4,8,9,27, \ldots . .\}$
So, complement of $\mathrm{A} \cup \mathrm{B}$ is infinite set. Hence, $\mathrm{A} \cup \mathrm{B} \neq \mathrm{N}$
$\mathrm{A}=\{1,4,9,16, \ldots . .\}$
and $\mathrm{B}=\{1,8,27,64, \ldots \ldots\}$
Now, $\mathrm{A} \cap \mathrm{B}=\{1\}$ which is a finite set.
Also, $\mathrm{A} \cup \mathrm{B}=\{1,4,8,9,27, \ldots . .\}$
So, complement of $\mathrm{A} \cup \mathrm{B}$ is infinite set. Hence, $\mathrm{A} \cup \mathrm{B} \neq \mathrm{N}$
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