Subjective Type

If $$ \xi=\{1,2,3, \ldots .9\}, A=\{1,2,3,4,6,7,8\} $$ and $$ B=\{4,6,8\}, $$ then find. (i) $$ A^{\prime} $$ (ii) $$ B^{\prime} $$ (iii) $$ A \cup B $$ (iv) $$ \mathrm{A} \cap \mathrm{B} $$ (v) $$ A-B $$ (vi) $$ B-A $$ (vii) $$ (A \cap B)^{\prime} $$ (viii) $$ A^{\prime} \cup B^{\prime} $$ Also verify that: (a) $$ (A \cap B)^{\prime}=A^{\prime} \cup B^{\prime} $$ (b) $$ n(A)+n\left(A^{\prime}\right)=n(\xi) $$ (c) $$ n(A \cap B)+n\left((A \cap B)^{\prime}\right)=n(\xi) $$ (d) $$ n(A-B)+n(B-A)+n(A \cap B)=n(A \cup B) $$

Solution

From the question it is given that, $$ \xi=\{1,2,3,4,5,6,7,8,9\} $$
$$ A=\{1,2,3,4,6,7,8\} $$
$$ B=\{4,6,8\} $$
(i) $$ \mathrm{A}^{\prime}=\xi-\mathrm{A}=\{1,2,3,4,5,6,7,8,9\}-\{1,2,3,4,6,7,8\} $$
$$ \mathrm{A}^{\prime}=\{5,9\} $$
(ii) $$ \mathrm{B}^{\prime}=\xi-\mathrm{B}=\{1,2,3,4,5,6,7,8,9\}-\{4,6,8\} $$
$$ B^{\prime}=\{1,2,3,5,7,9\} $$
(iii) $$ A \cup B=\{1,2,3,4,6,7,8\} \cup\{4,6,8\} $$
$$ A \cup B=\{1,2,3,4,6,7,8\} $$
(iv) $$ \mathrm{A} \cap \mathrm{B}=\{1,2,3,4,6,7,8\} \cap\{4,6,8\} $$
$$ A \cap B=\{4,6,8\} $$
(v) $$ \mathrm{A}-\mathrm{B}=\{1,2,3,4,6,7,8\}-\{4,6,8\} $$
$$ A-B=\{1,2,3,7\} $$
(vi) $$ \mathrm{B}-\mathrm{A}=\{4,6,8\}-\{1,2,3,4,6,7,8\} $$
$$ B-A=\{\} $$
(vii) $$ (\mathrm{A} \cap \mathrm{B})^{\prime}=\xi-(\mathrm{A} \cap \mathrm{B}) $$
$$ (\mathrm{A} \cap \mathrm{B})^{\prime}=\{1,2,3,4,5,6,7,8,9\}-\{4,6,8\} $$
$$ (\mathrm{A} \cap \mathrm{B})^{\prime}=\{1,2,3,5,7,9\} $$
(viii) $$ \mathrm{A}^{\prime} \cup \mathrm{B}^{\prime}=\{5,9\} \cup\{1,2,3,5,7,9\} $$
$$ \mathrm{A}^{\prime} \cup \mathrm{B}^{\prime}=\{1,2,3,5,7,9\} $$
Then, $$ n(\xi)=9 $$
$$ n(A)=7 $$
$$ n\left(A^{\prime}\right)=2 $$
$$ n\left(B^{\prime}\right)=6 $$
$$ n(A \cap B)=3 $$
$$ n\left((A \cap B)^{\prime}\right)=6 $$
$$ n\left(A^{\prime} \cup B^{\prime}\right)=6 $$
$$ n(A-B)=4 $$
$$ n(B-A)=0 $$
$$ n(A \cup B)=7 $$
(a) $$ (A \cap B)^{\prime}=A^{\prime} \cup B^{\prime} $$
$$ (\mathrm{A} \cap \mathrm{B})^{\prime}=\{1,2,3,5,7,9\} $$
$$ \mathrm{A}^{\prime} \cup \mathrm{B}^{\prime}=\{1,2,3,5,7,9\} $$
By comparing the results, $$ (A \cap B)^{\prime}=A^{\prime} \cup B^{\prime} $$
(b) $$ n(A)+n\left(A^{\prime}\right)=n(\xi) $$
$$ 7+2=9 $$
$$ 9=9 $$
Therefore, by comparing the results, $$ n(A)+n\left(A^{\prime}\right)=n(\xi) $$
(c) $$ n(A \cap B)+n\left((A \cap B)^{\prime}\right)=n(\xi) $$
$$ 3+6=9 $$
$$ 9=9 $$
Therefore, by comparing the results, $$ n(A \cap B)+n\left((A \cap B)^{\prime}\right)=n(\xi) $$
(d) $$ n(A-B)+n(B-A)+n(A \cap B)=n(A \cup B) $$
$$ 4+0+3=7 $$
$$ 7=7 $$
Therefore, by comparing the results, $$ n(A-B)+n(B-A)+n(A \cap B)=n(A \cup B) $$

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