Question

If $U=\{1,2,3,4\},A=\{2,3\},B=\{0,3\}$ then prove $(A\cup B{)}^{{}^{\prime }}=A^{{}^{\prime }}\cap B^{{}^{\prime }}$

Answer 1

We have,

$U=\{1,2,3,4\}$

$A=\{2,3\}$

$B=\{0,3\}$

L.H.S

${(A\cup B)}^{\prime }=U-(A\cup B)$

${(A\cup B)}^{\prime }=\{1,2,3,4\}-(\{2,3\}\cup \{0,3\})$

${(A\cup B)}^{\prime }=\{1,2,3,4\}-\{0,2,3\}$

${(A\cup B)}^{\prime }=\{1,4\}$

R.H.S

$A^{\prime }\cap B^{\prime }=(U-A)\cap (U-B)$

$A^{\prime }\cap B^{\prime }=(\{1,2,3,4\}-\{2,3\})\cap (\{1,2,3,4\}-\{0,3\})$

$A^{\prime }\cap B^{\prime }=\{1,4\}\cap \{1,2,4\}$

$A^{\prime }\cap B^{\prime }=\{1,4\}$

Hence, proved.