Class 9th math Review Exercise 3 solution english PCTB

Question 7: Verify De Morgan’s Laws for the following sets:

\(U=\{1,2,3, \ldots, 20\}\), \(A=\{2,4,6, \ldots, 20\}\) and \(B=\{1,3,5, \ldots, 19\}\).

De Morgan’s Laws:

\( (i)\;\) \({(A\cup B)}’={A}’\cap {B}’ \)

\( (ii)\;\) \({(A\cap B)}’={A}’\cup {B}’ \)


\( (i)\;\) \({(A\cup B)}’={A}’\cap {B}’ \)

\(A\cup B=\{2,4,6, \ldots, 20\}\cup \{1,3,5, \ldots, 19\}\)

\(\ \ \ \ \ \ \ \ \ \ \ =\{1,2,3,\ldots, 20\}\)

\(\text{L.H.S.}={(A\cup B)}’\)

\(\ \ \ \ \ \ \ \ \ \ \ =U-(A\cup B)\)

\(\ \ \ \ \ \ \ \ \ \ \ =\{1,2,3,\ldots, 20\}\)\(\ -\ \)\(\{1,2,3,\ldots, 20\}\)

\(\ \ \ \ \ \ \ \ \ \ \ =\{\}\)

\(\text{R.H.S.}={A}’\cap {B}’\)

\({A}’=U-A\)

\(\ \ \ \ \ \ \ \ \ \ \ =\ \)\(\{1,2,3,\ldots, 20\}\)\(\ -\ \)\(\{2,4,6, \ldots, 20\}\)

\(\ \ \ \ \ \ \ \ \ \ \ =\ \)\(\{1,3,5, \ldots, 19\}\)

\({B}’=U-B\)

\(\ \ \ \ \ \ \ \ \ \ \ =\ \)\(\{1,2,3,\ldots, 20\}\)\(\ -\ \)\(\{1,3,5, \ldots, 19\}\)

\(\ \ \ \ \ \ \ \ \ \ \ =\ \)\(\{2,4,6, \ldots, 20\}\)

\(\text{R.H.S.}={A}’\cap {B}’\)

\(\ \ \ \ \ \ \ \ \ \ \ =\{1,3,5, \ldots, 19\}\)\(\cap\)\(\{2,4,6, \ldots, 20\}\)

\(\ \ \ \ \ \ \ \ \ \ \ =\{\}\)

Therefore, \({(A\cup B)}’={A}’\cap {B}’ \)


\( (ii)\;\) \({(A\cap B)}’={A}’\cup {B}’ \)

\(A\cap B=\{2,4,6, \ldots, 20\}\cap \{1,3,5, \ldots, 19\}\)

\(\ \ \ \ \ \ \ \ \ \ \ =\{\}\)

\(\text{L.H.S.}={(A\cap B)}’\)

\(\ \ \ \ \ \ \ \ \ \ \ =U-(A\cap B)\)

\(\ \ \ \ \ \ \ \ \ \ \ =\{1,2,3,\ldots, 20\}\)\(\ -\ \)\(\{\}\)

\(\ \ \ \ \ \ \ \ \ \ \ =\{1,2,3,\ldots, 20\}\)

\(\text{R.H.S.}={A}’\cup {B}’\)

\({A}’=U-A\)

\(\ \ \ \ \ \ \ \ \ \ \ =\ \)\(\{1,2,3,\ldots, 20\}\)\(\ -\ \)\(\{2,4,6, \ldots, 20\}\)

\(\ \ \ \ \ \ \ \ \ \ \ =\ \)\(\{1,3,5, \ldots, 19\}\)

\({B}’=U-B\)

\(\ \ \ \ \ \ \ \ \ \ \ =\ \)\(\{1,2,3,\ldots, 20\}\)\(\ -\ \)\(\{1,3,5, \ldots, 19\}\)

\(\ \ \ \ \ \ \ \ \ \ \ =\ \)\(\{2,4,6, \ldots, 20\}\)

\(\text{R.H.S.}={A}’\cup {B}’\)

\(\ \ \ \ \ \ \ \ \ \ \ =\{1,3,5, \ldots, 19\}\)\(\cup\)\(\{2,4,6, \ldots, 20\}\)

\(\ \ \ \ \ \ \ \ \ \ \ =\{1,2,3,\ldots, 20\}\)

Therefore, \({(A\cap B)}’={A}’\cup {B}’ \)

Leave a comment