Answer:
[tex]\text{Solution:}\\\text{U = \{1,2,...,20\}}\\\text{A = \{1,3,5,7,9\}}\\\text{B = \{1,2,3,...,20\}}\\\therefore\ \text{A}'=\text{U}-\text{A}=\{1,2,...,20\}-\{1,3,5,7,9\}=\{2,4,6,8,10,11,12,...,20\}\\\therefore\ \text{B}'=\text{U}-\text{B}=\{1,2,3,...,20\}-\{1,2,3,...,10\}=\{11,12,13,...,20\}[/tex][tex]\text{Now,}\\\text{According to Demorgan's law,}\\\text{1. A}'\cup\text{B}'=(\text{A}\cap\text{B})'\\\text{2. A}'\cap\text{B}'=(\text{A}\cup\text{B})'[/tex]
[tex]\text{Now we will prove them one by one.}\\\text{1. L.H.S. = A}'\cup\text{B}'=\{2,4,6,8,10,11,12,...,20\}\cup\{11,12,13,...,20\}\\\text{}\ \ \ \ \ \ \ \ \ \ \ \ =\{2,4,6,8,11,12,...,20\}\\\text{}\ \ \ \text{R.H.S. = }(\text{A}\cap\text{B})'=\{1,3,5,7,9\}=\text{U}-(\text{A}\cap\text{B})=\{2,4,6,8,10,11,...,20\}\\\therefore\ \text{A}'\cup\text{B}'=(\text{A}\cap\text{B})'[/tex]
[tex]\text{2. L.H.S. = A}'\cap\text{B}'=\{2,4,6,8,10,11,12,...,20\}\cap\{11,12,13,...,20\}\\\text{}\ \ \ \ \ \ \ \ \ \ \ \ =\{11,12,...,20\}\\\text{}\ \ \ \text{R.H.S. = }(\text{A}\cup\text{B})'=\{1,2,...,10\}=\text{U}-(\text{A}\cup\text{B})=\{11,12,...,20\}\\\therefore\ \text{A}'\cap\text{B}'=(\text{A}\cup\text{B})'[/tex]
