Question

Let A = {a| a is a letter in the word 'college'} and
B = {b| b is a letter in the word 'luggage'}
and U = {a,b,c,d,e,f,g,l,o,u}, verify De Morgan's law.

Answer 1

A = {c,o,l,l,e,g,e} = {c,o,l,e,g}
B = {l,u,g,g,a,g,e} = {l,u,g,a,e}
∴ U = { a,b,c,d,e,f,g,l,o,u}

We have to verify De Morgan 's law.

$$\begin{gathered} \mathrm{We}\mathrm{have}\left(\mathrm{A}\cup \mathrm{B}\right)=\{a,c,e,g,l,o,u\} \\ \mathrm{Then}\left(\mathrm{A}\cup \mathrm{B}\right)\text{'}=\{b,d,f\}...\left(1\right) \\ \mathrm{A}\text{'}=\{a,b,d,f,u\} \\ \mathrm{B}\text{'}=\{b,c,d,f,o\} \\ \therefore \mathrm{A}\text{'}\cap \mathrm{B}\text{'}=\{b,d,f\}...\left(2\right) \\ \mathrm{From}\left(1\right)\mathrm{and}\left(2\right),\mathrm{we}\mathrm{have}: \\ \left(\mathrm{A}\cup \mathrm{B}\right)\text{'}=\mathrm{A}\text{'}\cap \mathrm{B}\text{'} \end{gathered}$$

$$\begin{gathered} \mathrm{Now},\left(\mathrm{A}\cap \mathrm{B}\right)=\{e,g,l\} \\ \mathrm{Then}\left(\mathrm{A}\cap \mathrm{B}\right)\text{'}=\{a,b,c,d,f,o,u\}...\left(3\right) \\ \mathrm{Also},\mathrm{A}\text{'}\cup \mathrm{B}\text{'}=\{a,b,c,d,f,o,u\}...\left(4\right) \\ \mathrm{From}\left(3\right)\mathrm{and}\hspace{0.17em}\left(4\right),\mathrm{we}\mathrm{have}: \\ \left(\mathrm{A}\cap \mathrm{B}\right)\text{'}=\mathrm{A}\text{'}\cup \mathrm{B}\text{'} \end{gathered}$$

Hence, De Morgan's law is verified.