Let ${\mathit{B}}_{\mathbf{n}}\mathbf{=}\left\{ak|k is a multiple of n\right\}$Show that for each$\mathit{n}\mathbf{⩾}\mathbf{1}\mathit{n}$, the language Bis regular.

$k=ni$, where is any positive integer.so, on is the smallest positive integer and the greatest positive integer.

Now, To Increase the value of n keeping the value of i equal 1 to. When $i=1$and $n=2$

$\begin{array}{rcl}{B}_2& =& \left\{a^k\right\}\\ & =& \left\{a^{ni}\right\}\\ & =& \left\{a^{2\times 1}\right\}\\ & =& \left\{aa\right\}\\ & & \end{array}$

Then, the String comes out to be $\left\{aa\right\}$.

Again, Increase the value of n keeping the value of i equal to 1. When $i=1$ and$n=3$

$\begin{array}{rcl}{B}_3& =& \left\{a^k\right\}\\ & =& \left\{a^{ni}\right\}\\ & =& \left\{a^{3\times 1}\right\}\\ & =& \left\{aaa\right\}\end{array}$

Then, the String comes out to be$\left\{aaa\right\}$ and so on

Finite automation of the regular expression is as shown

In the above finite automaton,$q_0$is the initial and final state and $q_1,q_2,q_{3}and{q}_{n-1}$are the subsequent states.

The language B is the regular language. According to the closure property of the regular expression, it is clearly seen that the specific expression is a regular expression when the value of n is greater than and equal to 1.

Closure property includes various operations such as union, intersection, set complement, set reversal, set difference and many more. Assume that$B_n$ is regular.

Union of$B_1$ and$B_2$ results in the third string and it is also a regular expression.

Similarly, if user applies any property of closure, then the result is the regular expression.

Hence, it is proved that the above expression is the regular expression.