Question 25

Given $L_1 = \{a^n \mid n \ge 0\}$ and $L_2 = \{b^n \mid n \ge 0\}$.

1. Language Definitions:

   • $L_1 = a^*$ (all strings of zero or more $a$'s).
   • $L_2 = b^*$ (all strings of zero or more $b$'s).
   • Both $L_1$ and $L_2$ are regular languages.

2. Statement (I): $L_1 \cdot L_2$ is a regular language:

   • Concatenation of $L_1$ and $L_2$ produces $L_1 \cdot L_2 = a^* b^*$.
   • Because the class of regular languages is closed under concatenation, $a^* b^*$ is a regular language.
   • Hence, Statement (I) is CORRECT.

3. Statement (II): $L_1 \cdot L_2 = \{a^n b^n \mid n \ge 0\}$:

   • In $L_1 \cdot L_2$, the number of $a$'s and the number of $b$'s are independently chosen: $a^m b^k$ with $m, k \ge 0$.
   • For example, the string $a^2 b^3 = aabbb$ is in $L_1 \cdot L_2$, but it is not in $\{a^n b^n \mid n \ge 0\}$ (which requires equal counts of $a$ and $b$).
   • Furthermore, $\{a^n b^n \mid n \ge 0\}$ is non-regular, whereas $L_1 \cdot L_2$ is regular.
   • Hence, Statement (II) is INCORRECT.

4. Therefore, Only (I) is CORRECT (option A).