Let's address each part step-by-step: --- ### (a) Let \( L = \{a^n : \exists k \geq 1 \text{ such that } n = k^2\} \). Then \( L^0 \cup L^+ = \{a^n : n \text{ is a non-negative integer}\} \). **Step 1: Understanding the sets** - \( L \) is the set of strings of 'a's whose length is a perfect square (\(1, 4, 9, 16, \ldots\)). - \( L^0 \) is the set containing just the empty string, i.e., \( \{\epsilon\} \). - \( L^+ \) is the set of all strings that can be formed by concatenating one or more strings from \( L \). - \( L^0 \cup L^+ \) is then the set of all strings that can be formed by concatenating zero or more strings from \( L \), which is actually \( L^* \). **Step 2: The right side** - \( \{a^n : n \text{ is a non-negative integer}\} \) is the set of all strings consisting of zero or more 'a's (i.e., \( a^* \)). **Step 3: Are the two sets equal?** - \( L^0 \cup L^+ = L^* \). - \( L^* \) is the set of all strings that are concatenations of perfect square-length strings. - The question is: Can every non-negative integer \( n \) be written as a sum of one or more perfect squares? **No!** **Counterexample:** - For example, \( n=2 \): The only perfect squares less than or equal to 2 are 1. So \( 2 = 1+1 \), which is possible. - For \( n=3 \): \( 3 = 1+1+1 \), which is possible. - For \( n=5 \): \( 5 = 4+1 \), possible. - For \( n=7 \): \( 7 = 4+1+1+1 \), possible. - For \( n=6 \): \( 6 = 4+1+1 \), possible. But let's check for generality. Actually, by Lagrange's Four Square Theorem, **every** natural number is the sum of at most four perfect squares. So, every \( n \geq 0 \) can be written as a sum of perfect squares. So, **every possible \( n \) can be written as a sum of perfect squares** (including zero, which is given by the empty sum). **Therefore, the statement is TRUE.** --- #### **Proof:** By Lagrange's Four Square Theorem, every non-negative integer \( n \) can be written as a sum of four or fewer perfect squares. Since each element of \( L \) is \( a^{k^2} \) for some \( k \geq 1 \), concatenating \( m \) such strings gives \( a^{k_1^2 + k_2^2 + \cdots + k_m^2} \), covering all \( n \geq 0 \). Thus, \( L^* = \{a^n : n \text{ is a non-negative integer}\} \). --- ### (b) Suppose that no string from \( L_1 \) is a prefix of a string in \( L_2 \cup L_3 \). Suppose in addition that \( L_1L_2 = L_1L_3 \). Then \( L_2 = L_3 \). Let’s see if this is true. #### **Step 1: What does \( L_1L_2 = L_1L_3 \) mean?** - This says: For any \( x \in L_1 \), \( y \in L_2 \), \( xy \) is in \( L_1L_3 \). Similarly, for any \( x \in L_1 \), \( z \in L_3 \), \( xz \) is in \( L_1L_2 \). - In other words, for every \( x \in L_1 \), \( xL_2 = xL_3 \). #### **Step 2: The prefix condition** - No string from \( L_1 \) is a prefix of a string in \( L_2 \cup L_3 \). #### **Step 3: Is it true that \( L_2 = L_3 \)?** Suppose \( w \in L_2 \). Since \( L_1L_2 = L_1L_3 \), for any \( x \in L_1 \), \( xw \in L_1L_3 \), so there is some \( x' \in L_1 \), \( z \in L_3 \), such that \( xw = x'z \). - If \( x = x' \), then \( w = z \in L_3 \). - If \( x \neq x' \), then \( xw = x'z \) implies that one of \( x, x' \) is a prefix of the other, or one is shorter and matches the beginning of the other. - But, by the prefix condition, no string from \( L_1 \) is a prefix of a string from \( L_2 \cup L_3 \), so \( w \) must be equal to \( z \), and \( x = x' \). Therefore, every \( w \in L_2 \) is in \( L_3 \), and by symmetry, every \( z \in L_3 \) is in \( L_2 \). **Thus, the statement is TRUE.** --- #### **Proof:** Suppose \( w \in L_2 \). For any \( x \in L_1 \), \( xw \in L_1L_2 = L_1L_3 \), so \( xw = x'z \) for some \( x' \in L_1 \), \( z \in L_3 \). If \( x \neq x' \), then \( x \) is a prefix of \( x' \) or vice versa, or \( x' \) is a prefix of \( xw \), contradicting the prefix condition. Therefore, \( x = x' \) and \( w = z \in L_3 \). Similarly, \( L_3 \subseteq L_2 \), so \( L_2 = L_3 \). --- ## **Final Answers** **(a) TRUE.** *Proof:* By Lagrange’s Four Square Theorem, any non-negative integer \( n \) can be written as a sum of perfect squares, so \( L^* = \{a^n : n \text{ is a non-negative integer}\} \). **(b) TRUE.** *Proof:* The prefix condition ensures \( L_1L_2 = L_1L_3 \) only if \( L_2 = L_3 \); otherwise, the prefix property would be violated.