Let's solve both problems **step-by-step using the Pumping Lemma for Regular Languages**. --- ## **1. Prove \( L = \{ a^k b^\ell : k, \ell \geq 0 \text{ and } k \geq 3\ell \} \) is NOT regular** ### **Step 1: State the Pumping Lemma** If \(L\) is regular, there exists a pumping length \(p\) such that for any string \(w \in L\) with \(|w| \geq p\), \(w\) can be split into \(w = xyz\) such that: - \(|xy| \leq p\) - \(|y| > 0\) - For all \(i \geq 0\), \(xy^iz \in L\) --- ### **Step 2: Choose an appropriate string \( w \in L \)** Pick \( \ell = p \). Then \( k \) must be at least \( 3\ell = 3p \). Let \( k = 3p \), so the string is: \[ w = a^{3p} b^{p} \] Clearly, \( w \in L \) since \( k = 3p \geq 3p = 3\ell \). --- ### **Step 3: Split \( w = xyz \) with \( |xy| \leq p \), \( |y| > 0 \)** Since \( |xy| \leq p \) and the first \( 3p \) symbols are all \( a \)'s, \( y \) consists only of \( a \)'s. Let \( x = a^r, y = a^s \), where \( r + s \leq p \), \( s > 0 \). So, \( x = a^r \), \( y = a^s \), \( z = a^{3p - r - s} b^p \). --- ### **Step 4: Pump \( y \) (try \( i = 0 \))** \[ xy^0z = a^r a^{3p - r - s} b^p = a^{3p - s} b^p \] Now, the number of \( a \)'s is \( 3p - s \), and the number of \( b \)'s is \( p \). For the string to be in \( L \), we need: \[ 3p - s \geq 3p \implies -s \geq 0 \implies s \leq 0 \] But \( s > 0 \), so this is **impossible**. --- ### **Step 5: Conclusion** We found a string \( w \in L \) such that for any splitting \( w = xyz \) with Pumping Lemma constraints, there is some \( i \) (specifically \( i=0 \)) for which \( xy^iz \notin L \). **Thus, \( L \) is NOT regular.** --- ## **2. Prove \( L = \{ a^{2025n + n^{2025}} : n \geq 0 \} \) is NOT regular** ### **Step 1: State the Pumping Lemma** (As above.) --- ### **Step 2: Choose a string \( w \in L \) with length at least \( p \)** Let \( n \) be large so \( 2025n + n^{2025} \geq p \). Let \( n = t \) where \( t \) is large enough so that \( w = a^{2025t + t^{2025}} \) and \( |w| \geq p \). --- ### **Step 3: Split \( w = xyz \) with \( |xy| \leq p \), \( |y| > 0 \)** Since \( |xy| \leq p \) and \( w \) is a long string of \( a \)'s, \( y = a^j \) for some \( j > 0 \), and \( x = a^i \), \( z = a^{2025t + t^{2025} - i - j} \) with \( i + j \leq p \). --- ### **Step 4: Pump \( y \) (try \( i = 2 \))** \[ xy^2z = a^{i} a^{2j} a^{2025t + t^{2025} - i - j} = a^{2025t + t^{2025} + j} \] Now, the length of the string is \( 2025t + t^{2025} + j \). Can this be written as \( 2025n' + (n')^{2025} \) for some integer \( n' \)? **No**, because \( j \) is a small positive number (since \( j \leq p \)), and for large \( t \), adding a small \( j \) cannot be expressed as \( (n' - t)^{2025} + 2025(n' - t) \) unless \( j = 0 \), which is not possible as \( |y| > 0 \). Thus, for sufficiently large \( t \), \( 2025t + t^{2025} + j \) does **not** fit the form \( 2025n' + (n')^{2025} \). --- ### **Step 5: Conclusion** We found a string \( w \in L \) such that for any splitting \( w = xyz \) with Pumping Lemma constraints, there is some \( i \) (specifically \( i=2 \)) for which \( xy^iz \notin L \). **Thus, \( L \) is NOT regular.** --- ## **Final Answers:** 1. \( L = \{ a^k b^\ell : k, \ell \geq 0, k \geq 3\ell \} \) is **not regular** (by the pumping lemma). 2. \( L = \{ a^{2025n + n^{2025}} : n \geq 0 \} \) is **not regular** (by the pumping lemma).