Let's solve it **step by step** as requested. --- ### **Step 1: Understanding the Dedekind Cut \( L \)** The Dedekind cut \( L \) is defined as: \[ L = \{ r \in \mathbb{Q} : r \geq 0 \text{ and } r^2 < 2 \} \cup \{ r \in \mathbb{Q} : r < 0 \} \] - This is the set of all negative rationals, and all non-negative rationals whose square is less than 2. - The real number determined by this cut, say \( s \), is \( \sqrt{2} \), since it is the least upper bound of all rationals whose square is less than 2. **Explanation:** A Dedekind cut is a way to construct real numbers by partitioning the rationals into two sets, with the lower set containing all rationals less than a certain real number. Here, the cut partitions at \( \sqrt{2} \). --- ### **Step 2: Proving \( L_x^2 = L_2 \) ** - \( L_x \) is the cut corresponding to the real number \( x = \sqrt{2} \). - \( L_2 \) is the cut corresponding to the real number 2. - \( L_x^2 \) is the set of all rationals \( q \) such that \( q < x^2 \). But \( x = \sqrt{2} \implies x^2 = 2 \), so \( L_x^2 \) is the set of all \( q \in \mathbb{Q} \) such that \( q < 2 \). But that is **exactly** the Dedekind cut for 2, i.e., \( L_2 \). **Explanation:** The square of the real number defined by \( L \) is the supremum of all rationals whose square is less than 2, which is 2. So the cut for \( x^2 \) is the same as the cut for 2. --- ## **Final Answer:** \[ L_x^2 = L_2 \] **Thus, the real number corresponding to \( L \) has square 2.**