91Ó°ÊÓ

First, let's define the set of integers S as follows: \[S = \{n = 2^k \cdot m | m \text{ is an odd integer, } k \text{ is a nonnegative integer}\}\] The goal is to show that for any given integer \(n \geq 1\), it belongs to the set S.

The well-ordering principle tells us that there exists a smallest element in the set S. Let m be an odd integer and k be a nonnegative integer such that for every integer, \(n = 2^k \cdot m\). Let's consider the set \(T = \{ t = 2^k | n = 2^k \cdot m\}\), where m is an odd integer, and t is the smallest possible power of 2 that satisfies the equation. According to the well-ordering principle, T must have a smallest element. Let \(t_0 = 2^{k_0}\) be the smallest element in the set T.

Now consider the integer \(n = 2^{k_0} \cdot m_0\), where \(m_0\) is an odd integer, and \(k_0\) is a nonnegative integer. We have: \[n - t_0 = (2^{k_0} \cdot m_0) - 2^{k_0}\] Factor out \(2^{k_0}\): \[n - t_0 = 2^{k_0} (m_0 - 1)\] Since \(m_0\) is odd, and we subtract 1 from it, the result (m_0 - 1) will be even. Let's denote \(m_1 = \frac{m_0 - 1}{2}\). Now, we can rewrite the equation as: \[n - t_0 = 2^{k_0 + 1} \cdot m_1\] From this equation, we can see that, by choosing \(k_1 = k_0 + 1\), we have: \[n = 2^{k_1} \cdot (m_1 + 1)\] Since \(m_1\) is even, the quantity \((m_1 + 1)\) is odd, so we found an odd integer and nonnegative integer for which the equation \(n = 2^k \cdot m\) holds. Thus, the given integer n belongs to the set S.

Using the well-ordering principle, we have proved that for any integer \(n \geq 1\), there exists an odd integer m and a nonnegative integer k such that \(n = 2^k \cdot m\). This result applies to all integers greater than or equal to 1, and therefore, the statement is true.