91Ó°ÊÓ

We need to check if \(\log_{2}1 \leq 1\) is true. Recall that \(\log_{a}1 = 0\) for any base \(a \neq 1\). Therefore, for our base case: \[ \log_{2}1 = 0 \] Since \(0 \leq 1\), the inequality holds for the base case.

Now let's assume that the inequality is true for some arbitrary integer \(k \geq 1\), so we have: \[ \log_{2}k \leq k \] We want to prove that the inequality also holds for \(k+1\). To achieve that we need to show that: \[ \log_{2}(k+1) \leq (k+1) \] Let's manipulate the inductive hypothesis by noticing that since \(\log_{2}(k+1)\) is an increasing function, if \(\log_{2}k \leq k\), then \(\log_{2}(k+1) \leq (k+1) \leq (k+2)\). Now, since \(k \geq 1\), we have \((k+2) \geq 3\) and \(k+1 \geq 2\), so we can use the property of logarithms to rewrite our inequality. We have: \[ \log_{2}(k+1) \leq k+1 \leq k+2 \] By the property of logarithms: \[ \log_{2}(k+1) \leq \log_{2}(k+1) + \log_{2}(1+\frac{1}{k}) \] \[ \log_{2}(k+1) \leq \log_{2}(k+1) + \log_{2}(2) \] since \(\log_{2}(1+\frac{1}{k})<1\) for all \(k \geq 1\). \[ \log_{2}(k+1) \leq \log_{2}(2(k+1)) \] Since logarithm functions are increasing functions, we have: \[ k+1 \leq 2(k+1) \] \[ k+1 \leq 2k+2 \] Thus, the inequality \(\log_{2}(k+1) \leq (k+1)\) holds. By the principle of mathematical induction, we have proven that \(\log_{2}n \leq n\) is true for all integers \(n \geq 1\).