91Ó°ÊÓ

In mathematical induction, the base case serves as the starting point for the proof. We begin by validating the given inequality for the smallest possible value of the variable. In our example, this smallest value is \( n = 5 \). By substituting \( n = 5 \) into the inequality \( 2n - 3 \leq 2^{n-2} \), we check if it holds true. This results in \( 2 \times 5 - 3 = 7 \) on the left-hand side and \( 2^{5-2} = 8 \) on the right-hand side. As the calculation shows, 7 \( \leq \) 8, confirming the base case. Therefore, our inequality is valid when \( n = 5 \). Once the base case is verified, this foundational step allows us to continue with the induction process.

The induction hypothesis is an assumption about the truth of the statement for an arbitrary but fixed natural number. It is the bridge that connects the base case to the next steps of the proof. For our inequality \( 2n - 3 \leq 2^{n-2} \), the induction hypothesis assumes that it holds true for some arbitrary number \( k \), where \( k \geq 5 \). This means, we assume \( 2k - 3 \leq 2^{k-2} \) is true. This assumption is crucial because it forms the basis for proving the statement for \( n = k + 1 \). Remember that in mathematical induction, we do not need to prove the hypothesis itself. Instead, we use it as a stepping stone to establish the statement for the subsequent case.

The induction step is the part of the proof where we demonstrate that if the statement holds for an arbitrary case \( k \), then it logically holds for the next case, \( k + 1 \). For our example, this involves proving the inequality \( 2(k + 1) - 3 \leq 2^{(k+1) - 2} \). Simplifying the left-hand side, we get \( 2k - 1 \). According to our induction hypothesis, \( 2k - 3 \leq 2^{k-2} \). Therefore, \( 2k - 1 \leq 2^{k-2} + 2 \). We need to show that the right-hand side, \( 2^{k-1} \), is indeed greater than or equal to \( 2^{k-2} + 2 \). Since doubling a number is greater than simply adding two, it follows that \( 2^{k-1} \geq 2^{k-2} + 2 \). This concludes our proof, showing the statement is true for \( n = k + 1 \) whenever it is true for \( n = k \). Thus, by the principle of mathematical induction, the inequality is valid for all \( n \geq 5 \).