Problem 27 - A Graphical Approach to Precalculus with Limits

Question

Prove each statement by mathematical induction. \(2^{n}>2 n,\) if \(n \geq 3\)

Step-by-Step Solution

Answer

The inequality holds for all integers \(n \geq 3\) by induction.

Step 1: Base Case

We first test the inequality for the smallest integer greater than or equal to 3. Substitute \(n = 3\) into the expression: \[2^{n} > 2n\]\[2^{3} > 2 imes 3\]\[8 > 6\]The base case is true as 8 is greater than 6.

Step 2: Inductive Hypothesis

Assume the statement is true for some arbitrary integer \(k\) such that \(k \geq 3\). That is, assume:\[2^{k} > 2k\]

Step 3: Inductive Step

We must show that if the statement is true for \(n = k\), then it is also true for \(n = k + 1\). So we need to prove:\[2^{k+1} > 2(k+1)\]Start by expressing \(2^{k+1}\) in terms of \(2^k\):\[2^{k+1} = 2 imes 2^k\]Now using the inductive hypothesis that \(2^k > 2k\), we multiply both sides by 2:\[2 imes 2^k > 2 imes 2k\]\[2^{k+1} > 4k\]

Step 4: Simplifying the Inductive Step

To complete the step, we need to show \(4k > 2(k+1)\):\[4k > 2k + 2\]Simplifying:\[2k > 2\]\[k > 1\]This inequality \(k > 1\) is always true for \(k \geq 3\). Thus, the statement holds for \(n = k+1\).

Step 5: Conclusion

Since the base case and the inductive step are both verified, by mathematical induction, the given inequality \(2^{n} > 2n\) holds for all integers \(n \geq 3\).

Key Concepts

Inequality Proofs

In mathematics, inequality proofs are crucial for establishing relationships between expressions that are not equal but rather greater than or less than one another. The process involves verifying a mathematical statement that indicates one side of an inequality is consistently greater or less than the other under defined conditions. In our case, we aimed to show that \(2^n > 2n\) for all integers \(n\geq 3\).
This type of proof is essential as it helps underscore the behavior of expressions as they grow larger and ensures that certain relational properties hold universally within the specified domain. It often provides insight into the growth rates of functions which is vital in fields like computational complexity and algorithm analysis.
Inequality proofs using mathematical induction require careful establishment of a base case, an inductive hypothesis, and an inductive step, ensuring that the statement is verified across its intended range.

Base Case

The base case is the starting point in mathematical induction. It sets the foundation for the induction process by demonstrating the truth of the statement for the initial step, often known as the smallest value for which the statement is to be proven.
In proving the inequality \(2^n > 2n\), we began with \(n=3\). By substituting \(n = 3\) into the inequality, the calculation \(2^3 > 2 \, \cdot \, 3\) resulted in \(8 > 6\), which is true. Successfully proving the base case ensures that our inductive process starts on solid ground.
Without confirming the base case, the entire proof could be based on an unsupported assumption, making it untrustworthy.

Inductive Hypothesis

The inductive hypothesis is a pivotal step that drives the proof forward. It involves assuming that the statement or inequality is true for a particular case, usually denoted as \(n = k\). This assumption helps bridge the transition to the next step, allowing the proof to extend from known instances to unknown ones.
For our inequality, we assumed the truth of \(2^k > 2k\) for an arbitrary integer \(k\) with \(k \geq 3\). This step does not require proof itself but lays the groundwork for proving it for \(k+1\).
Overall, the inductive hypothesis is a placeholder that enables the procession of logic from one instance to another, guiding the proof towards the general case.

Inductive Step

The inductive step involves proving that if the statement holds true for an arbitrary case \(n = k\), it must also hold for the next case, \(n = k+1\). This step completes the framework to cover all possible values within the domain.
For the inequality \(2^n > 2n\), we had to show that \(2^{k+1} > 2(k+1)\) given that \(2^k > 2k\). By rewriting \(2^{k+1}\) as \(2 \cdot 2^k\), and utilizing our inductive hypothesis, we established \(2 \cdot 2^k > 2 \cdot 2k\), leading to the simplified inequality \(4k > 2k + 2\). Further simplification showed \(2k > 2\), which holds true for \(k > 1\).
The successful conclusion of the inductive step solidifies the statement for all applicable cases. It connects the initial truth of the base case to all sequential instances covered by the induction.