Problem 21 - Precalculus Mathematics for Calculus

Question

Prove that \(n<2^{n}\) for all natural numbers \(n .\)

Step-by-Step Solution

Answer

The inequality \(n < 2^n\) holds for all natural numbers by induction.

Step 1: Prove the Base Case

To use mathematical induction, we first need to verify the base case where \(n = 1\). For \(n = 1\), the inequality \(1 < 2^1\) holds true since \(1 < 2\). Therefore, the base case is satisfied.

Step 2: Inductive Hypothesis

Assume that the inequality holds for some arbitrary natural number \(k\), i.e., assume \(k < 2^k\) is true. This assumption is known as the inductive hypothesis.

Step 3: Inductive Step

Show that if the statement is true for \(n = k\), then it must also be true for \(n = k+1\). We need to prove that \(k + 1 < 2^{k+1}\).

Step 4: Manipulate Inductive Step

We start from the assumption, \(k < 2^k\). Add 1 to both sides to get \(k + 1 < 2^k + 1\). Next, note that \(2^k + 1 \leq 2 \times 2^k = 2^{k+1}\) for all \(k\geq 1\), which implies \(k + 1 < 2^{k+1}\). Thus, the inequality \(k + 1 < 2^{k+1}\) holds.

Step 5: Conclusion by Inductive Principle

Since the base case is true, and assuming \(k < 2^k\) implies \(k+1 < 2^{k+1}\), by mathematical induction, the statement \(n < 2^n\) holds for all natural numbers \(n\). Thus, the proof is complete.

Key Concepts

Base Case

The process of mathematical induction begins by establishing the base case, which serves as the foundation for the logical argument. In simpler terms, the base case is the initial step where we prove that the statement holds true for the first element in our context, typically starting with the lowest natural number.

• We begin by checking a specific value, often where the simplest form of the statement can be assessed.
• In the original problem, the base case involved verifying the inequality for \( n = 1 \), which is demonstrated as true because \( 1 < 2^1 \) simplifies to \( 1 < 2 \), a true statement.
• This step is essential as it sets the stage for all natural numbers to follow, confirming the statement's validity at the starting point.

By verifying the base case, we establish a solid starting ground. This ensures that any subsequent applications of the statement are based on a proven initial truth.

Inductive Hypothesis

The inductive hypothesis is a key part of the mathematical induction process. It involves assuming that the statement or inequality is valid for a particular case. This assumption allows us to make further deductions later in the proof.

• In this method, we assume that the statement holds true for some arbitrary number, commonly denoted as \( k \).
• In our example, the hypothesis assumes \( k < 2^k \), relying on this condition to explore possibilities for the subsequent step.
• The beauty of this assumption is that it helps us bridge to future scenarios within the same logical framework.

While the inductive hypothesis may feel like a leap of faith, it's a strategic step. It enables the proof to explore continuity and transition from specific cases to more general applicability.

Inductive Step

The inductive step involves proving that if the statement is true for an arbitrary number \( k \), it must also be true for the next consecutive number \( k+1 \). This step essentially links each case to the next through logical reasoning.

• In practical terms, we extend our assumption from the inductive hypothesis to demonstrate its truth in the next instance.
• For the given problem, we start with the known \( k < 2^k \) and strive to prove \( k + 1 < 2^{k+1} \).
• The manipulation is clever and involves simple arithmetic and exponential properties: adding 1 to both sides and showing it fits within the exponential expansion.

This step solidifies the chain reaction effect, ensuring the pattern holds true beyond our initial assumption. By confirming this link, we assure the statement's validity for all natural numbers.