Problem 19 - Precalculus: Mathematical for Calculus

Question

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

Step-by-Step Solution

Answer

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

Step 1: Base Case

Start by checking the base case where \( n = 1 \). For \( n = 1 \), we need to verify that \( 1 < 2^1 \), which simplifies to \( 1 < 2 \). This is true, so the base case holds.

Step 2: Inductive Hypothesis

Assume that the statement \( n < 2^n \) is true for some natural number \( n = k \). So, we have \( k < 2^k \) as our inductive hypothesis.

Step 3: Inductive Step

Using the inductive hypothesis, show that the statement also holds for \( n = k + 1 \). We need to prove that \( k + 1 < 2^{k+1} \).

Step 4: Induction Proof Conclusion

We know from our hypothesis that \( k < 2^k \). Adding 1 to both sides gives \( k + 1 < 2^k + 1 \). Since \( 2^k + 1 \leq 2^k + 2^k = 2 \times 2^k = 2^{k+1} \), it follows that \( k + 1 < 2^{k+1} \). This completes the induction step.

Step 5: Conclusion

By the principle of mathematical induction, since the base case holds and the inductive step is valid, the inequality \( n < 2^n \) is true for all natural numbers \( n \).

Key Concepts

Inequality Proof

An inequality proof is a type of mathematical demonstration to show that one quantity is consistently less than or greater than another across a range of cases. In our exercise, we are tasked to prove that, for all natural numbers \(n\), the inequality \(n < 2^n\) holds true. This statement suggests that, starting from minimal values, \(n\) will always remain smaller than the exponential counterpart \(2^n\).

The key to proving this inequality is ensuring that it works for a starting point and then demonstrating that if it holds for one number, it will also hold for the next. That's where the principles of mathematical induction become crucial tools in our proof.

Base Case

The base case forms the foundation of a proof by mathematical induction. It involves showing that the statement is true for the initial value of the natural numbers. For the inequality \(n < 2^n\), we start by checking the smallest natural number, which is usually \(n=1\).

For \(n = 1\), we need to verify the inequality: \(1 < 2^1\). This simplifies to \(1 < 2\), which is clearly true. Establishing the base case is essential because it is the starting point that guarantees the rest of the proof can be built upon it reliably. Without a valid base case, the entire premise of inductive reasoning would crumble.

Inductive Hypothesis

The inductive hypothesis is an assumption that the statement holds true for some arbitrary natural number \(n = k\). In our exercise, we assume that \(k < 2^k\) is correct.

This step might feel a bit counterintuitive because we are assuming the statement is true to ultimately prove it. However, this assumption allows us to explore the implications and extend our proof further. This deductive step sets the stage for the inductive step, where we show how the truth for one case implies the truth for the next.

Inductive Step

With the inductive hypothesis in hand, the inductive step involves demonstrating that if \(k < 2^k\) is true, then \(k+1 < 2^{k+1}\) must also be true.

• Start by knowing \(k < 2^k\) based on our hypothesis.
• Add 1 to both sides, giving \(k + 1 < 2^k + 1\).
• Observe that \(2^k + 1 \leq 2^k + 2^k\), which simplifies to \(2 \times 2^k = 2^{k+1}\).
• This implies \(k + 1 < 2^{k+1}\).

Bridging this gap from \(k < 2^k\) to \(k + 1 < 2^{k+1}\) securely fastens each link in the chain of our induction, ensuring the overall statement holds true for any natural number. Thus, the principle of mathematical induction allows us to conclude that \(n < 2^n\) for all natural numbers \(n\).