Problem 21 - Algebra and Trigonometry

Question

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

Step-by-Step Solution

Answer

By induction, \(n < 2^n\) for all natural numbers \(n\).

Step 1: Establish the Base Case

To prove inequalities involving all natural numbers using mathematical induction, first establish the base case. For the inequality \(n < 2^n\), check it for the smallest natural number, \(n = 1\):\[1 < 2^1\]Simplifying, \(1 < 2\), which is true. Therefore, the base case holds.

Step 2: State the Induction Hypothesis

Assume that \(n < 2^n\) holds true for some natural number \(k\). This is our induction hypothesis:\[k < 2^k\]We will use this hypothesis to prove that the inequality also holds for \(k + 1\).

Step 3: Inductive Step

We need to show that if \(k < 2^k\), then \(k+1 < 2^{k+1}\). Starting from the hypothesis, add 1 to both sides:\[k + 1 < 2^k + 1\]Since \(2^k + 1 < 2^k + 2^k = 2 \cdot 2^k = 2^{k+1}\), we can conclude:\[k + 1 < 2^k + 1 < 2^{k+1}\]Thus, \(k+1 < 2^{k+1}\) is true.

Step 4: Conclusion

By the principle of mathematical induction, since both the base case and the inductive step have been verified, the statement \(n < 2^n\) holds for all natural numbers \(n\).

Key Concepts

Natural Numbers

Natural numbers are the set of positive integers that start from 1 and go on infinitely: 1, 2, 3, and so on. These numbers form the basic building blocks of number theory and are fundamental to various branches of mathematics. In simple terms, natural numbers are used for counting and ordering.

• They do not include zero in traditional definitions, but sometimes zero is included, depending on the mathematical context.
• Natural numbers are denoted by the symbol \(\mathbb{N}\).
• There is no upper limit to the values in the natural numbers set.

Understanding natural numbers is crucial not only in arithmetic but also in understanding broader mathematical concepts like mathematical induction.

Inductive Step

The inductive step is a crucial part of a mathematical proof by induction. It involves proving that if a statement holds for one natural number, it also holds for the next one. This step builds the bridge that allows a statement to be true for all natural numbers once it’s true for the first one. Here’s how it works:

• Start by assuming that a statement holds for an arbitrary natural number \(k\). This is called the induction hypothesis.
• Then, under this assumption, demonstrate that the statement is also true for \(k+1\).
• If these steps succeed, it establishes a chain reaction ensuring that the statement holds for all numbers in the set.

The inductive step is like climbing a ladder. Once you prove one step is safe, you confirm that all following steps are equally safe. This ensures the statement applies to every natural number.

Base Case

In mathematical induction, the base case acts as the foundation of the proof. It’s the starting point, verifying the truth of the statement for the smallest value in the natural numbers set. Here’s why it matters:

• The base case ensures there’s no gap in the logic and that the statement is definitely true at least once.
• Typically, the base case uses the smallest natural number, which is often 1.
• Successfully proving the base case anchors the rest of the inductive proof.

Think of the base case as the first link in a chain, where each link's strength needs to be established to rely on the entire chain's integrity. If the first link is strong, then with each successful inductive step, the whole proof holds together throughout the infinite set of natural numbers.