Problem 23 - Algebra and Trigonometry

Question

Prove that if \(x>-1,\) then \((1+x)^{n} \geq 1+n x\) for all natural numbers \(n\)

Step-by-Step Solution

Answer

The inequality \((1+x)^{n} \geq 1+n x\) is proven by induction for \(x > -1\) and all natural \(n\).

Step 1: Understand the Inequality

We need to prove that for a real number \( x > -1 \) and for any natural number \( n \), the expression \( (1 + x)^n \) is greater than or equal to \( 1 + nx \). This is a form of the Binomial inequality.

Step 2: Use Mathematical Induction

To prove this inequality, we can use the principle of mathematical induction. This process involves two main steps: the base case and the inductive step.

Step 3: Base Case

For \( n = 1 \), the inequality becomes \((1 + x)^1 \geq 1 + 1 \cdot x\). This simplifies to \(1 + x \geq 1 + x\), which is obviously true for all \( x \). Thus, the base case is correct.

Step 4: Inductive Hypothesis

Assume that the inequality holds for some natural number \( k \), i.e., \((1 + x)^k \geq 1 + kx\). This assumption is what we call the inductive hypothesis.

Step 5: Inductive Step

We need to prove that if the inequality holds for \( k \), then it also holds for \( k+1 \), i.e., \((1 + x)^{k+1} \geq 1 + (k+1)x\).

Step 6: Establish \( (1 + x)^{k+1} \) using \( (1 + x)^k \)

Firstly, note that \((1 + x)^{k+1} = (1 + x)^k(1 + x)\). Using the inductive hypothesis, \((1 + x)^k \geq 1 + kx\). Therefore, \((1 + x)^{k+1} = (1 + x)^k(1 + x) \geq (1 + kx)(1 + x)\).

Step 7: Expand the Product

Expand \((1 + kx)(1 + x)\):\[(1 + kx)(1 + x) = 1 + x + kx + kx^2 = 1 + (k+1)x + kx^2.\]Since \( kx^2 \geq 0 \) for any \( k \geq 0 \) and \( x > -1 \), we have:\((1 + (k+1)x + kx^2) \geq 1 + (k+1)x\).

Step 8: Conclude the Inductive Step

Since \((1 + x)^{k+1} \geq (1 + (k+1)x + kx^2) \geq 1 + (k+1)x\), the inductive step is concluded. If the inequality holds for \( k \), it holds for \( k+1 \) as well.

Step 9: Conclusion

By mathematical induction, since the base case is true and the inductive step holds, the inequality \( (1 + x)^n \geq 1 + nx \) is proven for all natural numbers \( n \) when \( x > -1 \).

Key Concepts

Mathematical Induction

Mathematical induction is a powerful proof technique, often used to establish the truth of an infinite number of statements. It works similar to a chain reaction. This method consists of two key parts:

• Base Case: We prove the initial step is true. Usually, this involves checking the property for the smallest number in the set, such as 0 or 1. If the base case holds, the foundations of our chain are sound.
• Inductive Step: Assume the statement is true for some arbitrary case, let's call it "k". Then show that if this assumption (inductive hypothesis) is true, the statement must hold for the "next" case, which is often "k+1". This step demonstrates that every link in the chain is firmly connected to the previous one.

By proving these two steps, we "cover" all natural numbers. The base case shows the property is true at the starting point, and the inductive step confirms that the property propagates from one number to the next. This is what makes mathematical induction incredibly useful for proving inequalities and formulas involving natural numbers.

Inequalities

Inequalities are mathematical expressions involving the symbols \( >, <, \geq, \leq \), which show the relative size or order of two values. Understanding inequalities is fundamental in mathematics because they can be used to compare numbers and expressions.

When working with inequalities, there are several essential properties to keep in mind:

• Addition and Subtraction: You can add or subtract the same number from both sides of an inequality without changing its direction.
• Multiplication and Division: If you multiply or divide both sides of an inequality by a positive number, the direction of the inequality stays the same. However, if you use a negative number, the inequality sign flips.
• Transitivity: If \( a > b \) and \( b > c \), then \( a > c \).

In the given exercise, we are dealing specifically with the binomial inequality \((1 + x)^n \geq 1 + nx\) for \( x > -1 \), which is proven using mathematical induction. This inequality helps analyze the minimum growth bound of powers of expressions, making it a fundamental tool in algebra and calculus.

Natural Numbers

Natural numbers are the set of positive integers starting from 1 and incrementing without bound: 1, 2, 3, etc. They are denoted by \( \mathbb{N} \) and are considered the most basic number set in mathematics.

The use of natural numbers is common in proofs involving counting, sequences, and induction, because they provide a straightforward, infinite sequence of numbers to work with. Natural numbers have several core properties that are crucial in mathematical induction:

• Well-Ordering Principle: Any non-empty set of natural numbers has a least element. This concept underpins the use of base cases in induction.
• Closure under addition and multiplication: Adding or multiplying two natural numbers results in another natural number, which ensures that arithmetic operations within induction remain within the set of natural numbers.
• Inductive Nature: The defining feature that allows proofs like mathematical induction to operate effectively, following a predictable and infinite sequence.

In the context of the exercise, the natural numbers (\( n \)) provide the steps over which the inequality \((1 + x)^n \geq 1 + nx\) is proven, illustrating how these numbers naturally extend the proof's validity over an infinite domain.