Problem 23 - Precalculus: Mathematics for Calculus

Question

Use mathematical induction to prove that the formula is true for all natural numbers \(n\) If \(x>-1,\) then \((1+x)^{n} \geq 1+n x\) for all natural numbers \(n\)

Step-by-Step Solution

Answer

The inequality \((1+x)^n \geq 1+nx\) holds for all natural numbers \(n\) and \(x > -1\) by induction.

Step 1: Base Case

To prove the statement by induction, we start with the base case. For the base case, we test the formula with the smallest natural number, which is \( n = 1 \). Substitute \( n = 1 \) into the inequality: \((1+x)^n = (1+x)^1 = 1+x\) and compare this with \( 1 + 1 \times x = 1 + x \). Both sides are equal, hence the base case holds true.

Step 2: Induction Hypothesis

Assume the formula is true for some arbitrary natural number \( k \). That is, assume that \( (1+x)^k \geq 1+kx \) holds true for \( n = k \). This is our induction hypothesis and will be used to show the formula is true for \( n = k+1 \).

Step 3: Inductive Step

For the induction step, we need to show that if the inequality holds for \( n = k \), then it must also hold for \( n = k+1 \). Start with the expression for \( n = k+1 \):\((1+x)^{k+1} = (1+x)^k \cdot (1+x) \).By the induction hypothesis, we know that \( (1+x)^k \geq 1+kx \). Substituting, we have:\((1+x)^{k+1} = (1+x)^k (1+x) \geq (1+kx)(1+x) \).Expand this product:\( 1+kx + x + kx^2 = 1 + (k+1)x + kx^2 \).Since \( x > -1 \) implies \( kx^2 \geq 0 \), we have:\( 1 + (k+1)x + kx^2 \geq 1 + (k+1)x \).Hence, \((1+x)^{k+1} \geq 1 + (k+1)x \), which completes the inductive step.

Step 4: Conclusion of Induction

By mathematical induction, since the base case is true and the inductive step has been validated, the formula \((1+x)^n \geq 1 + nx\) is true for all natural numbers \( n \) given \( x > -1 \). This concludes the proof.

Key Concepts

Base Case

The base case in mathematical induction is like the starting point. We begin the proof by checking if the statement is true for the smallest possible value of our variable. Here, the smallest natural number is 1. For our specific exercise, we need to verify if the inequality

• \((1 + x)^n \geq 1 + nx\)

holds true when \(n = 1\).

By substituting \(n\) with 1, the expression

• \((1+x)^1 = 1+x\)

can be directly compared to \(1 + 1 \times x = 1 + x\). Both sides are indeed equal, confirming that the base case works. This gives us a solid foundation to step into the next part of induction.

Inductive Step

Once the base case is confirmed, the inductive step shows that if the statement works for one value, it should work for the next. Think of it like climbing a ladder: when you step on one rung and it's stable, the next one should support you, too.

For this part of the process, you assume that our statement is true for \(n = k\), which is the induction hypothesis. The hypothesis says

• \((1+x)^k \geq 1+kx\)

holds. Next, we use this to prove it's true for \(n = k+1\).

We start by analyzing

• \((1+x)^{k+1} = (1+x)^k \cdot (1+x)\)

and apply the induction hypothesis:

• \((1+x)^k \cdot (1+x) \geq (1+kx)(1+x)\)

which expands to

• \(1 + (k+1)x + kx^2\).

Since \(kx^2\) is non-negative (because \(x > -1\)), the final expression

• \(1 + (k+1)x + kx^2 \geq 1 + (k+1)x\)

holds as true. This seals the inductive step, ensuring the statement is true for the next number, \(n = k+1\).

Natural Numbers

Natural numbers are the cornerstone of mathematical induction. These are the numbers we count with: 1, 2, 3, and so on. Importantly, there is no zero, fractions, or negative numbers in natural numbers.

In our exercise, we only consider these positive integers for proving our statement. When you start with the smallest natural number, which is 1, and show the ladder climb works through mathematical induction, you've shown your statement applies to the entire infinite set of natural numbers.

Each step in induction respects the properties of natural numbers, like every natural number having a successor (the next number). This is essential for building our induction logic from one number to the next.

Inequality

Inequality is a mathematical statement about the relative size or order of two objects. In our exercise, the inequality

• \((1+x)^n \geq 1+nx\)

is what we aim to prove using induction.

The symbol \(\geq\) means "greater than or equal to." In our case, it establishes a comparison between two expressions involving \(x\) and natural numbers \(n\).

Understanding how inequalities work is crucial for verifying the statements during both the base case and the inductive step. During the inductive step, manipulating inequalities helps us extend the proof from \(n = k\) to \(n = k+1\), ensuring that our induction hypothesis applies smoothly. This work with inequalities affirms the logical flow of our proof in establishing the robustness of our statement for all natural numbers.