Problem 21 - 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

By mathematical induction, \((1+x)^n \geq 1+nx\) holds for all natural numbers \(n\) when \(x > -1\).

Step 1: Understanding the Induction Base Case

First, verify the inequality for the smallest natural number, which is typically the base case of induction. Let's take \(n = 1\). Substitute \(n = 1\) into both sides of the inequality:\((1 + x)^1 \geq 1 + 1 \cdot x\).This simplifies to:\(1 + x \geq 1 + x\).The statement clearly holds for \(n = 1\) as it is an equality.

Step 2: Express the Inductive Hypothesis

Assume the inequality holds for some natural number \(k\), i.e.,\((1 + x)^k \geq 1 + kx\).This forms the inductive hypothesis.

Step 3: Prove Inductive Step

Prove that the statement holds for \(k + 1\). We need to show:\((1 + x)^{k+1} \geq 1 + (k+1)x\).Using the induction hypothesis:\[(1 + x)^{k+1} = (1 + x)^k (1 + x) \geq (1 + kx)(1 + x).\]Expand the right-hand side:\[1 + kx + x + kx^2 = 1 + (k+1)x + kx^2.\]Since \(kx^2 \geq 0\) for \(x > -1\), it follows that:\[1 + (k+1)x + kx^2 \geq 1 + (k+1)x.\]Thus, \((1 + x)^{k+1} \geq 1 + (k+1)x\). The inequality holds for \(k + 1\).

Step 4: Conclusion of Mathematical Induction

Using the principle of mathematical induction, the given inequality \((1 + x)^n \geq 1 + nx\) holds for all natural numbers \(n\) where \(x > -1\). We've verified the base case and the inductive step to confirm this.

Key Concepts

Inductive Hypothesis

In mathematical induction, the inductive hypothesis assumes that a statement is true for a particular case. It's like saying, "If it works up to this point, let's use it to prove the next point." This step bridges our base case with the general case.For the exercise, we imagine it's true that \((1+x)^k \geq 1+kx\) for some natural number \(k\). This is our inductive hypothesis.

• We assume the inequality holds true for \(n=k\).
• This assumption helps us prove the statement for \(n=k+1\).

By having this foundation, we can logically step forward to the next case.

Base Case

The base case is the foundation of our mathematical induction proof. Think of it as the starting point of our domino effect. We show that the formula is true for the very first step. For natural numbers, this often means starting with \(n=1\).In this exercise, by substituting \(n = 1\) into the inequality \((1+x)^1 \geq 1+x\), we show that both sides are equal:

• On the left: \(1+x\)
• On the right: \(1+x\)

Clearly, the base case holds since both sides are equal when \(x > -1\). It starts the chain reaction for induction.

Inequality Proof

To prove an inequality through induction, we ensure that any assumptions or operations performed maintain the inequality's truth. The exercise gives us \((1+x)^n \geq 1+nx\).When trying to prove the inductive step, we progress from \(n=k\) to \(n=k+1\). Our job is to show: \((1+x)^{k+1} \geq 1+(k+1)x\).Through expansion and simplification, \((1+x)^k (1+x) \geq (1+kx)(1+x)\).Upon expanding, \( 1+kx+x+kx^2 \geq 1+(k+1)x\).And since \(kx^2 \geq 0\), the inductive step is complete, ensuring it holds for \((k+1)\). This confirms the inequality for every subsequent \(n\) once the base case is established.

Natural Numbers

Natural numbers are the positive integers starting from 1. In mathematical induction, they form the sequential steps upon which the proof builds. When we say a statement holds for "all natural numbers \(n\)", it means you can start at 1 and keep going indefinitely.They are denoted by \(\mathbb{N}\) and are used because of their inherent order:

• They start with 1 and increase by an increment of 1 each time.
• They provide a framework for incrementally proving statements.

In this proof, natural numbers help us establish a stepwise pattern, where verifying the statement at one level helps confirm it for the next. This characteristic is key in understanding why induction is effective. It systematically addresses each possible value of \(n\).