Problem 23 - Precalculus Mathematics for Calculus

Question

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

Step-by-Step Solution

Answer

The inequality holds by mathematical induction for all natural \(n\) with \(x > -1\).

Step 1: Understand the Problem

We are asked to prove the inequality \((1+x)^{n} \geqq 1+n x\) for all natural numbers \(n\) under the condition that \(x > -1\). This suggests we will use induction since the inequality must hold for any natural number \(n\).

Step 2: Base Case

Check the base case when \(n = 1\). Substitute \(n = 1\) into the inequality to get \((1 + x)^1 \geqq 1 + 1\cdot x\), which simplifies to \(1 + x \geqq 1 + x\). This is clearly true, so the base case holds.

Step 3: Inductive Step Assumption

Assume the inequality holds for \(n = k\), that is \((1 + x)^k \geqq 1 + k x\). This is our induction hypothesis.

Step 4: Inductive Step - Prove for n = k+1

We need to show that \((1+x)^{k+1} \geqq 1 + (k+1)x\). Start with the left side of the inequality using the induction hypothesis:\[(1+x)^{k+1} = (1+x)^{k} imes (1+x)\]\( ext{{Using the induction hypothesis, }} (1+x)^{k} \geqq 1 + kx\), substitute it into the expression:\[(1+x)^{k+1} \geqq (1 + kx)(1+x) = 1 + kx + x + kx^2\]Simplify the right side: \[1 + (k+1)x + kx^2\]Since \(x > -1\), we have \(kx^2 \geqq 0\). Thus:\[1 + (k+1)x + kx^2 \geqq 1 + (k+1)x\]This shows \((1+x)^{k+1} \geqq 1 + (k+1)x\), completing the inductive step.

Step 5: Conclusion

Since the base case holds and the inductive step has been verified, by mathematical induction, the inequality \((1+x)^{n} \geqq 1+n x\) holds for all natural numbers \(n\) when \(x > -1\).

Key Concepts

Inequalities

Inequalities play a crucial role in mathematics. They allow us to compare values and express relationships between different mathematical expressions. In our problem, the inequality \((1+x)^{n} \geqq 1+n x\) is the focal point. This particular inequality compares the exponential expression \((1+x)^n\) with the linear expression \(1+n x\).

It's important to understand why inequalities are useful. Here are some points:

• Inequalities help us determine ranges where certain properties hold.
• They can be used to establish limits and bounds in optimization problems.
• Understanding inequalities enhances our problem-solving skills.

In this inequality,

• \((1+x)^{n}\) represents an exponential growth.
• \(1+n x\) is linear, growing only linearly with \(n\).

If \(x > -1\), the inequality holds for all natural numbers \(n\), showing the exponential expression often grows faster than the linear one as \(n\) increases.

Natural Numbers

Natural numbers are the building blocks of our numerical system. They're the set of positive integers starting from 1, 2, 3, and so on. In mathematical induction, the property or truth we are interested in applying typically applies to natural numbers. Here's why they are significant in our exercise:

• Natural numbers provide a great starting point for induction since they follow a successive order.
• They are countable, making it easy to track logical processes from one case to the next.
• The step from one number to the next is always the same: simply add 1.

For our problem, the variable \(n\) must be a natural number. By proving the inequality for all natural numbers, we can generalize the conclusion to any number in this set, affirming that as we add 1 to \(n\), the inequality still holds due to their simple structural nature. Thus, the dependency on natural numbers ensures the logical flow in mathematical induction.

Inductive Proof

An inductive proof is a mathematical technique used to prove statements involving natural numbers. It is like climbing a ladder: first, you need to step onto the first rung (base case), then prove you can move from one rung to the next (inductive step). Let's explore the components of an inductive proof:

• Base Case: You begin by proving the statement is true for the initial value, usually \(n = 1\) or \(n = 0\).
• Inductive Hypothesis: Assume the statement is true for some arbitrary positive integer \(k\).
• Inductive Step: Prove the statement holds for \(k + 1\), relying on the truth of the inductive hypothesis.

In our example, we started by demonstrating that \((1+x)^{1} \geqq 1+1x\) is true. For the inductive step, we assumed the truth of \((1+x)^{k} \geqq 1+kx\) and showed it implies \((1+x)^{k+1} \geqq 1+(k+1)x\). This confirms that if you "step" is true for any one natural number \(k\), then it must also be true for the next integer \(k+1\), proving the statement for all natural numbers by induction.