Problem 32 - College Algebra

Question

In Exercises 25-34, use mathematical induction to prove that each statement is true for every positive integer \(n.\) If \(0 < x < 1,\) then \(0 < x^{n} < 1\).

Step-by-Step Solution

Answer

The inequality \(0 < x^{n} < 1\) is true for every positive integer \(n\), assuming that \(0 < x < 1\).

Step 1: Base Case

The first step in a mathematical induction proof is to verify the given statement for the lowest possible value, which here is \(n=1\). If you plug in \(1\) for \(n\) in \(x^{n}\), you obtain \(x^1\). We know that if \(0 < x < 1\), then \((x^1) = x\) and thus \(0 < x < 1\). Thereby, the base case is proved.

Step 2: Assumption Step

For the induction step, we assume the statement is true for \(n=k\). Namely, we assume that if \(0 < x < 1\), then \(0 < x^{k} < 1 \). This is our inductive hypothesis.

Step 3: Inductive Step

We want to show that the statement is then true for \(n = k+1\). Meaning if \(0 < x < 1\), we have to prove that \(0 < x^{k+1} < 1\). Consider: \(0 < x^{k+1} = x^k * x < 1* x = x\). Because \(0 < x < 1\) (given), and \(0 < x^k < 1\) (our inductive hypothesis), we conclude that \(0 < x^{k+1} < 1\). This completes the inductive step.

Key Concepts

Base Case

The concept of the base case is the starting point in a mathematical induction proof. It serves as the foundation upon which subsequent steps are built. In this exercise, the base case involves verifying the statement for the smallest positive integer, which is usually \(n=1\).
When we substitute \(n=1\) into the expression \(x^n\), it simplifies to \(x^1 = x\). Given the initial condition \(0 < x < 1\), it follows naturally that \(x\) remains between 0 and 1, thus \(0 < x^1 < 1\).
This verification means the statement holds true for \(n=1\), establishing a solid fundamental step, and verifying our hypothesis for the simplest scenario, often termed as the 'base case'. From here, the journey of induction can begin.

Inductive Hypothesis

The inductive hypothesis is a crucial middle step that forms the bridge between the base case and the inductive step. It involves the assumption that the statement is true for some arbitrary positive integer, usually denoted as \(n=k\).
In this case, our statement is that if \(0 < x < 1\), then \(0 < x^k < 1\). While it might feel like taking a leap of faith, this assumption does not mean guessing. Instead, it helps to establish a framework from which we can logically extend the truth of the statement to the next integer, \(n = k+1\).
This strategic assumption is vital as it sets the stage for proving the further progression to the next number by ensuring continuity in our progression of truths.

Inductive Step

The inductive step is the climax of the mathematical induction process where we extend the validity of our statement from \(n=k\) to \(n=k+1\). This step tests the strength of our inductive hypothesis.
In the context of this exercise, we need to show that if \(0 < x < 1\), then \(0 < x^{k+1} < 1\). We start by considering that \(x^{k+1} = x^k \times x\).
Since our inductive hypothesis ensures \(0 < x^k < 1\), and knowing \(0 < x < 1\) from the problem's given condition, multiplying \(x^k\) by \(x\) keeps the product within the same bounds: \(0 < x^{k+1} < 1\).

• This multiplication exhibits a property of numbers between 0 and 1: the multiplication of such numbers is commutative and results in a product that also stays between 0 and 1.

This elegant completion of the inductive step upholds and extends the truth established in the base case by successfully leveraging the inductive hypothesis. Thus, the mathematical induction is complete, affirming that the statement holds for all positive integers \(n\).