91Ó°ÊÓ

Sometimes we cannot use mathematical induction to prove a result we believe to be true, but we can use mathematical induction to prove a stronger result. Because the inductive hypothesis of the stronger result provides more to work with, this process is called inductive loading. We use inductive loading in Exercise \(74-76\) . Suppose that we want to prove that $$ \frac{1}{2} \cdot \frac{3}{4} \cdots \frac{2 n-1}{2 n}<\frac{1}{\sqrt{3 n}} $$ for all positive integers \(n .\) a) Show that if we try to prove this inequality using mathematical induction, the basis step works, but the inductive step fails. b) Show that mathematical induction can be used to prove the stronger inequality $$ \frac{1}{2} \cdot \frac{3}{4} \dots \frac{2 n-1}{2 n}<\frac{1}{\sqrt{3 n+1}} $$ for all integers \(n\) greater than \(1,\) which, together with a verification for the case where \(n=1\) , establishes the weaker inequality we originally tried to prove using mathematical induction.

Step by step solution

01

Basis Step for the Weaker Inequality

Verify the basis step for the weaker inequality \[ \frac{1}{2} \cdot \frac{3}{4} \cdots \frac{2 \cdot 1 - 1}{2 \cdot 1} < \frac{1}{\sqrt{3 \cdot 1}}. \] For \(n = 1\), this simplifies to \( \frac{1}{2} < \frac{1}{\sqrt{3}}. \) This is true since \(0.5 < 0.577.\)

02

Inductive Step for the Weaker Inequality

Assume \( \frac{1}{2} \cdot \frac{3}{4} \cdots \frac{2k-1}{2k} < \frac{1}{\sqrt{3k}} \) holds for some positive integer \(k\). We need to show that \( \frac{1}{2} \cdot \frac{3}{4} \cdots \frac{2k+1}{2(k+1)} < \frac{1}{\sqrt{3(k+1)}}. \) However, this step fails because it does not provide a valid pathway to the proof.

03

Basis Step for the Stronger Inequality

Verify the basis step for the stronger inequality \[ \frac{1}{2} \cdot \frac{3}{4} \cdots \frac{2 \cdot 1 - 1}{2 \cdot 1} < \frac{1}{\sqrt{3 \cdot 1 + 1}}. \] For \(n = 1\), this simplifies to \( \frac{1}{2} < \frac{1}{2}. \) This is true since \(0.5 < 0.707.\)

04

Inductive Step for the Stronger Inequality

Assume \( \frac{1}{2} \cdot \frac{3}{4} \cdots \frac{2k-1}{2k} < \frac{1}{\sqrt{3k + 1}} \) holds for some positive integer \(k\). We need to show that \( \frac{1}{2} \cdot \frac{3}{4} \cdots \frac{2k+1}{2(k+1)} < \frac{1}{\sqrt{3(k+1) + 1}}. \) Using the inductive hypothesis: \[ \frac{1}{2} \cdot \frac{3}{4} \cdots \frac{2k-1}{2k} < \frac{1}{\sqrt{3k + 1}}, \] multiply both sides by \( \frac{2k+1}{2(k+1)} \): \[ \frac{1}{2} \cdot \frac{3}{4} \cdots \frac{2k+1}{2(k+1)} < \frac{2k+1}{2(k+1)} \cdot \frac{1}{\sqrt{3k + 1}}. \] Simplify: \[ \frac{2k+1}{2(k+1)} < \sqrt{\frac{3k+1}{3k+4}}. \] Since \( \frac{2k+1}{2(k+1)} < 1 \) and \( \sqrt{\frac{3k+1}{3k+4}} < 1 \), \( \frac{2k+1}{2(k+1)} < \sqrt{\frac{3k+1}{3k+4}} \) holds, thus proving the inductive step.

05

Concluding the Proof

Since we have proven the basis step and the inductive step for the stronger inequality, \( \frac{1}{2} \cdot \frac{3}{4} \cdots \frac{2n-1}{2n} < \frac{1}{\sqrt{3n+1}} \) for all \(n > 1\), and verified the case for \(n = 1\), the weaker inequality \( \frac{1}{2} \cdot \frac{3}{4} \cdots \frac{2n-1}{2n} < \frac{1}{\sqrt{3n}} \) holds for all positive integers \(n\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Inductive Loading

Inductive loading is a technique used to prove a mathematical statement by demonstrating a stronger version of the original statement. This method enhances the proof by adding more assumptions or constraints, making it easier to establish the desired result. In the exercise provided, inductive loading is used to prove a stronger inequality which indirectly validates the original weaker inequality. Think of it as generating more 'tools' to work with in your proof, ensuring you can tackle the problem more effectively.

Proof by Induction

Proof by induction is a powerful method often used in mathematics to prove statements involving natural numbers. It's like a domino effect. If you can show the first statement (basis step) is true, and then prove that if any one statement is true, the next statement is also true (inductive step), you can conclude that the statement holds for all natural numbers. The main components are:

• Basis Step: Show that the statement is true for the initial value, often n = 1.
• Inductive Step: Assume the statement is true for n = k (inductive hypothesis), and then prove it’s true for n = k + 1.

This method provides a logical pathway to cover infinite cases using a finite process.

Inequality Proofs

Inequality proofs are used to show that a particular mathematical inequality holds true for certain values. In the given exercise, inequality proofs are employed to demonstrate that: \[ \frac{1}{2} \times \frac{3}{4} \times \frac{2n-1}{2n} < \frac{1}{\frac{3n}} \]
Using induction makes this proof easier by tackling a stronger statement initially: \[ \frac{1}{2} \times \frac{3}{4} \times \frac{2n-1}{2n} < \frac{1}{\frac{3n+1}} \]
When proving inequalities, be sure to clearly identify the assumptions and transformations at each step, ensuring each inequality is logically justified.

Basis Step

The basis step is the initial step in a proof by induction where you verify that the statement holds for the smallest value in the set, typically n = 1. For the weaker inequality: \[ \frac{1}{2} < \frac{1}{\frac{3}} \]
This simplifies to: \[ 0.5 < 0.577 \]
Which is true. Similarly, for the stronger inequality wherein: \[ \frac{1}{2} < \frac{1}{\frac{3}+1} \]
This simplifies to: \[ 0.5 < 0.707 \] and this is also true. Verifying the basis step is crucial as it acts as the foundation for your inductive proof.

Inductive Step

The inductive step is where you assume that the statement is true for a particular case (n = k), called the inductive hypothesis, and then prove it's true for the next case (n = k + 1).
For the stronger inequality, assume: \[ \frac{1}{2} \times \frac{3}{4} \times \frac{2k-1}{2k} < \frac{1}{\frac{3k+1}} \]
To prove for k + 1:
\[ \frac{1}{2} \times \frac{3}{4} \times \frac{2k+1}{2(k+1)} < \frac{1}{\frac{3(k+1) + 1}} \]
The process involves working through logical steps and algebraic manipulation. Successfully demonstrating the truth of n = k + 1 based on the assumption that n = k completes the inductive process, proving the statement for all natural numbers.