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Chapter 7: Problem 24

Suppose the real numbers \(r\) and \(s\) satisfy \(0

Short Answer

Expert verified
By using the method of induction, it can be shown that the sequence \(\left(a_{0}, a_{1}, a_{2}, \ldots\right)\) of real numbers fulfills the inequalities \(r^{n} a_{0}<a_{n}<s^{n} a_{0}\) for any natural number \(n\).

Step by step solution

01

Understand the problem

The problem is about a sequence of real numbers \(\left(a_{0}, a_{1}, a_{2}, \ldots\right)\) which satisfy the inequality \(r a_{n-1}<a_{n}<s a_{n-1}\) for \(n \geq 1\), where \(r\) and \(s\) are real numbers that satisfy \(0<r<s\). The task is to prove for any natural number \(n\), that \(r^{n} a_{0}<a_{n}<s^{n} a_{0}\).
02

Base case for the Induction

To prove the given statement, we will use induction. First let's verify for the base case where \(n = 1\). By substituting \(n = 1\) in the statement, \(r a_{0}<a_{1}<s a_{0}\). This holds true as its given in the problem that \(r a_{n-1}<a_{n}<s a_{n-1}\). Thus, our base case is true.
03

Inductive Hypothesis

Next, lets assume that the given statement holds true for \(n = k\). This implies that \(r^{k} a_{0}<a_{k}<s^{k} a_{0}\).
04

Inductive Step

Now proved for \(n = k + 1\). By replacing \(n\) with \(k+1\) in the the initial inequality, we get \(r a_{k}<a_{k+1}<s a_{k}\). Multiply the inequality \(r^{k} a_{0}<a_{k}<s^{k} a_{0}\) by \(r\) and \(s\) respectively, we get \(r^{k+1} a_{0}<r a_{k}<s^{k+1} a_{0}\) and \(r^{k+1} a_{0}<s a_{k}<s^{k+1} a_{0}\). Combining these inequalities, we get \(r^{k+1} a_{0}<a_{k+1}<s^{k+1} a_{0}\). Hence our inductive step is proved.
05

Conclusion of the proof

Hence by principle of mathematical induction, our statement holds true for all \(n=1,2,3,....\)

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Most popular questions from this chapter

Suppose \(A\) is an \(n \times n\) matrix and \(r\) is a real number. Find a simple formula for \(\operatorname{det}(r A)\) in terms of \(r\) and \(\operatorname{det} A\). Prove your conjecture.

Prove that the intersection of any collection of inductive subsets of \(\mathbb{R}\) is an inductive set.

Prove Theorem 7.16, part f: \(\mathbf{v} \times \mathbf{v}=\mathbf{0}\).

Prove Theorem 7.16, part g: \(\mathbf{v} \cdot(\mathbf{w} \times \mathbf{x})=(\mathbf{v} \times \mathbf{w}) \cdot \mathbf{x}=\operatorname{det}\left[\begin{array}{lll}v_{1} & v_{2} & v_{3} \\\ w_{1} & w_{2} & w_{3} \\ x_{1} & x_{2} & x_{3}\end{array}\right]\).

Prove Theorem 7.17, part a: \(\mathbf{v} \cdot(\mathbf{v} \times \mathbf{w})=0\).
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