Can You Prove \(x_n < x_{n+1}\) Using Induction?

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Discussion Overview

The discussion revolves around proving the inequality \(x_n < x_{n+1}\) for the sequence defined by \(x_1=\sqrt{2}\) and \(x_{n+1}=\sqrt{2+x_n}\) using mathematical induction. Participants explore different approaches to this problem, including the necessity of induction and the properties of the sequence.

Discussion Character

  • Debate/contested

Main Points Raised

  • One participant suggests that induction may not be necessary and proposes showing that the expression under the radical for \(x_{n+1}\) is greater than \(x_n\) based on the positivity of \(x_n\).
  • Another participant argues that induction is indeed required, providing a counterexample to illustrate that without induction, the inequality may not hold.
  • This participant outlines a potential inductive proof structure, starting with showing \(x_1 < x_2\), assuming \(x_k < x_{k+1}\), and then proving \(x_{k+1} < x_{k+2}\) by expressing them in terms of \(x_k\) and leveraging the increasing nature of the function defined by \(f(a) = a^{1/2}\).

Areas of Agreement / Disagreement

Participants express differing views on the necessity of induction for this proof. While some believe induction is essential, others propose alternative methods that do not rely on it. The discussion remains unresolved regarding the best approach to take.

Contextual Notes

Some assumptions about the properties of the sequence and the function involved are not fully explored, and the discussion does not clarify the implications of the counterexample provided.

MrBailey
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Hi, all.
I'm working on some proof by induction problems. While I understand the concept, this one threw me for a loop.
Let [tex]x_1=\sqrt{2}[/tex] and [tex]x_{n+1}=\sqrt{2+x_n}[/tex]
Show that [tex]x_n < x_{n+1}[/tex]
I'd greatly appreciate help with this.
Thanks,
bailey
 
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sure you need to use induction? i would show that the stuff under the radical for [tex]x_{n+1}> x_n[/tex] we know this because [tex]x_n>0[/tex]. and 2 plus some other positive number will always be greater than two, and therefore the sq rt of that sum will be greater eh?
 
Uh, you certainly need induction. If xn = 98, then xn+1 = (2 + 98)1/2 = 1001/2 = 10 < 98 = xn.

Show that x1 < x2
Assume that xk < xk+1
Use this to prove that xk+1 < xk+2
Write out xk+1 and xk+2 in terms of xk. Then there xk+1 < xk+2 will follow immediately from xk < xk+1 as long as you know that the function f defined by f(a) = a1/2 is an increasing function.
 
thanks...much clearer now

Bailey
 

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