Proving x_n+1 > x_n with Mathematical Induction | Real Number Sequence

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Homework Help Overview

The discussion revolves around proving that the sequence defined by x1 = 1 and xn+1 = √(1 + 2xn) satisfies the inequality xn+1 > xn for all n ≥ 1 using mathematical induction.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the initial conditions and the implications of the sequence definition, questioning the necessity of the proof given the apparent obviousness of the inequality.
  • Some participants suggest performing algebraic manipulations to demonstrate the inequality, while others express uncertainty about the validity of their approaches.
  • Questions arise regarding the role of the initial value x1 = 1 in the proof and how it influences subsequent terms in the sequence.

Discussion Status

The discussion is ongoing, with participants sharing their thoughts on the proof structure and exploring various algebraic methods. There is a recognition of the need to establish the base case and the induction step, but no consensus has been reached on the best approach yet.

Contextual Notes

Participants express concerns about the limitations of their reasoning and the clarity of the sequence's properties, particularly in relation to the initial term x1 = 1 and its implications for proving the inequality.

AerospaceEng
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Homework Statement



Consider the sequence of real numbers x1, x2, x3,... defi ned by the relations x1 = 1

and xn+1 =\sqrt{1 + 2x<sub>n</sub>}

1. Use mathematical induction to show that xn+1 > xn
for all n \geq 1.

The Attempt at a Solution



I'm a bit thrown off by this question because it seems very obvious that it would be greater. if x1=1 is it safe to assume that xn=n? and if so I feel like this proof is stupid cause it shows right in front of you that its greater. any help is great thanks.
 
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AerospaceEng said:

Homework Statement



Consider the sequence of real numbers x1, x2, x3,... defi ned by the relations x1 = 1

and xn+1 =\sqrt{1 + 2x_n}

1. Use mathematical induction to show that xn+1 > xn
for all n \geq 1.

The Attempt at a Solution



I'm a bit thrown off by this question because it seems very obvious that it would be greater. if x1=1 is it safe to assume that xn=n? and if so I feel like this proof is stupid cause it shows right in front of you that its greater. any help is great thanks.
The obviousness (or not) is irrelevant to what you need to do, which is to prove this statement by induction.

You are given x1 = 1. What is the value for x2? If x2 > x1, then use that for your base case.

Next, assume that xk > xk - 1 (the induction hypothesis), and use that assumption to show that xk + 1 > xk. If you can do this, you will have proved that the statement is true for all n >= 1.
 
The proof is not stupid! Lol engineers!

Perform the substraction x_{n+1}- x_n then show the substraction is > 0. Hint: rationalise and use the fact that a^2 >0 for an real number a.
 
I'm I allowed to say that xn+1 is the same thing as xn + x1 cause i feel I can't and if I can't I feel so limited.
 
AerospaceEng said:
I'm I allowed to say that xn+1 is the same thing as xn + x1 cause i feel I can't and if I can't I feel so limited.

But that's not true.

xn+1 is clearly shown to be \sqrt{1+2x_n}.

So...

x_{n+1}-x_n = \sqrt{1+2x_n} - x_n

So you need to prove that the above equation is always greater than 0, given x1=1.
 
okay so I think i see where this is going. what I've done so far is I've taken the RHS and multiplied it by itself just with a positive between the terms, to give me 1+2xk-xk2

But i feel like I've done something wrong because i haven't used x1=1 as of yet..
 
Last edited:
AerospaceEng said:
okay so I think i see where this is going. what I've done so far is I've taken the LHS and multiplied it by itself just with a positive between the terms, to give me 1+2xk+xk2
and i can factor this to give me (x+1)2
Is that good enough to show that it's always greater than 0. and then ill just plug this mini proof into my main proof.

But i feel like I've done something wrong because i haven't used x1=1 as of yet..

Considering that x_k itself is always greater than 0, that should be enough. But I'll wait for a second opinion on that.
 
now if i work with the left hand side I get xk+12-xk2 and i can cancel the -xk2 on both sides to give me:

xk+12=2k+1

and i feel I am stuck again and where does the x1=1 come into play?
 
the x_1 comes into play when you show that x_2>x_1 which is the initial step when you should have done first.
 
  • #10
AerospaceEng said:
now if i work with the left hand side I get xk+12-xk2 and i can cancel the -xk2 on both sides to give me:

xk+12=2k+1

and i feel I am stuck again and where does the x1=1 come into play?
See post #2.
 

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