Proof by induction for inequalities

Click For Summary
SUMMARY

The discussion focuses on proving by induction that the sequence defined by X1=3 and Xn+1=(6Xn+1)/(2Xn+5) satisfies the inequality Xn-1 > 0 for all n ∈ N. The initial cases for n=1 and n=2 are confirmed to be true, leading to the assumption that Xk > 1 holds for an arbitrary k. The inductive step demonstrates that if Xk > 1, then Xk+1 > 1 can be derived through algebraic manipulation, confirming the proposition for all natural numbers n.

PREREQUISITES
  • Understanding of mathematical induction
  • Familiarity with sequences and recursive definitions
  • Basic algebraic manipulation skills
  • Knowledge of inequalities in mathematics
NEXT STEPS
  • Study mathematical induction techniques in detail
  • Explore recursive sequences and their properties
  • Practice algebraic manipulation involving inequalities
  • Learn about convergence and divergence of sequences
USEFUL FOR

Students in mathematics, particularly those studying sequences, inequalities, and mathematical induction. This discussion is beneficial for anyone looking to strengthen their proof-writing skills in mathematical contexts.

Easty
Messages
19
Reaction score
0

Homework Statement



A sequence (Xn) is defined by X1=3 and Xn+1= (6Xn+1)/(2Xn+5) for all n\in N.

Prove by induction or otherwise that Xn-1 > 0 for all n \in N.


Homework Equations





The Attempt at a Solution



I'm not sure with what to do when dealing with inequalities in an induction proof. Initial i tried subing in the recursion formula when attempting the inductive step but i don't think it gets me anywhere. I'd really appreciate any guidance on where to start.

Thanks
 
Physics news on Phys.org
The proposition is clearly true for n = 1 and n = 2, so suppose it's true for n = k. I.e., that xk > 1.

For the induction step, you have to show that xk + 1 > 1.

xk + 1 = (6xk + 1)/(2xk + 5)

Carry out the division to get something that you can show is greater than 1. Is that enough of a start?
 
It is easier if you write it as x_n > 1 instead of x_n - 1 >0.

Suppose x_n > 1 [Show that x_(n+1) > 1]

Mult both sides by 4.
Add 2x to both sides.
Add 1 to both sides.
Divide both sides by the right side quantity (which is > 0, why?)

And you should see x_(n+1) > 1.
 
Thanks a lot for the help. Its such a simple solution, no wonder i didnt get it
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
5
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 13 ·
Replies
13
Views
2K
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K