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Proof by induction for inequalities

  • Thread starter Easty
  • Start date
  • #1
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Homework Statement



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

Prove by induction or otherwise that Xn-1 > 0 for all n [tex]\in[/tex] 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 dont think it gets me anywhere. I'd really appreciate any guidance on where to start.

Thanks
 

Answers and Replies

  • #2
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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?
 
  • #3
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.
 
  • #4
19
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Thanks alot for the help. Its such a simple solution, no wonder i didnt get it
 

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