Mathematical induction inequality

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Homework Help Overview

The discussion revolves around proving the inequality n^2 > n + 1 for n = 2, 3, 4 using mathematical induction. Participants are exploring the validity of the inequality and the steps involved in the induction process.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to establish the base case and induction step for the inequality. Questions arise regarding the validity of the initial assumptions and the logical flow from one step to the next. There is also discussion about the choice of variable in the induction hypothesis.

Discussion Status

The discussion is ongoing, with participants providing hints and guidance on how to apply the induction hypothesis. There is recognition of the need to clarify assumptions and the implications of the inequality being true for certain values of n.

Contextual Notes

Some participants note that the inequality is supposed to hold for n > 1, and there is a focus on ensuring that the induction step is correctly formulated without assuming what needs to be proven.

Julia Maria
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Homework Statement



Prove; n^2 > n+1 for n = 2,3,4 by Induction

Homework Equations




The Attempt at a Solution



p(n)= P(2) 2^2> 2+1 --> 4>3

Induction step:
P(n+1): (n+1)^2 > (n+1) +1
(n+1)^2> n+2
n^2 + 2n + 1 > n+2 | -n
n^2 +n + 1> 2 | -1
n^2 +n > 1

Is this correct, and how do I go from here?
 
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Hint: for what n is the inequality suppose to be valid?
 
For n>1. But I Still struggle to get further..
 
You're now supposed to apply your assumption that n^2>n+1 is true. If we know that n^2>n+1 then is n^2+n>1 ?
 
Julia Maria said:

Homework Statement



Prove; n^2 > n+1 for n = 2,3,4 by Induction

Homework Equations




The Attempt at a Solution



p(n)= P(2) 2^2> 2+1 --> 4>3
Better said "for n= 2, n^2= 2^2= 4> 3= 2+ 1"

Induction step:
You have to first say "assume that for some n, n^2> n+1".
(Personally I prefer to use another letter, k, say, so as not to confuse it with the general n.

P(n+1): (n+1)^2 > (n+1) +1
No, You are asserting what you want to prove.
Instead look at just the left side: (n+1)^2= n^2+ 2n+ 1.
By the "induction hypothesis", n^2> n+ 1 so (n+1)^2> (n+1)+ 2n+ 1= 3n+ 2> n+ 2= (n+1)+1

(n+1)^2> n+2
n^2 + 2n + 1 > n+2 | -n
n^2 +n + 1> 2 | -1
n^2 +n > 1

Is this correct, and how do I go from here?
 

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