Mathematical induction inequality

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?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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