Mathematical induction inequality

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
4 replies · 2K views
Julia Maria
Messages
4
Reaction score
0

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?
 
on Phys.org
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 [itex]n^2>n+1[/itex] is true. If we know that [itex]n^2>n+1[/itex] then is [itex]n^2+n>1[/itex] ?
 
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?