# Don't know if I got this right. Prove n^2>n+1

1. Apr 23, 2008

### torquerotates

1. The problem statement, all variables and given/known data
The principal of mathematical induction can be extended as follows. A list P(m),P(m+1)... of propositions is true provided 1)P(m) is true, 2) P(n+1) is true whenever P(n) is true and n>(or =) m

I have to use the above to prove that n^2>n+1 for n>(or equal to) 2

2. Relevant equations
n^2>n+1 for n>(or equal to) 2

3. The attempt at a solution

so I said m=n+1

Then since I assume that the original statement implies that I hold m constant and increase n by 1

Inductive step (n+1)^2>(n+1)
=> n+1>1 True b/c {1,2,3,...n}=N

2. Apr 23, 2008

### rock.freak667

Assume true for n=k

$k^2>k+1 for k \geq 2$

+(2k+1)
$k^2+2k+1>k+1+2k+1$

and $k^2+2k+1=(k+1)^2$ which is what you need on the left side. Deal with the right side now.

3. Apr 23, 2008

### torquerotates

Right side: (k+1+2k+1)=(2k+k+2)>(k+2)

Right?

4. Apr 23, 2008

### rock.freak667

2k+k+2>k+2 => 3k>k which is true so that 2k+k+2>k+2 is true

and now you have

(k+1)^2>2k+k+2>k+2

5. Apr 23, 2008

### torquerotates

Why do you have to show that many steps?

6. Apr 23, 2008

### rock.freak667

Because that is to show how P(k) => P(k+1) instead of putting n=k+1 in the formula and showing it is true.