# Thread: Proof of mathematical induction inequality

1. ## Proof of mathematical induction inequality

Hi,

Im having trouble trying to prove the following:

n^2 > 2n; n 3

2. Make first an equation out of it and try to understand how these numbers relate

3. ## Or you can, do this

In order to prove it by induction u have to prove that it is true for n=3
and then assume that it is true for some n and that it then follows for n+1
So
Step one:
Is it true for 3?
3^2>2*3......True!
Good not we assume that it is true for n and prove that it then follows for n+1.

$n^2>2*n$
(n+1)^2>2*(n+1)
n^2+2n+1>2n+1
and almost done now =).
n^2>2n by our first assumtion.
so we can subtract that from the inequality.
then we are left with 2n+1>1
and since n=positive number clearly this is true...
Hope I was clear enough can you take it from here
?
Regards Henry

4. hi
let $S=\left \{ n\in \mathbb{N}\mid n\geq 3,\text{and},n^2> 2n \right \}$
Clearly $3\in S$ since $9> 6$,now assume $m\in S$ to prove $m+1\in S$ we must prove $(m+1)^2> 2(m+1)$...
i'll let you do that .

5. Thanks for the replies.

when I tried to do it I got as far as:

$
k^2 + 2k + 1 > 2k + 2
$

I wasn't aware I could subtract $k^2 > 2k$ from the inequality.

6. Originally Posted by blackhug
Thanks for the replies.

when I tried to do it I got as far as:

$
k^2 + 2k + 1 > 2k + 2
$

I wasn't aware I could subtract $k^2 > 2k$ from the inequality.
In that case you can simply write

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

2k+1 is common to both sides so it is redundant,
hence we can eliminate it.

$k^2>1$ ?

This is true for k>1