Hi,

Im having trouble trying to prove the following:

n^2 > 2n; n ≥ 3

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- Mar 15th 2010, 07:30 AMblackhugProof of mathematical induction inequality
Hi,

Im having trouble trying to prove the following:

n^2 > 2n; n ≥ 3 - Mar 15th 2010, 07:55 AMMarNie
Make first an equation out of it and try to understand how these numbers relate

- Mar 15th 2010, 08:43 AMHenryt999Or 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.

$\displaystyle n^2>2*n$

(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 - Mar 15th 2010, 08:54 AMRaoh
hi :)

let $\displaystyle S=\left \{ n\in \mathbb{N}\mid n\geq 3,\text{and},n^2> 2n \right \}$

Clearly $\displaystyle 3\in S$ since $\displaystyle 9> 6$,now assume $\displaystyle m\in S$ to prove $\displaystyle m+1\in S$ we must prove $\displaystyle (m+1)^2> 2(m+1)$...

i'll let you do that :). - Mar 15th 2010, 01:10 PMblackhug
Thanks for the replies.

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

$\displaystyle

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

$

I wasn't aware I could subtract $\displaystyle k^2 > 2k$ from the inequality. - Mar 15th 2010, 03:34 PMArchie Meade