Hi,
Im having trouble trying to prove the following:
n^2 > 2n; n ≥ 3
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
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 .