# Thread: Induction n^2 > 2n for n>2

1. ## Induction n^2 > 2n for n>2

Prove n^2 > 2n for n>2 by induction

Check n = 3
9 > 6
Say it is true for n
now check for n+1

This is where I am worried that what I'm doing isn't enough proof

(n+1)^2 > 2(n+1)

n^2+2n+1 > 2n +2

by cancelling

n^2 > 1

we know that n > 2 so n^2 is always going to be greater than 1.

Does this suffice?

2. Hi

The idea behind what you use is a non-inductive proof, which is:

Let $\displaystyle n>2$ be an integer, $\displaystyle n^2>2n \Leftrightarrow n>2$ (the equivalence is true because n positive integer). Since we assumed $\displaystyle n>2,$ we have what we want.

In a usual proof by induction, you assume your induction hypothesis (case $\displaystyle n$) and try to show the statement for $\displaystyle n+1,$ using the case $\displaystyle n.$

So for your proof, writing $\displaystyle (n+1)^2=n^2+2n+1$ is a good thing. What you have to do now is to obtain $\displaystyle n^2+2n+1>2(n+1),$ using the induction hypothesis: $\displaystyle n^2>2n$

3. Ok how bout this

Step 1 as above

Step 2 (I need to prove (n+1)^2 > 2(n+1)

(n+1)^2=n^2 +2n +1
>2n + 2n +1 (Using induction)
= 4n +1
= 2(n+1) +2n -1 (2n-1) is always positive for n>2
>2(n+1)

Should i be using k instead?

4. What you've done is correct, no need to choose another letter.