Thread: 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?

Hi

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

Let be an integer, (the equivalence is true because n positive integer). Since we assumed we have what we want.


In a usual proof by induction, you assume your induction hypothesis (case ) and try to show the statement for using the case

So for your proof, writing is a good thing. What you have to do now is to obtain using the induction hypothesis:

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?

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