Extremely Simplistic Proof Check

[Hotsuma]

Doing this simplistic proof for one of my computer math classes. I've already taken Abstract Algebra and I'm having trouble with this lol. Actually, I just need someone to verify this is correct for me. It seems far too simple.

Homework Statement

Prove the following $$n^3 > n^2$$.

Homework Equations

None of importance...

The Attempt at a Solution

Well, I really just think it is something as simple as:

$$\frac{n^3}{n^2}>\frac{n^2}{n^2} \Rightarrow n > 1$$

...unless I'm really missing something

 

[lanedance]

could be wrong i think you might have used the original theorem

as
$$\frac{1}{n^2} = \frac{1}{n^2}$$
then you effectively multiply the expression by
$$n^2 > n^3$$

however it only a slight tweak to change the proof, start with the assumption:
1<n
then assuming you could use the following property:
if a<b and c>0, then a.c < b.c

otherwise this could be done by induction no worries

 

[Hotsuma]

Oh right, lol, I started with the wrong assumption! Yeah, so, is it really that simple? He he, awesome.

 

[lanedance]

if you can assume the property

i think induction would only need to assume 1<n

 

[Landau]

It would be convenient to explain what kind of number n is. A natural number greater than 1?

 

[statdad]

Induction is possible but not needed: if $$n > 1$$ (which it must be for the desired inequality to be true)
simply look at

$$n^3 - n^2$$

think how you show this difference is > 0.