1. ## Prooving inequalities

I'm supposed to proove the proposition that for all integers k>=2, that k^2<k^3. The book says you can use induction or another method. I can't seem to get either fully proved. Any help would be great. Thanks!!

2. Originally Posted by nataliemarie
I'm supposed to proove the proposition that for all integers k>=2, that k^2<k^3. The book says you can use induction or another method. I can't seem to get either fully proved. Any help would be great. Thanks!!
$k^2 is clearly equivalent to $k^3-k^2=k^2(k-1)>0$.
Can you prove that if $k\ge 2?$

3. Ok... well I had started doing it, but I'm not sure.

k>=2 so (k-2) > 0

I can multiply it by k^2 because k^2 >0

k^3 - 2k^2 > 0
so
k^3 > 2k^2
but I'm missing something

4. Originally Posted by nataliemarie
Ok... well I had started doing it, but I'm not sure.

k>=2 so (k-2) > 0

I can multiply it by k^2 because k^2 >0

k^3 - 2k^2 > 0
so
k^3 > 2k^2
but I'm missing something
If I were you, I would say if $k=2$ then $2^2<2^3.$

If $k>2$ then $k-1>0$ so $k^2(k-1)>0$
So $k^3>k^2$.

5. This is actually easier without induction

Basically, $1 since $n>2$ and we "know" that for $p>0$, if $m then $pm

Also, $n^2>0$ since $n>2$

So letting $n^2=p$ and $1, then $n^2\cdot 1

Now of course, you might not acutally "know" the proposition that I referenced if you haven't proven it yet

6. Thank you both very much. I can't believe I missed the fact that k >1.

7. You may also do it this way:

$k^2 is equivalent to $\frac{k^2}{k^3} < 1 \Rightarrow \frac{1}{k}<1$ (we can do that since $k\neq 0$)

Now, $\frac{1}{k}<1$ is obviously true for any $k>1$...