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!!
$\displaystyle k^2<k^3$ is clearly equivalent to $\displaystyle k^3-k^2=k^2(k-1)>0$.
Can you prove that if $\displaystyle k\ge 2?$

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 $\displaystyle k=2$ then $\displaystyle 2^2<2^3.$

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

5. This is actually easier without induction

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

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

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

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:

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

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