1. ## Induction: Look correct?

5.2 Prove by induction on $m$ that $m^{3} \leq 2^{m}$ for $m \geq 10$.
Proof: We use induction on $m$. Base case: For $n = 10$, $m^{3} = 1000$ and $2^{10} = 1024$ and so $m^{3} \leq 2^{m}$. Inductive step: Suppose now as inductive hypothesis that $k^{3} \leq 2^{k}$ for some $k \geq 10$. Then $2^{k+1} = 2 \times 2^{k} \geq 2k^{3}$ (by induction hypothesis). So we will have proved that $2^{k+1} \geq (k+1)^{3}$ if we can prove that $2k^{3} \geq (k+1)^{3}$. But $2k^{3} \geq (k+1)^{3} \Leftrightarrow 2k^{3} \geq k^{3} + 3k^{2} + 3k + 1 \Leftrightarrow k^{3} \geq 3k^{2} + 3k + 1$, and since $k \geq 10$, $k^{3} \geq 10k^{2} \geq 3k^{2} + 3k + 2 \geq 3k^{2} + 3k + 1$ so that $k^{3} \geq 3k^{2} +3k+1$. Hence $2k^{3} \geq (k+1)^{3}$ and so we have deduced that $(k+1)^{3} \leq 2^{k+1}$ as required to complete the inductive step. Conclusion: Hence, by induction, $m^{3} \leq 2^{m}$ for all $m \geq 10$. $\square$

2. Originally Posted by tukeywilliams
$m^{3} \leq 2^{m}$.
You wish to show this for $m\geq 10$.

Say it is true for $m=k$, i.e. $k^3 \leq 2^k$.

Then, multiply both sides by 2,
$2k^3 \leq 2 \cdot 2^m = 2^{m+1}$
Now if you can show that,
$(k+1)^3 \leq 2k^3$

3. didnt I do that. And is that correct?

4. Originally Posted by tukeywilliams
didnt I do that. And is that correct?
Yes it is correct.

5. Originally Posted by tukeywilliams
5.2 Prove by induction on $m$ that $m^{3} \leq 2^{m}$ for $m \geq 10$.
Proof: We use induction on $m$. Base case: For $n = 10$, $m^{3} = 1000$ and $2^{10} = 1024$ and so $m^{3} \leq 2^{m}$. Inductive step: Suppose now as inductive hypothesis that $k^{3} \leq 2^{k}$ for some $k \geq 10$. Then $2^{k+1} = 2 \times 2^{k} \geq 2k^{3}$ (by induction hypothesis). So we will have proved that $2^{k+1} \geq (k+1)^{3}$ if we can prove that $2k^{3} \geq (k+1)^{3}$. But $2k^{3} \geq (k+1)^{3} \Leftrightarrow 2k^{3} \geq k^{3} + 3k^{2} + 3k + 1 \Leftrightarrow k^{3} \geq 3k^{2} + 3k + 1$, and since $k \geq 10$, $k^{3} \geq 10k^{2} \geq 3k^{2} + 3k + 2 \geq 3k^{2} + 3k + 1$ so that $k^{3} \geq 3k^{2} +3k+1$. Hence $2k^{3} \geq (k+1)^{3}$ and so we have deduced that $(k+1)^{3} \leq 2^{k+1}$ as required to complete the inductive step. Conclusion: Hence, by induction, $m^{3} \leq 2^{m}$ for all $m \geq 10$. $\square$
Yes, you are correct. It's just that as written, your proof is kind of hard to read. Try formatting it better so that the reader does not get lost in the sea of words and inequalities. I believe that's why TPH's first post was like that. He probably read through your post once and realized that he had to read it again to really get what you were saying--but thought it easier to just tell you what to do than to read through it again. Your second comment kind of forced him to read through it carefully to see that you did, in fact, do what he suggested in the first place.