Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

enter image description here

My question is regarding the question posed at the end of the proof.

My answer is that the result does not hold for all $m \ge 7$ because when $m=7$, the result is $343 \le 128$, which is false.

Is there another way to answer this sort of question besides directly trying different values until you get a false result?


Another question I have regarding this solution is:

In the proof for the inductive step, we start by assuming $k \ge 10$. But along the way, the author mentions $k \ge 1$ and $k \ge 7$ to justify the inequality.

Why do we bother to do this instead of just sticking with $k \ge 10$?

share|cite|improve this question
Your argument is correct. Having proved that : "if $P(k)$ holds with $k \ge 7$, then it holds for $P(k+1)$" it is not enough to conclude with $P(m)$ holds, for all $m \ge 7$. To do this, we have also to prove that $P(7)$ holds, which is not. – Mauro ALLEGRANZA Jun 26 '14 at 14:38
See also this post: – Ragnar Jun 26 '14 at 14:38
up vote 2 down vote accepted

Even if the case $m=7$ worked, $\textit{this proof}$ would not prove the result for $m=7$ because the base case is at $10$ and we apply induction to prove only for all $m\ge 10$.

share|cite|improve this answer
So, even if the case $m=7,8,9$ worked, we still cannot accept that the result works for $m \ge 7$ based on this proof alone? So, in order to be sure, we need to prove the statement "$m^3 \le 2^m$ for $m\ge 7$". Is this what you meant? – mauna Jun 26 '14 at 14:57
@mauna That's exactly what I meant. – Peter Woolfitt Jun 26 '14 at 14:58

"We only need $k \geq 7$ for the inductive step" is referring strictly to assertion that $k^3 + 7k^2 \leq k^3 + k^3$ within the argument of the inductive step. (See the chain of inequalities.) When $k = 7, \;k^3 + 7k^2 = 7^3 + 7\cdot 7^2 = 7^3 + 7^3$, which is true

Essentially, since the proposition we are aiming to prove, which applies to $k \geq 10$, assures that since $10>7$, we know that the step $k^3 + 7k^2 \leq k^3 + k^3$, because that mini-claim holds for any $k\geq 7$, so that mini-claim certainly holds for $k\geq 10$.

RE: you edit

The author states those claims along the way in the inductive step as a way of stating the minimal requirement at the respective claims for the claim to be true. And since $k\geq 10 \implies k\geq 7$, so any claim that is true for $k \geq 7 is certainly true for $k \geq 10. Likewise, $k \geq 10 \implies k \geq 1$. That's all.

share|cite|improve this answer

That $k \ge 7$ is only used to establish the inequality that $ 7k^2 \le k^3$, which is just one step in the proof. That $k \ge 7$ is stated only for clarification: the initial assumption is that $k \ge 10$ which also leads to $ 7k^2 \le k^3$, but is not quite so clear.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.