Proving 2^n > n^3 for all n > 9 (WRONG)

by Nibir Sankar, Jun 26, 2025

Prove $2^{n} > n^{3}$ for all $n > 9$

Proof:

By mathematical induction.

Let $A (n)$ denote $2^{n} > n^{3}$ .

For $n = 10$ (base step)

\begin{array}{r} 2^{10} > 10^{3} \\ 1024 > 1000 \end{array}

Thus, $A (10)$ is true.

Now, assuming $A (n)$ is true, we get:

\begin{array}{r} 2^{n} > n^{3} \end{array}

Taking the $\log$ of both sides:

\begin{aligned} \log (2^{n}) > \log (n^{3}) \\ n \log 2 > 3 \log n (*) \end{aligned}

Now, for $A (n + 1)$ :

\begin{aligned} (n + 1) \log 2 > 3 \log (n + 1) \\ ⟹ n \log 2 + \log 2 > 3 [\log n + \log 1] \\ ⟹ n \log 2 + \log 2 > 3 \log n \end{aligned}

This is true since $(*)$ - induction hypothesis.

Which follows that $A (n)$ must be true.

Therefore, by the principle of mathematical induction, we have proven that $2^{n} > n^{3}$ for all $n > 9$ .