2 to the n is Greater than n Cubed when n is 10 and above

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Theorem

$\forall n \in \Z, n \ge 10: 2^n > n^3$


Proof

Proof by induction:

For all $n \in \Z$ such that $n \ge 10$, let $\map P n$ be the proposition:

$2^n > n^3$


We note that:

$2^9 = 512 < 729 = 9^3$

so when $n < 10$ the proposition does not hold.


Basis for the Induction

$\map P {10}$ is the case:

$2^{10} = 1024 > 1000 = 10^3$

so $\map P {10}$ is seen to hold.

This is our basis for the induction.


Induction Hypothesis

Now we need to show that, if $\map P k$ is true, where $k \ge 10$, then it logically follows that $\map P {k + 1}$ is true.


So this is our induction hypothesis:

$2^k > k^3$


We need to show that:

$2^{k + 1} > \paren {k + 1}^3$


Induction Step

This is our induction step:

We note that when $k \ge 10$:

\(\ds \paren {1 + \frac 1 k}^3\) \(\le\) \(\ds \paren {1 + \frac 1 {10} }^3\)
\(\ds \) \(=\) \(\ds 1.331\)
\(\text {(1)}: \quad\) \(\ds \) \(<\) \(\ds 2\)


Thus:

\(\ds 2^{k + 1}\) \(=\) \(\ds 2 \times 2^k\)
\(\ds \) \(>\) \(\ds \paren {1 + \dfrac 1 k}^3 2^k\) from $(1)$
\(\ds \) \(>\) \(\ds \paren {1 + \dfrac 1 k}^3 k^3\) Induction Hypothesis
\(\ds \) \(=\) \(\ds \paren {k + 1}^3\) Product of Powers

So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.

$\blacksquare$


Sources

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