# 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

- 1997: Donald E. Knuth:
*The Art of Computer Programming: Volume 1: Fundamental Algorithms*(3rd ed.) ... (previous) ... (next): $\S 1.2.1$: Mathematical Induction: Exercise $10$