Factorial Greater than Cube for n Greater than 5
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Theorem
Let $n \in \Z$ be an integer such that $n > 5$.
Then $n! > n^3$.
Proof 1
We note that:
\(\ds 1!\) | \(=\) | \(\ds 1\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 1^3\) | ||||||||||||
\(\ds 2!\) | \(=\) | \(\ds 2\) | ||||||||||||
\(\ds \) | \(<\) | \(\ds 8\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 2^3\) | ||||||||||||
\(\ds 3!\) | \(=\) | \(\ds 6\) | ||||||||||||
\(\ds \) | \(<\) | \(\ds 27\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 3^3\) | ||||||||||||
\(\ds 4!\) | \(=\) | \(\ds 24\) | ||||||||||||
\(\ds \) | \(<\) | \(\ds 64\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 4^3\) | ||||||||||||
\(\ds 5!\) | \(=\) | \(\ds 120\) | ||||||||||||
\(\ds \) | \(<\) | \(\ds 125\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 5^3\) |
The proof then proceeds by induction.
For all $n \in \Z_{\ge 6}$, let $\map P n$ be the proposition:
- $n! > n^3$
Basis for the Induction
$\map P 6$ is the case:
\(\ds 6!\) | \(=\) | \(\ds 720\) | ||||||||||||
\(\ds \) | \(>\) | \(\ds 216\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 6^3\) |
Thus $\map P 6$ is seen to hold.
This is the basis for the induction.
Induction Hypothesis
Now it needs to be shown that if $\map P k$ is true, where $k \ge 6$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
- $k! > k^3$
from which it is to be shown that:
- $\paren {k + 1}! > \paren {k + 1}^3$
Induction Step
This is the induction step:
\(\ds \paren {k + 1}!\) | \(=\) | \(\ds \paren {k + 1} \times k!\) | ||||||||||||
\(\ds \) | \(>\) | \(\ds \paren {k + 1} \times k^3\) | Induction Hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {k - 3} \times k^3 + 3 \times k^3 + k^3\) | ||||||||||||
\(\ds \) | \(>\) | \(\ds 1 + 3 \times k + 3 \times k^2 + k^3\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {k + 1}^3\) |
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
- $\forall n \in \Z_{>5}: n! > n^3$
$\blacksquare$
Proof 2
For $n > 5$, notice that the following inequalities hold:
- $2 \paren {n - 1} = 2 n - 2 > n + 5 - 2 > n$
- $3 \paren {n - 2} = 3 n - 6 > n + 10 - 6 > n$
And thus:
\(\ds n!\) | \(\ge\) | \(\ds n \paren {n - 1} \paren {n - 2} \paren {3!}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds n \paren {2 \paren {n - 1} } \paren {3 \paren {n - 2} }\) | ||||||||||||
\(\ds \) | \(>\) | \(\ds n \paren n \paren n\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds n^3\) |
$\blacksquare$
Sources
- 1980: David M. Burton: Elementary Number Theory (revised ed.) ... (previous) ... (next): Chapter $1$: Some Preliminary Considerations: $1.1$ Mathematical Induction: Problems $1.1$: $6$