Factorial Greater than Square for n Greater than 3
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Theorem
Let $n \in \Z$ be an integer such that $n > 3$.
Then $n! > n^2$.
Proof
We note that:
\(\ds 1!\) | \(=\) | \(\ds 1\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 1^2\) | ||||||||||||
\(\ds 2!\) | \(=\) | \(\ds 2\) | ||||||||||||
\(\ds \) | \(<\) | \(\ds 4\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 2^2\) | ||||||||||||
\(\ds 3!\) | \(=\) | \(\ds 6\) | ||||||||||||
\(\ds \) | \(<\) | \(\ds 9\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 3^2\) |
The proof then proceeds by induction.
For all $n \in \Z_{\ge 4}$, let $\map P n$ be the proposition:
- $n! > n^2$
Basis for the Induction
$\map P 4$ is the case:
\(\ds 4!\) | \(=\) | \(\ds 24\) | ||||||||||||
\(\ds \) | \(>\) | \(\ds 16\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 4^2\) |
Thus $\map P 4$ 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 $k \ge 4$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
- $k! > k^2$
from which it is to be shown that:
- $\paren {k + 1}! > \paren {k + 1}^2$
Induction Step
This is the induction step:
\(\ds \paren {k + 1}!\) | \(=\) | \(\ds \paren {k + 1} \times k!\) | ||||||||||||
\(\ds \) | \(>\) | \(\ds \paren {k + 1} \times k^2\) | Induction Hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {k - 2} \times k^2 + 2 \times k^2 + k^2\) | ||||||||||||
\(\ds \) | \(>\) | \(\ds 1 + 2 \times k + k^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {k + 1}^2\) |
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
- $\forall n \in \Z_{>3}: n! > n^2$
$\blacksquare$
Sources
- 1980: David M. Burton: Elementary Number Theory (revised ed.) ... (previous) ... (next): Chapter $1$: Some Preliminary Considerations: $1.1$ Mathematical Induction: Problems $1.1$: $6$