# Stirling Number of the Second Kind of Number with Greater/Proof 2

## Theorem

Let $n, k \in \Z_{\ge 0}$ such that $k > n$.

Let $\displaystyle {n \brace k}$ denote a Stirling number of the second kind.

Then:

- $\displaystyle {n \brace k} = 0$

## Proof

The proof proceeds by induction.

For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:

- $\displaystyle k > n \implies {n \brace k} = 0$

### Basis for the Induction

$\map P 0$ is the case:

- $\displaystyle {0 \brace k} = \delta_{0 k}$

from Stirling Number of the Second Kind of 0.

So by definition of Kronecker delta:

- $\forall k \in \Z_{\ge 0}: k > 0 \implies \displaystyle {0 \brace k} = 0$

and so $\map P 0$ is seen to hold.

This is the basis for the induction.

### Induction Hypothesis

Now it needs to be shown that, if $\map P r$ is true, where $0 \le r$, then it logically follows that $\map P {r + 1}$ is true.

So this is the induction hypothesis:

- $\displaystyle k > r \implies {r \brace k} = 0$

from which it is to be shown that:

- $\displaystyle k > r + 1 \implies {r + 1 \brace k} = 0$

### Induction Step

This is the induction step:

\(\displaystyle {r + 1 \brace k}\) | \(=\) | \(\displaystyle k {r \brace k} + {r \brace k - 1}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle r \times 0 + {r \brace k - 1}\) | Induction Hypothesis | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle r \times 0 + 0\) | Induction Hypothesis: $k > r + 1 \implies k - 1 > r$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle 0\) |

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

Therefore:

- $\displaystyle \forall n \in \Z_{\ge 0}: k > n \implies {n \brace k} = 0$

$\blacksquare$