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

Then:

$\ds {n \brace k} = 0$

Proof

The proof proceeds by induction.

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

$\ds k > n \implies {n \brace k} = 0$

Basis for the Induction

$\map P 0$ is the case:

$\ds {0 \brace k} = \delta_{0 k}$

from Stirling Number of the Second Kind of 0.

So by definition of Kronecker delta:

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

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:

$\ds k > r \implies {r \brace k} = 0$

from which it is to be shown that:

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

Induction Step

This is the induction step:

\(\ds {r + 1 \brace k}\)

\(=\)

\(\ds k {r \brace k} + {r \brace k - 1}\)

\(\ds \)

\(=\)

\(\ds r \times 0 + {r \brace k - 1}\)

Induction Hypothesis

\(\ds \)

\(=\)

\(\ds r \times 0 + 0\)

Induction Hypothesis: $k > r + 1 \implies k - 1 > r$

\(\ds \)

\(=\)

\(\ds 0\)

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

Therefore:

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

$\blacksquare$