> [!NOTE] Theorem > For all $k,l\in \mathbb{N}^{+},$ $R(k,l)\leq {k+l-2 \choose k-1}.$ ###### Proof by induction on $k+l$: For all $n\geq 5,$ let $P(n)$ be the following statement: for all $k,l\in \mathbb{N}^{+}$ with $k+l \leq n,$ $R(k,l)\leq {k+l-2 \choose k-1}.$ By [[Ramsey Numbers on Two Colours in First Row or Column]] and [[Ramsey Numbers on Two Colours in Second Row or Column]], $P(5)$ is true since either $k$ or $l$ must equal $2$ or $1$ when $k+l \leq 5.$ Suppose there exists $n-1\geq 5$ so that $P(n-1)$ is true. By [[Recurrence Inequality for Ramsey Numbers on Two Colours]], for all $k,l\in \mathbb{N}^{+}$ $R(k,l)\leq R(k-1,l)+R(k,l-1).$ Thus for all $k+l \leq n,$ we have $\begin{align} R(k,l)& \leq R(k-1,l)+R(l-1,k) \\ &\leq {k+l-3 \choose k-2} + {k+l-3 \choose l-2} & \text{by inductive hypothesis} \\ &= {k + l -2 \choose k-1} \end{align}$that is, $P(n)$ is true. Therefore, by [[Induction Principle|induction principle]], the statement of the theorem is true. # Applications **Generalisations**: Closed form formula for recurrence inequalities.