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If a graph $H$ has minimum degree $d$, then its graph Ramsey number satisfies $R(H)\ge 2^{d/2}$. The probabilistic proof chooses a red-blue edge colouring and applies the local lemma to its monochromatic copies of $H$.

The graph Ramsey number $R(G,H)$ is the smallest $N$ such that every red-blue coloring of the edges of $K_N$ has a red copy of $G$ or a blue copy of $H$. These copies are subgraphs, not necessarily induced. Equivalently every graph on $N$ vertices contains $G$ or its complement contains $H$.

The graph Ramsey number $R(G,H)$ is the least $N$ such that every red-blue coloring of the edges of $K_N$ contains a red copy of $G$ or a blue copy of $H$. Copies are ordinary subgraphs, not necessarily induced. Equivalently every graph on $N$ vertices contains $G$ or its complement contains $H$.

Define the multicolour graph Ramsey number $R_k(s_1,\dots ,s_k)$ as the least $n$ such that every colouring of the edges of the complete graph $K_n$ with $k$ colours contains, for some $i$, a complete subgraph on $s_i$ vertices all of whose edges have colour $i$.

Prove finiteness by induction on ${\sum }_i{s}_i$, temporarily allowing $s_i=1$. If some $s_i=1$, a single vertex supplies the required vacuous monochromatic clique, so the Ramsey number is one. Otherwise put $N_i=R_k(s_1,\dots ,s_i-1,\dots ,s_k)$, which is finite by induction, and takeFix a vertex $v$ and partition its other $n-1$ vertices according to the colour of their edge to $v$. Some colour-$i$ class has at least $N_i$ vertices: if every class had at most $N_i-1$, their total would be at most ${\sum }_i{N}_i-k=n-2$. In that class the inductive property either gives a colour-$j$ clique of size $s_j$ with $j\ne i$, or a colour-$i$ clique of size $s_i-1$. In the latter case add $v$. ThusIn particular the requested two-colour diagonal number is $R(s)=R_2(s,s)$, so it exists for every $s\ge 2$.

Now colour each positive integer by its partition class, say there are $k$ classes, and take $m=R_k(3,\dots ,3)$. Colour the edge $ij$, $1\le i<j\le m$, by the class of $j-i$. There is a monochromatic triangle $a<b<c$. Its three differences $x=b-a$, $y=c-b$, $z=c-a$ are positive integers in one class, andThis is the finite-Ramsey proof of the partition conclusion. The statement does not require $x$ and $y$ to be distinct.

The graph Ramsey number $R(s,t)$ is the least $N$ for which every graph on $N$ vertices has a clique of size $s$ or an independent set of size $t$; $R(s)=R(s,s)$ is the diagonal Ramsey number. The base cases are $R(2,t)=t$ and $R(s,2)=s$.

Assume the smaller parameters already have finite graph Ramsey numbers, and take $N=R(s-1,t)+R(s,t-1)$. For any vertex, either its neighbourhood has at least $R(s-1,t)$ vertices, or its nonneighbourhood has at least $R(s,t-1)$ vertices. In the first case it contains an independent $t$-set, or an $(s-1)$-clique which extends with the chosen vertex. In the second it contains an $s$-clique, or a $(t-1)$-independent set which extends with that vertex. Induction proves existence and

A graph Ramsey number $R(s,t)$ is the least $N$ such that every red-blue edge colouring of the complete graph $K_N$ has a red $K_s$ or a blue $K_t$. At a vertex in $K_{R(s-1,t)+R(s,t-1)}$, either at least $R(s-1,t)$ neighbours have red incident edges or at least $R(s,t-1)$ have blue incident edges. In the first case find a red $K_{s-1}$ and adjoin the vertex, or a blue $K_t$; the second case is symmetric. ThusStarting with $R(2,t)=t$ and $R(s,2)=s$, induction gives the binomial upper bound for a Ramsey numberFor the lower construction, partition $2t-2$ vertices into two classes of size $t-1$, colour inside-class edges red and between-class edges blue. Red cliques have at most $t-1$ vertices, and the blue graph is bipartite, so has no odd cycle.

Conversely, a graph without an odd cycle is bipartite: in each connected component colour vertices by the parity of the length of a path from a root; differing path parities would give an odd closed walk, hence an odd cycle. On $2t-1$ vertices one of the two blue bipartition classes has at least $t$ vertices. Every edge within it is red, giving a red $K_t$. Thus $2t-1$ is the exact threshold for this alternative.

Let $R=R(C_5)$ be the fixed graph Ramsey number of the five-cycle. Partition $\lfloor n/R\rfloor R$ vertices into $\lfloor n/R\rfloor$ disjoint blocks of size $R$. For any one block, the probability that $G(n,1/2)$ induces a complete graph isThese events are independent for the disjoint blocks. Therefore the probability that none of them induces $K_R$ isWith probability tending to one, $G$ contains a copy of $K_R$. Every red-blue colouring of this copy contains a monochromatic $C_5$ by the definition of $R(C_5)$. Hence

The graph Ramsey number $R(G)$ is the least positive integer $n$ for which every red-blue colouring of $K_n$ contains a monochromatic copy of $G$. If $g=|V(G)|$, the Ramsey theorem gives a monochromatic $K_g$ in every colouring of $K_{R(g)}$, and that clique contains $G$. Hence

The graph Ramsey number $R(H)$ is the least $N$ such that every red-blue colouring of the edges of $K_N$ contains a monochromatic copy of $H$.

The minimum-degree Ramsey lower bound states that a graph of minimum degree $d$ admits an $H$-free colouring on every integer $N<2^{d/2}$. Its probabilistic proof colours edges independently and applies the local lemma to the events that a labelled copy of $H$ is monochromatic. Sincethe exponential cost $2^{1-e(H)}$ of a monochromatic copy dominates the number of compatible copies through every fixed edge. Hence such a colouring exists, and therefore

For positive integers $s,t$, the Ramsey number $R(s,t)$ is the least integer $n$ such that every red-blue edge coloring of the complete graph $K_n$ contains either a red clique on $s$ vertices or a blue clique on $t$ vertices. The complete graph $K_n$ has all $\left(\genfrac{}{}{0ex}{}{n}{2}\right)$ unordered pairs as edges; every edge gets exactly one color, with no requirement that incident edges have different colors. The definition and its graph interpretation appear in D. Yogeshwaran's Discrete Mathematics notes, section 5.3.

For example, $R(3,3)$ asks how many vertices force a monochromatic triangle in a graph. Swapping colors gives $R(s,t)=R(t,s)$. Also $R(2,t)=t$: either a red edge exists, or all edges are blue. On $t-1$ vertices, an all-blue coloring avoids both a red edge and a blue $t$-clique. Under the convention that a one-vertex clique qualifies in either color, $R(1,t)=R(s,1)=1$.

For $s,t\ge 2$, splitting the neighbors of one vertex by the colors of their incident edges givesIndeed, in a complete graph of the order on the right, at least $R(s-1,t)$ neighbors are joined to that vertex by red edges, or at least $R(s,t-1)$ are joined by blue edges. In the red case, the defining property of the smaller Ramsey number supplies either a blue $t$-clique among the neighbors or a red $(s-1)$-clique that extends by the chosen vertex; the blue case is symmetric. Starting from the one-vertex base cases, this recursively proves finite existence. The binomial upper bound for a Ramsey number packages the resulting estimate. ^[ref] present the neighbor argument and the sharp Ramsey number for triangles.

Equivalently, every graph on at least $R(s,t)$ vertices has an $s$-clique or a $t$-element independent set: color its edges red and the edges of its complement graph blue. This is the Ramsey theorem in graph theory. Larger monochromatic patterns have their own thresholds, such as a graph Ramsey number, matching Ramsey number or triangle-packing Ramsey number; these ask different questions from finding one clique.