Proof.Suppose p R(r-1,s)and g=R(r,s-1)are both even.Let t=p+q-1 and consider a two-coloured graph of t vertices.If di is degree of i-th vertex in the graph,then,according to the Handshaking lemma,is even.Given that tis odd,there must be an evenAssume dis even,M and Nare the vertices incident to vertex 1 in the blue and red subgraphs,respectively.Then both M=d and N=t-1-d are even.According to the Pigeonhole principle,either M>p-1,or N >g.Since M]is even,while p-1 is odd,the first inequality can be strengthened,so either M2p or N2g =R(r-1,).Then either the M subgraph has a red K and the p lete,or has a blue K-1 which along with vertex 1 makes a blue Kr.The case=R(r,s-1)is treated similarly. Case of more colours Lemma 2.If c>2,then R(m...n)s R(n..n.R(nc-1.nc)). Proof.Consider a complete graph of R(n n-2,R(n-,n)vertices and colour its edges with c colours and are th olou Thus the 1) chromatically coloured with colour ifor some 1sisc-2or aKg cooured in the blurred colourIn the former cas ecover our sight again and seefrom the definitionof either a(c 1)-mon chrome either case the proo is complete. Lemma 1 impli that any R(rs)is finite.The right hand side of the inequality in Lemma 2 expresses a Ramsey number for c colours in terms of Ramsey numbers for fewer colours.Therefore any R(n,...,n)is finite for any number of colours.This proves the theorem. Ramsey numbers The numbers R(r,s)in Ramsey's theorem (and their extensions to more than two colours)are known as Ramsey numbers.The ramsey number.R(m.n).gives the solution to the party problem.which asks the minimum number of guests.R(m,n).that must be invited so that at least m will know each other or at least n will not know each other.In the language of graph theory,the Ramsey number is the minimum number of vertices,v=R(m,n),such that all undirected simple graphs of order v,contain a clique of order m,or an independent set of order n.Ramsey's theorem states that such a number exists for all m and n. By symmetry,it is true that R(m,n)=R(n,m).An upper bound for R(r,s)can be extracted from the proof of the theorem,and other arguments give lower bounds.(The first exponential lower bound was obtained by Paul Erdos using the probabilistic method.)However,there is a vast gap between the tightest lower bounds and the tightest upper bounds.There are also very few numbers r and s for which we know the exact value of R(r.s). Computing a lower bound L forR(s) usually requires exhibiting a blue/red colouring of the graph KL-1 with no blue Kr subgraph and no red Ks subgraph.Such a counterexample is called a Ramsey graph. Brendan McKay maintains a list of known Ramsey graphs.[5]Upp oer bounds are often considerably more difficult to establish one gither has to check all nossible co olourings to confirm the absence of a counterexample,or to present a mathematical argument for its absence.A sophisticated c ter program does not need to look at all colourings individually in order to eliminate all of them:nevertheless it is a verv difficult computational task that existing software can only manage on small sizes.Each complete graph KProof. Suppose p = R(r − 1, s) and q = R(r, s − 1) are both even. Let t = p + q − 1 and consider a two-coloured graph of t vertices. If is degree of -th vertex in the graph, then, according to the Handshaking lemma, is even. Given that t is odd, there must be an even . Assume is even, M and N are the vertices incident to vertex 1 in the blue and red subgraphs, respectively. Then both and are even. According to the Pigeonhole principle, either , or . Since is even, while is odd, the first inequality can be strengthened, so either or . Suppose . Then either the M subgraph has a red and the proof is complete, or it has a blue which along with vertex 1 makes a blue . The case is treated similarly. Lemma 2. If c>2, then R(n1 , …, nc ) ≤ R(n1 , …, nc−2 , R(nc−1 , nc )). Proof. Consider a complete graph of R(n1 , …, nc−2 , R(nc−1 , nc )) vertices and colour its edges with c colours. Now 'go colour-blind' and pretend that c − 1 and c are the same colour. Thus the graph is now (c − 1)- coloured. Due to the definition of R(n1 , …, nc−2 , R(nc−1 , nc )), such a graph contains either a Kni mono￾chromatically coloured with colour i for some 1 ≤ i ≤ c − 2 or a KR(nc − 1 , nc ) -coloured in the 'blurred colour'. In the former case we are finished. In the latter case, we recover our sight again and see from the definition of R(nc−1 , nc ) we must have either a (c − 1)-monochrome Knc−1 or a c-monochrome Knc . In either case the proof is complete. Lemma 1 implies that any R(r,s) is finite. The right hand side of the inequality in Lemma 2 expresses a Ramsey number for c colours in terms of Ramsey numbers for fewer colours. Therefore any R(n1 , …, nc ) is finite for any number of colours. This proves the theorem. The numbers R(r, s) in Ramsey's theorem (and their extensions to more than two colours) are known as Ramsey numbers. The Ramsey number, R(m, n), gives the solution to the party problem, which asks the minimum number of guests, R(m, n), that must be invited so that at least m will know each other or at least n will not know each other. In the language of graph theory, the Ramsey number is the minimum number of vertices, v = R(m, n), such that all undirected simple graphs of order v, contain a clique of order m, or an independent set of order n. Ramsey's theorem states that such a number exists for all m and n. By symmetry, it is true that R(m, n) = R(n, m). An upper bound for R(r, s) can be extracted from the proof of the theorem, and other arguments give lower bounds. (The first exponential lower bound was obtained by Paul Erdős using the probabilistic method.) However, there is a vast gap between the tightest lower bounds and the tightest upper bounds. There are also very few numbers r and s for which we know the exact value of R(r, s). Computing a lower bound L for R(r, s) usually requires exhibiting a blue/red colouring of the graph KL−1 with no blue Kr subgraph and no red Ks subgraph. Such a counterexample is called a Ramsey graph. Brendan McKay maintains a list of known Ramsey graphs. [5] Upper bounds are often considerably more difficult to establish: one either has to check all possible colourings to confirm the absence of a counterexample, or to present a mathematical argument for its absence. A sophisticated computer program does not need to look at all colourings individually in order to eliminate all of them; nevertheless it is a very difficult computational task that existing software can only manage on small sizes. Each complete graph Kn Case of more colours Ramsey numbers