- #1

MathematicalPhysicist

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## Main Question or Discussion Point

I have this next question which I am trying to resolve.

Let G=(V,E) be some graph (usually in this context threre aren't loops nor directed edges),assume that every red-blue colouring of the edges of G contains a red copy of [tex]K_s[/tex] or blue copy of [tex]K_t[/tex]. show that [tex] R(s,t)\le\chi(G) [/tex], where R(s,t) is ramsey number and [tex]\chi(G)[/tex] is vertex colouring minimum number of G.

Now I thought of proving that [tex](1)\binom{s+t-2}{t-1}\le\chi(G)=\chi'(L(G))[/tex]

where L(G) is the graph in which you identify each edge of G as a vertex and each vertex in G which is common with two edges in G as an edge in L(G); [tex]\chi'(L(G))[/tex] is the edges colouring index of L(G).

How do I show (1), I am not sure?

Let G=(V,E) be some graph (usually in this context threre aren't loops nor directed edges),assume that every red-blue colouring of the edges of G contains a red copy of [tex]K_s[/tex] or blue copy of [tex]K_t[/tex]. show that [tex] R(s,t)\le\chi(G) [/tex], where R(s,t) is ramsey number and [tex]\chi(G)[/tex] is vertex colouring minimum number of G.

Now I thought of proving that [tex](1)\binom{s+t-2}{t-1}\le\chi(G)=\chi'(L(G))[/tex]

where L(G) is the graph in which you identify each edge of G as a vertex and each vertex in G which is common with two edges in G as an edge in L(G); [tex]\chi'(L(G))[/tex] is the edges colouring index of L(G).

How do I show (1), I am not sure?