I have been asked to show (a homework problem) that the Ramsey number $R(mK_2,mK_2)=3m-1$. $mK_2$ is the graph of $m$ pairwise disjoint copies of $K_2$.
I know $K_2$ is just two vertices with one edge between them. My plan is:
First: Show that there cannot be $m$ different copies of $K_2$ in the complete graph $K_{3m-2}$. This means that there can never be $m$ copies of either colour which means that $R(mK_2,mK_2)>3m-2$.
Second: Show that there are always more than $4m$ edges which means that if we colour these edges with two colours in any way there will at least be $2m$ edges of one of the colours (according to pigeonhole principle).
Is this the correct idea? I do not know how to do the first step but the second should be something like $$ |E(K_{3m-1})|=\frac{(3m-1)(3m-2)}{2m}=\frac{9m^2-9m+2}{2}=\frac{9m(m-1)+2}{4m}=\frac{9}{4}(m-1)+\frac{1}{2m} $$ and since $\frac{9}{4}(m-1)>2$ for $m>2$ there will always be more than two groups of $2m$ edges.
Am I on the right track?
you've good ideas, but you are not exactly on the right track no. First a correction : the complete graph on $n$ vertices, has $\binom{n}{2}$ edges, so about $n^2$ edges, much more than in your calculation :
$$\vert E(K_{3m-1})\vert = \frac{(3m-1)(3m-2)}{2}\neq \frac{(3m-1)(3m-2)}{2m}$$
Then as you said the first step is showing that $R(mK_2,mK_2) > 3m−2$. However this does not mean that there can never be $m$ different copies of $K_2$ in the complete graph $K_{3m−2}$. It means that there exists (at least) one red/blue edge-colouring of $K_{3m−2}$ such that there is no monochromatic $mK_2$. So you "only" have to exhibit such a colouring.
Then, showing that $R(mK_2,mK_2) \geq 3m−2$ : Again it does not mean that if we colour with two colours in any way there will at least be 2m edges of one of the colours. You need your coloured edges to form $mK_2$ which is a disjoint union of $m$ edges.
You should look into perfect / maximal matching in complete graphs.