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Summary
| Description |
Proof without words of the two-colour case of Ramsey's theorem by CMG Lee. Due to the pigeonhole principle, there are at least 3 edges of the same colour (dashed purple) from an arbitrary vertex v. Calling 3 of the vertices terminating these edges r, s and t, if the edge rs, st or tr (solid black) had this colour, it would complete the triangle with v. But if not, each must be oppositely coloured, completing triangle rst of that colour. |
| Source | Own work |
| Author | Cmglee |
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| Date/Time | Thumbnail | Dimensions | User | Comment | |
|---|---|---|---|---|---|
| current | 17:33, 7 January 2024 | | 512 × 512 (521 bytes) | Cmglee | Align text and remove construction shapes |
| 17:28, 7 January 2024 |  | 512 × 512 (1 KB) | Cmglee | {{Information |Description=Proof without words of Ramsey's theorem by CMG Lee. Due to the pigeonhole principle, there are at least 3 edges from an arbitrary vertex ''v'' of the same colour (dashed purple). Calling 3 of the vertices terminating these edges ''r'', ''s'' and ''t'', if the edge ''rs'', ''st'' or ''tr'' (solid black) has this colour, it would complete the triangle with ''v''. But if not, each must be oppositely coloured, completing a triangle of that colour. |Source={{own}} |Date... |
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