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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 x, y and z, if the edge xy, yz or zx (solid black) had this colour, it would complete the triangle with v. But if not, each must be oppositely coloured, completing triangle xyz of that colour. |
| Source | Own work |
| Author | Cmglee |
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File history
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| Date/Time | Thumbnail | Dimensions | User | Comment | |
|---|---|---|---|---|---|
| current | 14:56, 10 November 2024 | | 512 × 512 (518 bytes) | Cmglee | Avoid cropped y descender |
| 14:54, 10 November 2024 |  | 512 × 512 (518 bytes) | Cmglee | Update vertex labels and line style | |
| 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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