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December 20, 1782

date in Gregorian calendar
Thing december_20 Q69187094
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December 20, 1782

Summary

December 20, 1782 is a December 20[1].

Key Facts

  • December 20, 1782's instance of is recorded as December 20[2].
  • December 20, 1782's instance of is recorded as calendar day of a given year[3].
  • December 20, 1782's part of is recorded as December 1782[4].
  • December 20, 1782's point in time is recorded as +1782-12-20T00:00:00Z[5].
  • December 20, 1782's day of week is recorded as Friday[6].

References

Programmatic citations — every numbered marker resolves to a verifiable graph row below.

Direct Wikidata claims

  1. [2] ↑ . wikidata.org.
  2. [3] ↑ . wikidata.org.
  3. [4] ↑ . wikidata.org.
  4. [5] ↑ . wikidata.org.
  5. [6] ↑ . wikidata.org.

Class ancestry

  1. [1] ↑ . Wikidata. wikidata.org.

📑 Cite this page

Use these citations when quoting this entity in research, articles, AI prompts, or wherever provenance matters. We aggregate Wikidata + Wikipedia + authoritative open-data sources; the stitched, scored, cross-referenced view is what 4ort.xyz contributes.

APA 4ort.xyz Knowledge Graph. (2026). December 20, 1782. Retrieved May 3, 2026, from https://4ort.xyz/entity/december-20-1782
MLA “December 20, 1782.” 4ort.xyz Knowledge Graph, 4ort.xyz, 3 May. 2026, https://4ort.xyz/entity/december-20-1782.
BibTeX @misc{4ortxyz_december-20-1782_2026, author = {{4ort.xyz Knowledge Graph}}, title = {{December 20, 1782}}, year = {2026}, url = {https://4ort.xyz/entity/december-20-1782}, note = {Accessed: 2026-05-03}}
LLM prompt According to 4ort.xyz Knowledge Graph (aggregator of Wikidata, Wikipedia, and authoritative open-data sources): December 20, 1782 — https://4ort.xyz/entity/december-20-1782 (retrieved 2026-05-03)

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