slinberg.net: Pythagorean Theorem

Steve Linberg

How did I make it to $adulthood * 2 without knowing that our old friend, the Pythagorean Theorem ($a^2 + b^2 = c^2$), works in any number of dimensions?

The Euclidean distance $d$ between the origin ($0, 0, \dots 0$) and a point defined by $n$ coordinates $x$ is just

\[d = \sqrt{x_1^2 + x_2^2 + \dots + x_n^2}\]

Add up all of the squares and take the square root. Dope-smack for me.

Comment on this article Share:

Corrections

If you see mistakes or want to suggest changes, please create an issue on the source repository.

Reuse

Text and figures are licensed under Creative Commons Attribution CC BY-NC 4.0. Source code is available at https://github.com/stevelinberg/distill-blog, unless otherwise noted. The figures that have been reused from other sources don't fall under this license and can be recognized by a note in their caption: "Figure from ...".

Citation

For attribution, please cite this work as

Linberg (2022, Feb. 2). slinberg.net: Pythagorean Theorem. Retrieved from https://slinberg.net/posts/2022-02-02-pythagorean-theorem/

BibTeX citation

@misc{linberg2022pythagorean,
  author = {Linberg, Steve},
  title = {slinberg.net: Pythagorean Theorem},
  url = {https://slinberg.net/posts/2022-02-02-pythagorean-theorem/},
  year = {2022}
}