Using the Pythagorean theorem, we can pretty much find the length of diagonals of 2d shapes(squares, rectangles, etc).

$$ c = \sqrt{a^2 + b^2}$$

But this property holds true for higher dimensions as well. All we need to do is take the root of total sum of squares of different dimensions.

Third Dimension: $$ d = \sqrt{x^2 + y^2 + z^2}$$ Any dimensions $$ (x_1^2, x_2^2,...x_n^2) = \sqrt{x_1^2 + x_2^2 + ....x_n^2}$$

The way I think of this is like vectors. This resultant vector(diagonal) produces the same effect of the combined different vectors.

But why is such the case? Are there any other explanations for the cause of this? And is there a limit to the dimensions this property holds true(like the 89th dimension)?

  • $\begingroup$ The Pythagorean theorem is "true" in special vector spaces only. Namely those that are metric vector spaces with the euclidean metric. (so called euclidean spaces). So I would not say there is a reason for this to be true. It is more a matter of definition. Moreover, in higher dimensions (e.g. 4dimensional space time) this is NOT true: c.f. Minkowski metric. $\endgroup$ – Caroline Sep 6 '19 at 9:14
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    $\begingroup$ Sticking to 3D Euclidean space, three points define a plane, so you are only really working in 2 dimensions. With one triangle corner at the origin and base in the x, y plane, the base has length $\sqrt{x^2+y^2}$. If the vertical has component z then the triangle hypothenus has squared length $(\sqrt{x^2+y^2})^2+z^2$. This idea scales up to 4D, 5D etc. It does not work in infinite dimensions though and not more generally, as @Caroline notes. $\endgroup$ – Paul Sep 6 '19 at 9:22
  • $\begingroup$ Hmmm, okay, thanks for explaining. $\endgroup$ – SreeLegend Sep 6 '19 at 9:46

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