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So for example when dealing with two dimensions you would use $a^2 + b^2 = c^2$ and for three dimensions you would use $a^2 + b^2 + c^2 = d^2$ (Say for example you are calculating the length of the diagonal of a box) but what about in the fourth dimension? (Not really sure if there is a fourth dimension or how that even works in math so correct me if that makes no sense to say) Wouldn't it be something like $a^2 + b^2 + c^2 + d^2 = e^2$ ?

Ben Sheller
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bigfocalchord
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  • In any finite dimensional inner product space (Euclidian space) you can use a formula that resemebles the 2d case, and it is easy to prove using induction and inner product.

    In infinite dimensional spaces, things get messy. The analogous formula is Parseval identity.

     – Ranc Dec 25 '15 at 17:43
  • A similar question with answers explaining the theorem with graphics: https://math.stackexchange.com/questions/2837230/pythagorean-theorem-in-higher-dimensions – Anton Tarasenko Jul 18 '19 at 09:42

4 Answers4

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Given two points in $\mathbb{R}^n$, $x = (x_1,x_2,\ldots,x_n)$ and $y = (y_1,y_2,\ldots,y_n)$ we can define an $m$-dimensional "hyperbox" where $m\leq n$:

$$Box(x,y) := \{z = (z_1, z_2, \ldots , z_n)\mid z_i = t_i\cdot x_i + (1-t_i)\cdot y_i,\ t_i\in [0,1]\}$$

Where the lengths of its edges are $a_1:=|x_1-y_1|,\ a_2:=|x_2 - y_2|,\ \ldots\ ,\ a_n:=|x_n-y_n|$.

The length of the largest diagonal of said box is precisely the distance between $x$ and $y$:

$$d = \sqrt{\sum_{k=1}^n(x_k-y_k)^2} = \sqrt{\sum_{k=1}^n a_k^2}$$

Darth Geek
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  • I don't understand the construction of your box. $t$ is a scalar, $x_i$ and $y_i$ are components, so $t \cdot x_i$ is a real number, as is $(1-t) \cdot y_i$, as is their sum. so $z$ is a sum of real numbers. So the box is a set of real numbers. How does that describe a box in $m$ dimensions? – user3578468 Sep 26 '17 at 21:04
  • @user3578468 sorry, that $(z_1 + z_2 + \ldots + z_n)$ was supposed to be $(z_1,z_2,\ldots,z_n)$. Fixed now. – Darth Geek Sep 27 '17 at 04:11
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That is right. One well-known application of this fact is the Euclidian norm $\| \mathbf x \|$ for a vector $\mathbf x \in \Bbb R^n$, which is defined as $$ \| \mathbf x \| = \sqrt{x_1^2 + x_2^2+\dots+ x_n^2}.$$ This is exactly the generalization of the Pythagorean theorem you are referring to.

molarmass
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Yes. Mathematicians are very happy in any number of dimensions - even infinite dimensions!
Another way to extend Pythagoras is to take a solid rectangular box, and slice off a corner. The corner has three sides that were part of the old box; and one diagonal side. The areas of these sides also obey $a^2+b^2+c^2=d^2$.

Empy2
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    The sliced-off-corner-of-a-rectangular-box version is de Gua's Theorem; importantly for this discussion, that result also generalizes to arbitrary dimensions in a natural way (with the dimensions of the "leg" and "hypotenuse" elements increasing at every stage). Non-Euclidean geometry, too, has counterpart relations, but I'm aware only of the ones in dimensions 2 and 3. (I discovered the $3$-dimensional case myself.) – Blue Dec 25 '15 at 20:01
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The length of a diagonal of a cube is square root of 3 times the length of a side. This is in three dimensions. I don't see how you arrive at this answer using a^2 + b^2 + c^2 though. Instead, you use a two dimensional slice with one side being square root of 2 times side and the other value the side length. Here d^2 = a^2 + b^2 which, with side being s equals sqrt((sqrt(2)s^2) + (s^2)) = sqrt(3)s = d. By extension, my guess of a four dimensional cube of length of side s would be e = sqrt( (sqrt(3)s^2) + (sqrt(2)s^2) + (s^2) ) which equals sqrt(6)s. I am not sure of this extension, however and unfortunately cannot prove this guess. After further analysis, a four dimensional cube of side equals 1 has diagonal of length sqrt(4) or 2.

PAUL FitzSimons
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