# Geometry Question

We all know the Pythagorean Theorem: x^2+y^2=z^2.
The example 3^2+4^2=5^2…9+16=25, I can verify geometrically; If I take a square 3 units on a side and place it on the x-axis, place a 4 unit square on the y-axis, a 5 unit square will perfectly fit in the Hypotenuse.

I discovered that 3^3+4^3+5^3=6^3…27+64+125=216

However, using theabove analogy of two-space in three-space,
the method no longer works. Three along the x-axis, four across the y-axis, five up the z-axis, the didtance back to the origin does not=6.

Why the incongruity?

Cause the distance from the origin in n-space is still determined the square root of the sums of squares of differences.

For your example, the distance from the origin of (3, 4, 5) is sqrt(3[sup]2[/sup] + 4[sup]2[/sup] + 5[sup]2[/sup]), which is 5sqrt(2).

As ultrafilter says, in Euclidean 3-space, the distance is determined by

d = d[sub]2[/sub] = (x[sup]2[/sup] + y[sup]2[/sup] + z[sup]2[/sup])[sup]1/2[/sup]

This is the Euclidean metric. (A metric is a way of defining the distance between two points.) This corresponds to real life, which is why it’s the one we use. However, you may be interested to know that there are other metrics. The one you used in your post, for instance, is

d[sub]3[/sub] = (x[sup]3[/sup] + y[sup]3[/sup] + z[sup]3[/sup])[sup]1/3[/sup]

And in general

d[sub]n[/sub] = (x[sup]n[/sup] + y[sup]n[/sup] + z[sup]n[/sup])[sup]1/n[/sup]

Different metrics are useful for different things. They also have some weird ones like:

d[sub]0[/sub], which is 0 if x = y = z = 0, and 1 otherwise. This is called the discrete metric.

d[sub][Sym]¥[/Sym][/sub], which is the limit as n approaches infinity of d[sub]n[/sub]. This turns out to be:

d[sub][Sym]¥[/Sym][/sub] = max(x, y, z)

One of the interesting diagrams I’ve seen related to this is the unit “circle” in each of these metrics. Unfortunately, I can’t seem to find it online.