MAM2000 (Essays/Dimension)

# Higher-Dimensional Spaces

The intuitions that students accumulate in dealing with coordinate pairs in the plane and coordinate triples in three-dimensional space lead naturally to coordinate geometry in higher dimensions. A thorough understanding of two and three dimensions provides an important foundation for the powerful generalizations of vector and matrix algebra in science and engineering, in economics and social science, and especially computer science and graphics. We illustrate this progression with two examples.

The vertices of a square can be given by four points (0,0), (1,0), (1, 1), and (0, 1). To obtain the vertices of a cube, we can take the points of a zero in the third coordinate and then move the square one unit in the third direction to obtain four more vertices, with a 1 in the last coordinate:

 (0, 0, 0), (1, 0, 0), (1, 1, 0), (0, 1, 0), (0, 0, 1), (1, 0, 1), (1, 1, 1), (0, 1, 1).

Thus we can describe either the square or the cube as having vertices that are either 0 or 1 in each coordinate.

Figure 32. Generalizing the Pythagorean theorem to three dimensions by applying it to two different triangles found in a rectangular box.

The procedure generalizes automatically: to obtain the vertices of a hypercube, we start with the eight vertices of a cube and put 0 in the final coordinate and then "move the cube in a fourth direction" to obtain eight more points with 1 the last coordinate:

 (0,0,0,0), (1,0,0,0), (1,1,0,0), (0,1,0,0), (0,0,1,0), (1,0,1,0), (1,1,1,0), (0,1,1,0), (0,0,0,1), (1,0,0,1), (1,1,0,1), (0,1,0,1), (0,0,1,1), (1,0,1,1), (1,1,1,1), (0,1,1,1).

We thus obtain the sixteen vertices of a hypercube, with 0 or 1 in each of four coordinates. It is this sort of representation that is ideal for communicating with a computer.

A second topic that generalizes in a very nice way is the Pythagorean theorem. If we think of this theorem as a way of calculating the length of the diagonal of a rectangle with given sides, then the extension to three dimensions is immediate: given a solid bounded by rectangular sides, we first apply the theorem to one side and then apply it to a rectangle built over the first diagonal (Figure 32). We easily get e2 = c2 + d2 = c2 + (a2 + b2), so the length of the diagonal of a rectangular prism with sides a, b, and c is (a2 + b2 + c2)1/2. The pattern is established, and the distance formula in four-dimensional space follows almost immediately. Students can then calculate the lengths of diagonals of the hypercube with the 0-1 coordinates. It turns out that the length of the major diagonal of a four-dimensional cube — say from (0,0,0,0) to (1,1,1,1) — is 41/2 = 2which is twice the length of a side. [an error occurred while processing this directive]