A
hypercube in
n dimensions, or
an
n cube, is the
n
dimensional analog of a cube. It has two
hyperfaces
on each axis; the hyperfaces are
n1 dimensional
hypercubes.
In general, we call the volume enclosed by a hypercube an
nvolume.
Ordinary "volume" (measured in things like quarts and liters)
is
3volume. Area (measured in things
like acres and square meters) is
2volume.
The magnitude of the volume of an ncube with edge length
r
is
r^{n}.
We can form such an
ncube by "sweeping"
an (
n1)cube with edge length
r
in a direction perpendicular to itself, through a distance of
r
units. The rest of this page just expands on, explains, and
attempts to justify those statements.
Some examples of "lowdimension"
ncubes:
 A 0cube is a point,
and has no defined volume.
 A 1cube is a line
segment; its 1volume is its length.
 A 2cube is a square;
its 2volume is its area.
 A 3cube is an
ordinary cube.
 A 4cube is a tesseract,
and is occasionally just called a "hypercube".
Figure
1: Sweeping out a cube

To construct an
n cube, we
start with an
n1 cube, and
sweep
it through space perpendicular to the hyperplane in which it lies.
In
figure 1 we
show this operation for the first three
dimensions. We start with a point, at the origin, which
sweeps
out a line segment; the line segment then moves perpendicularly to its
length, and sweeps out a square; the square then moves perpendicularly
to its surface, and sweeps out a cube. We have a hard time
drawing anything higher than 3 dimensional figures so we stopped there.
By paying attention to what happens as we do this, we can learn what we
need to know about how an
n cube
looks. For clarity, we'll usually call 2dimensional faces "
2faces",
and n1 dimensional faces, "
hyperfaces".
We want to know how many vertices, edges, and hyperfaces an
ncube has; each of those is described by a "recurrence relation",
based on the number of vertices, edges, and hyperfaces an (n1)cube
has, and by paying attention to the details as we build a cube we can
see what those relations must be.
 We started with a
point (0 dimensions). A point has 0 edges, 0 "2faces", and 1
vertex. It also has zero hyperfaces. It has no
volume.

Figure 2:
"1cube"

We sweep the point through space r
units to form a line segment, or 1cube,
with length r. It has 1 edge, zero
"2faces", and 2 vertices:
 The single vertex of the point was
replaced with two new vertices
when it swept along: one (old) vertex at its starting
position,
and one (new) vertex where it ended. We observe that the
single
vertex was doubled as a result. In
general, we will see that an ndimensional hypercube will have 2^{n}
vertices.
 In
general, just as the single point formed two images
of itself  one
where it started sweeping, one where it finished sweeping  we would
expect any hypercube to form two images of itself
if it were swept
through space perpendicular to the plane in which it lies.
Thus,
each edge will form two "image edges", each face will form two "image
faces", and so forth.
 The edge was created as the vertex
swept along. It seems clear that every vertex will
form an edge when it's swept through space.
 The 1volume of the line segment is r,
the distance which was "swept out" by the point.
 The line segment has two "hyperfaces", which are its
vertices; they lie on the axis along which the point swept.

Figure 3:
"2cube"

We sweep the line segment through space
to form a square, or 2cube.
The square has four vertices, four edges, and four
"hyperfaces" (the edges).
 The two vertices on the line segment each "doubled",
giving us four vertices. The number of vertices is,
again, 2^{n}.
 The line segment "doubles", leaving an edge where it
started, and forming another where it ended.
 Each
of the two vertices of the original line segment has swept out a new
edge. Thus, we have four edges altogether: two for
each
edge in the 1cube, plus one for each vertex in the 1cube.
 The 2volume of the square is the distance swept
times the "1volume" of the 1cube, or r^{2}.
 The square has four "hyperfaces", which are its edges.
The line segment moved along the y
axis, and left two images of itself along that axis. The two vertices
of the segment each swept out new edges, and they lie on the x
axis.
We
see that in general, (n1)cube forms two images of itself, which
become two hyperfaces of the ncube. In other words, the
ncube is the volume swept out by one of its hyperfaces.
In addition, each hyperface of the (n1)cube sweeps out a new
(n1)cube, which in turns becomes a face of the ncube.
So, we acquired one hyperface for each hyperface of the (n1)cube, plus
two new hyperfaces:
Since
we're "sweeping" along each axis in turn, this means we have one pair
of hyperfaces along each axis; the total number of hyperfaces is
Figure 4: "3cube"

We sweep the square through space to
form a cube, or 3cube.
The 3cube has eight vertices, 10 edges, and six "hyperfaces"
(in the case of a 3cube, they're the ordinary 2faces).
 The four vertices on the square each "doubled", giving us
eight vertices. Again, we have 2^{n}
vertices.
 Again, each edge of the square "doubled", giving us two
new edges in place of each old one.
 In addition, each vertex on the square swept out a new
edge as it moved. Thus, the number of edges on the 3cube is
equal to the number of vertices on the 2cube plus twice
the number of edges on the 2cube:
 The 3volume of the 3cube is the area of the square, times
the distance the square moved, or r^{3}.
 The
square formed two faces for our cube: one where it started, and one
where it ended. In addition, each edge (hyperface) of the
square
formed a new square. So the number of hyperfaces on the
3cube is
6, which is, as we expected, 2n.
Summary
Reflective Symmetry Along the Axes
There is an additional
item here worth emphasizing, which is that it does not make any
difference what axis we start with. We can sweep a point along
any axis, then sweep the resulting line segment along any of the
remaining axes, and so forth.
In other words, the hypercube which results is symmetric in swaps of the axes.
Furthermore
 and related  it is symmetric in reflections through a hyperplane
which passes through the center of the cube, and which is perpendicular
to one of the axes. Less obscurely, the cube is
r units on an edge and we formed it by sweeping along each axis in turn, then if we
reflect the cube across the plane
x_{i} =
r/2, for any coordinate
x_{i}, it will be unchanged.
To see this, consider first the
last axis, along which we swept an (
n1)cube to form the
ncube. We have two images of the (
n1)cube "lying on" that axis: One at 0, and one at
r. But think about what happens on the
other axes during the final sweep: Each of the faces of the (
n1)cube, which is, itself, an (
n2)cube, will sweep out a
new (
n1)cube, which will become a face of the newly formed
ncube. But because of the way we performed the earlier sweeps, there will be a
pair of these along each axis, and we'll end up with a
pair of faces along each axis.
This is, of course, directly related to the fact that there are 2
n faces on an
ncube.
This
argument could be formalized into an induction proof fairly easily, but
we won't be depending on it in any formal proofs so I won't go through
that here.
Coordinates of the Vertices
Let's consider the process of "sweeping" which we used to construct the hypercubes.
We start with a single point in a 0 dimensional space, which has no coordinates.
We then introduce a single coordinate (the
x
coordinate), in which our point is at the origin; when we "sweep" it
along the newly introduced axis, we create a second coordinate
r units from the origin. So, we have the vertices:
(0), (
r)
Next we introduce a second coordinate (the
y coordinate), and place each existing vertex at
0 on the
y axis:
(0, 0), (
r, 0)
That's still a line segment lying on the
x axis. When we
sweep this segment
r units along the
y axis, we introduce additional vertices which have
y values of
r. We now have
four vertices:
(0,0), (0,
r), (
r,0), (
r,
r)
And
so it goes. With each "sweep" operation, we add a new "0"
coordinate to each existing vertex
and we add new vertices which have
the same
old coordinate values as the old vertices,
but have "
r" for the new coordinate value instead
of 0. For the cube, we thus obtain the set:
(0,0,0), (0,
r,0), (
r,0,0), (
r,
r,0),
(0,0,
r), (0,
r,
r), (
r,0,
r), (
r,
r,
r)
If we do this again, the vertices of the
4 cube
will be:
(0,0,0,0), (0,
r,0,0), (
r,0,0,0), (
r,
r,0,0),
(0,0,
r,0), (0,
r,
r,0), (
r,0,
r,0),
(
r,
r,
r,0),
(0,0,0,
r), (0,
r,0,
r), (
r,0,0,
r),
(
r,
r,0,
r),
(0,0,
r,
r), (0,
r,
r,
r), (
r,0,
r,
r),
(
r,
r,
r,
r)
In general, we see (and we can prove by induction) that every possible combination of "r" and "0" is a vertex.
There are two possible values for each coordinate, and
n coordinates, so the number of vertices must be 2
^{n}, which matches the value we already determined, above.
The Set Enclosed Within an nCube
Each time we "sweep" along an axis as we construct the cube, we start at 0 and move
r units. Thus, the length spanned by the cube on each axis is
r units, and it includes all points in that region.
Since we sweep along
each axis, all the points along each axis which fall between 0 and
r must be included in the cube.
So, the set of points inside the cube must be all those points for which all coordinates are less than or equal to
r. (This, by the way, shows that the faces must be where one or more coordinates take on the values
r, and from that, we can in turn deduce that the coordinates of the vertices must all be 0 or
r, as we already pointed out in the last section.)
Formally, we can say:
Or, if we wish to include the "skin", we could say the volume occupied by the cube is the set:
Page first posted on 11/04/2007