4 dimension hypercube -- identifying its 8 "three dimensional sides faces)" Was reviewing a problem in 3rd edition of Gilbert Strang's Linear Algebra book.  It asks how many vertices, edges, and faces in a 4d hypercube.
I can understand the vertices (16) and the edge calculations (32).  But I cannot understand what they mean by "3-dimensional sides (faces)", of which there are eight.
What I mean is I cannot visualize what these 3d faces are on the typical drawing of a hypercube (the exploding cube)
 A: In a polyhedron, a face is a subset of the edges (or vertices) that form a polygon.
Similarly, a 3d-face is a subset of the edges that form a polyhedron.
In the 4d-hypercube (or "exploding cube"), you have 8 cubes.
To see that they are eight without having to draw it (which is hard and not very intuitive), you can label each vertex of the hypercube by a vector of $4$ bits. Two vertices are linked if they differ by exactly one bit (like $0010$ and $0110$). Then, if you fix one of the coordinates to $0$ or $1$, you are left with a cube when the other coordinates vary. That is why you have $8$.
You can also look at the 2d-faces of a 4d-figure if you want, or for any $n$-dimensional solid, you can look at its $k$-dimensional faces for any $k<n$.
A: Consider by analogy how we count the 2-dimensional faces of a cube, but based on looked at a cube as an "extruded" square, i.e. how we produce a cube by starting with a square, and "stretching" it along a unit interval in a perpendicular direction.
So where do we get each of the 2-dimensional faces from? Well, there are two types of 2-d faces that will arise. There are those that will be the ends of the extrusion, i.e. two copies of the original square. So we get two that way.
However, we also have another source: Any edge of the original square, when we stretch it, becomes a new square itself. Since there are four of these, this yields four additional squares, for a total of six.
So let's apply this to a 4-cube. As above, you can get a 4-cube $Y$ by starting with a 3-cube $X$ and "stretching it" along a unit interval $I = [0,1]$. That is, 4-cube will be $X \times [0,1]$. So we get two 3-d faces by looking at $X \times \{0\}$ and $X \times \{1\}$, the two ends of the cube.
However, we also get 3-d faces by taking any 2-d face (of which there are six, recall), and stretching it along the interval. So this yields a total of 8 3-d faces.
