Background

My physics professor at Washburn University, Dr. Steven Black, conducted a thermodynamic experiment that wanted to investigate the properties of molecular movement in a 4-D cube, sometimes referred to as a hypercube. He told me that he had to figure out how 4-D cubes would stack on one another, because he broke the space into smaller 4-D cubes for the computer. In other words, he did the equivalent of breaking a large 3-D cube into 27 smaller cubes as shown below.

You can see that there are 3x3x3 cubes, or 27 cubes, making up one larger cube. He peeked my curiosity, because I have an affinity for patterns, and told him to give me some time, because I wanted to try to come up with the least number of 4-D cubes needed to create one larger 4-D cube. In fact, I made it my goal to come up with an equation that would give you the least number of cubes no matter how many dimensions are involved. I do need to mention that not being a true mathematician, this proof won't have what mathematicians like to call true rigor. In other words, it will be made without defining every little detail. Being a physicist at heart, we tend to leave the rigor to the mathematicians, and use math for our purposes, and use it in a way that quite frankly makes mathematicians cringe, but that is OK, because they need something to do, so we leave all of the technicalities to them.

You will also notice that I will use lower dimensions (such as a 2-D square) as an example to illustrate a point that applies to higher dimensions. This is not a new trick, and is often done to help better understand what is otherwise more difficult to visualize in higher dimensions. So, my proof relies on the fact that logic used in lower dimensions applies in higher dimensions, and why shouldn't it? Pythagorean's Theorem is applicable to a 3-D cube (To find the diagonal of a cube for example) and on up: all you have to do is add one term for each dimension you add.

IMPORTANT NOTE: I looked at this from the perspective of the center cube. In other words, in a familiar 3-D cube cut into 27 equal smaller cubes, there is a center cube that is surrounded by all of the others. Therefore, the number I wanted to generate from my equation was the total number of cubes that would surround the center. In the case of 3-D, that number is 26 (26 plus the center cube = 27 total).

Getting Started

In order to figure this out, I had to create a chart and generate the number of corners, edges, and faces a cube would have as you add more dimensions.

   Table One

A point

In order to understand extra dimensions, and how you move into higher dimensions (and therefore generate table one), consider the following brief commentary:

A point is the smallest denominator in that it has zero dimensions (Although our rendering of a point does have dimensions, in the strictest mathematical sense, it has zero dimensions). Now, I considerd it to have one corner in table one only for the fact that the total yields one, or one object. In other words, since it has no dimensions, you can't surround it by more points.

A Line

Now imagine taking that point and tracing out a line as shown below:

There is an important thing to note: by taking the pre-existing Dimension and moving it in a perpindicular direction (which isn't clearly defined with this exampe since a point doesn't have any dimensions, but it will be important with the rest of the steps), a new dimension was created. A line has one dimension, because you can move left or right, but that is all. In table one, the line has two corners, and the edge has an ED. That stands for Extra-Dimension, because in a sense, you can't tell the line has an edge unless you were looking at it from the top, which is in an extra-dimension from the line's perspective (Although the line doesn't technically have an "edge", it is its own edge in a manner of speaking).

A Square

In order to create a square, all one has to do is move the line in a direction perpindicular to its right-left direction. Imagine if you will that you drag the line down, and in the process, leave a trail like the point left a trail to create the line.

Now you have an extra dimension (sometimes called degrees of freedom) to move: you have the left and right from before, but now you can move up or down on the page as well. Again, in table one, the square has 4 corners, 4 edges, and an ED face. the face is Extra-Dimensional, because you can't see the face unless you are looking down, or from the third dimension.

Note that in doing this excersize, the original shape is always there (in this case the original line, which is the top of the square), and a new shape is created when you stop "moving" it into the new dimension, namely the bottom of the square. In doing so, two new lines are created, which connect the original shape to the new shape, which is what gives us the other two edges for the square.

A Cube

In order to create a cube, all you have to do is take the entire square and move it into the third dimension, which is perpindicular to the already established two-dimensional space defined by the square. Technically, you would have to move the square off of the computer screen, but we'll use the trick to portray a 3-D cube on a 2-D surface.

Now we have one more degree of freedom, namely we can move left-right and up-down as before along with moving in-out of the screen (if it were truly 3-D). On careful examination, one can ascertain the numbers given in table one: 8 corners, 12 edges, and 6 faces.

Again, it is critical to point out that in order to create this new dimension, the square was moved in a direction perpindicular to its own two dimensions. As you can might already tell, this will create a problem with the next shape.

A HyperCube

OK, all we have to do to create a Hypercube is take the entire 3-D cube, and move it into a direction that is perpindicular to the three dimensions possessed by the cube.

It's kinda tough huh? How does one move into a direction that is perpindicular to three dimensions? Unfortunately, you can't, at least not with our brains. Our brains have evolved in a 3-D world, so it shouldn't be surprising that we cannot imagine or picture moving into a fourth dimensions perpindicular to the three we are used to. You might even want be tempted to say it is impossible, but not so fast! Just because we are unable to draw or construct a Hypercube, doesn't mean it can't be done.

Mathematically, it is easy to represent any number of extra dimensions, which you might argue is just "mathematics", and not the real world, and you would be justified in saying so. It is true that sometimes math can describe things that are literally not of this world, but extra-dimensions may not be one of them. Our Universe may contain 10 spatial dimensions! This is of course predicated by the mathematics in String Theory, which is a competing Theory (and the term Theory is used here loosely, because a true scientific theory must be tested and proven) for the "Theory of Everything." Since we need to get back to the goal at hand, long story short: String Theory demands that our Universe have 10 spatial dimensions. That is, the three we are used to, and seven that are curled up so tiny, we can't see or detect them, but that doesn't exclude their existance, which means extra-dimensional objects are not just a fancy mathematical trick. If you are interested, there are several websites and books dedicated to this fascinating topic. But I digress.

As table one shows, a hypercube has 16 corners, 32 edges, and 24 faces. Where do those numbers come from since we can't "see" a Hypercube? Good question. It is simply a matter of thinking: if I take a cube and move it into a higher dimension, its 8 original corners will remain, and 8 more will be added (as you can see, the number of corners simply doubles with each move into a higher dimension). Starting with that, you can with careful thought deduce that there will 32 edges and 24 faces.

Let's start with the edges. You will have the original 12 edges (again, when you move the cube into the higher dimension, the original cube will be left, and a new cube will be created when you stop "moving" it into the fourth dimension, just like our line into a square example). And, just like there were two more lines to connect the original and new line in the square, there will be lines to connect the cube's original corners to the new cube's corners, which exist in the fourth dimension. Therefore, you double the number of edges (from the two "new" cubes), and you have new edges that connect the two cubes, and since there are 8 corners, there will be 8 extra lines, bringing the total to 12 + 12 + 8 32. I'll leave the faces to you figure out.

Before we move on, I need to establish a few definitions:

n = number of dimensions

Total = number of corners + edges + faces

The Proof

As shown in table one, all you need to do to figure out the number of surrounding shapes (as seen from the middle shape) you just have to add the corners, edges, and faces. For example, in order to totally surround a square, you would need 8 squares. And 8 is what you get when you add the corners, edges, and faces of a square. Notice how the Total for a cube is 26, which is precisely the number of cubes surrounding the center cube as previously established (this is where a mathematician would rigorously prove this to be true, but I'm relying on pattern recognition and logic instead). For example, in the case of the square, you could logically argue it would take 8 squares to totally surround the center. You know that there are 4 egdes, which means there has to be at least four squares up agaisnt those edges.

Now, notice how the center square is not completely surrounded. That's because we haven't accounted for the four corners that are exposed, which means we need four more squares to "cover" those corners, which yields:

The point is, there had to be enough squares to "cover" the edges and corners. You might say what about the faces, which is a fair question, since the square didn't have any. We can extend this logic to a 3-D cube that does have faces. Since there are 6 faces, we know we need at least 6 cubes to surround the center. Six cubes surrounding a center cube would create a 3-D cross.

Notice that there are 12 edges still exposed. The exposed edges are right where two of the outer cubes meet, and occur all the way around this t-shaped structure.

So, add 12 cubes to "cover" the edges. Now you will have:

Now there are 8 exposed corners, which you can easily see from the diagram. The corners occur where three outside cubes meet with the center cube as shown below,

The point is, the 8 corners of the center cube are not completely covered, so adding 8 more cubes to fill in the corners yields: 6 + 12 + 8 = 26

So, although not a rigorous mathematical proof, the logic shown for the 2-D square and 3-D cube should suffice to show that the Total in table one does indeed give the number of cubes (no matter what the dimensions) needed to completely surround and cover the center cube. So, the trick is to find an equation that gives us this Total, which is the addition of the corners, edges, and faces. That means we really need to come up with the equations for those three variables.

Let's start with the corners since they are the easiest of the three.

The Corners

Since the number of corners doubles each progression, and begins with the number one, C can be written as,

Equation (1) simply says that the corners for the current cube equals the previous cube's corners times 2, which simplifies to a 2 raised to the nth power. Try it: for a 3-D cube, n = 3, and 2 to the third power equals 8, which is the correct number of corners for a 3-D cube.

The Edges

Here is where some pattern recognition comes in handy. If you study table one, you will see that the number of edges is equal to adding the previous dimension's corners to 2 times the previous dimensions's edges given by equation (2):

Here is where we need some more pattern recognition, or the logic that was used above to determine the number of edges. The number of edges increases by adding all of the previous number of edges, but with each one multiplied by two. For example, The Hypercube's edges is equal to 32, which is 2*4 + 2*12 = 8 + 24 = 32. Now the trick is to represent that mathematically.

Although that looks daunting, it is simply a matter of representing a series, which is shown by the elipses (...). The elipses simply means keep going until you get to 2 raised to the n-n. So, if you start with n = 3, equation (3) would look like this:

Which would become,

And that is precisely the number of edges for a cube. So, although it looks hard, it is really quite simple when you break it down.

We can actually simplify this further by distributing all of the 2's in the second part of equation (3). I'm going to do it all in one step, although one could use several since there are so many paranthesis (you have to make sure you follow the correct order of operations). Also, I'm going to make use of the property that when you multiply like bases (when you look at the second part of equation (3), you can see that each term has a base of 2) you add the powers. When you distribute the 2's, equation (3) becomes,

Pretty cool how they all reduce to the exact same term! (If you are wondering, the last term in the second row has 2 raised to the n-n power, which is 2 raised to the zero power, which equals just one). So, the question is how many of those terms exist with each given number of dimensions, because there is still the matter of the elipses. Since the distribution of 2's ends at n-1, that means the second term will always have n-1 number of terms. Add that to the first term and you get precisely n number of terms.

Almost there! One more to go: the faces.

The Faces

The number of faces can be found by adding the previous edges to 2 times the previous faces given by equation (5),

This one is going to be a little more complicated than the last, but the same logic is used. First, the previous number of edges is found by taking equation (4) and subracting 1 from each n, which becomes the first term in equation (5). By using the same logic process in the last section, we can show the previous number of faces, which equals the previous edges plus 2 times the previous faces (previous meaning two shapes from the current shape). For example, the cube has 6 faces. Take the previous edges, which is 4 and add that to 2 times the previous faces (although it is ED, it is still one), which is 2. 4 + 2 equals 6.

Since the last example had an ED, we can use the Hypercube as an example. The previous number of edges is 12, and add that to 2 times 6, you get 12 + 12 = 24. Table one confirms this is the correct number of faces for a Hypercube. Expressing that in a mathematical series yields,

Here's another place where I'll probably cause the mathematician to pull his/her hair out, but notice how the second term ends with 2 raised to the n-n power again. Notice too that each power of the 2's decrease by one: n-3, n-4, and so on. The n-n power at the end (and after the elipses, because that means in general we don't know what n is) only means you stop when the number that is in the power of the 2 equals the number of dimensions. Like I said, not very technical there, but it doesn't matter, because another magical thing will happen when we distribute the 2's in the second term of equation (5).

But first, to show that equation (5) does give the correct answer, consider n = 4, or a Hypercube. The answer should be 24.

Ok, so now let's distribute the 2's in the second term,

Notice another magical thing (the beauty of mathematics): every term has a 2 raised to the n-2 power! That means we can factor that out and get,

Equation 8 shows a summation formula. The terms inside the brackets actually reduce to 1 + 2 + 3 + 4 and so on, depending on the number of dimensions, or n. So, I'm going to make use of the famous summation formula Gauss supposedly used when he was in grade school to quickly add the numbers from 1 to 100. It looks like this in standard form:

That formula is used to add a succession of numbers. For example, if n equals 100, the formula gives an answer of 5050, which is the sum of the first 100 numbers starting at 1 (i = 1) and ending on 100 (n = 100). Instead of using n though, I'm going to use (n-1) since the summation begins with 1 and ends with (n-1). Therefore, equation nine would become,

Equation 8 then becomes,

As you can see, the top can easily reduce, and there is a 2 in the denominator. When combined with the 2 raised to the (n-2), the power becomes (n-3), because you subtract powers when dividing, so it would be (n-2) - 1, which is n-3.

We now have all three formulas! Let's put them together to make the Total which will be combining equation (1), (4), and (11):

I can easily combine the first two terms by factoring a 2 out of the 2 to the nth power. If I do that, the power would then become (n-1). Think of it like I took away one of the two's, so you have to subtract one from the power.

By doing that, you can see that the first and second term both share a 2 raised to the (n-1) power, so they can combine, and we are left with two terms. Now, combining that with the faces part of the equation is not as simple. The first step will require factoring a 2 to the negative one power from the first term, and a 2 to the negative three power from the second term. Doing this will cause the 2 raised to the n-1 to reduce to 2 to the nth power and the 2 raised to the n-3 to also reduce to 2 to the nth power. Since the new 2's have negative exponents, they move to the denominator, and the second 2 needs to be cubed. The details are shown below.

Almost there! All we need to do now is multiply the 2 in the denominator of the first term by 4 in order to make like denominators, but in doing so, we have to multiply the top by 4 so that we aren't changing the term itself (multiplying by 4/4 is multiplying by one). Once the denominators are the same, we can combine the two terms. After a few more steps of rearranging the numerator (factor out a 2 to the nth power and combine like terms). The details are shown below.

The last step moved the 8 back up with the 2, which changes the power back to n-3, and there is the equation! It never ceases to amaze me how you can go from a table of values and following the rules of mathematics, develop an equation that has the power to calculate the number of cubes needed to surround the center cube (thus creating a larger cube) with any desired dimension. If we plug in 3, the equation should give us 26. The following shows the answers with n equalling 3, 4 and 5. (Note: the equation works for n greater than or equal to 3. An n of 2 (a square) gives an answer of 9, which is one more than the prevously established 8 squares needed to surround the center square. The reason is the square's face is not considered since it is only two dimensional, although it is interesting that one more square on top of the center square would technically fulfill the rules. The equation was meant to apply for 3-D or greater cubes)

As you can see, the answers agree with the Totals in table one. The last calculation with n = 10 was used in my science fiction novel The Green Phantom. One of the characters describes the Universe as possibly having 10 spatial dimensions (As suggested by String Theory), but I took it one step further and suggested in the book that parallel Universes could be stacked like cubes, and since it is 10-dimensional, one Universe (if it were the center one) would be surrounded by 17,644 Universes. Although not correct, I thought it would be cool to use the equation and its result in the book.

I hope you enjoyed this little proof,

Timothy K. Loper, Ph.D. Curriculum & Instruction