4. Distances Within Basic Units
Another interesting phenomenon to observe in relation to our ascending sequence of multi-dimensional cubes is the expansion of internal distances.
Suppose we construct a series of basic units, starting from one dimension and working up to x dimensions, with the condition that all line segments (1-D component parts) used in the construction of all basic units have a uniformly fixed length. For the convenience of discussion, let us determine this length to be one foot.
Starting with the one-dimensional basic unit, we have only one thing to measure - a one-foot-long line segment. At the 2-D level, however, we find a new feature to measure. All four sides of the square measure one foot each, but the distance from any one corner to a diagonally opposing corner measures more than one foot:
The exact length of this diagonal can easily be determined by applying the Pythagorean theorem: namely, the square of the hypotenuse of a right triangle is equal to the sum of the squares of the other two sides. Thus, if we designate the length of the diagonal by "d", we have:
12 + 12 = d2 or d = SQRT(12 + 12) or d = SQRT(2)
Therefore we find that a 2-D basic unit of one foot contains a maximum internal length of SQRT(2) feet - roughly 1 foot 5 inches.
Moving up to three dimensions we find that a new measurement is possible, namely, the distance from any one corner to a 3-dimensionally opposed corner on the other side:
This distance can easily be represented on a two-dimensional plane by constructing a right triangle which it forms within the cube:
Again applying the Pythagorean theorem, we discover its length d2 to be SQRT(SQRT(22) + 12 ) or SQRT(3). Therefore, a 3-D basic unit of one foot contains a maximum internal length of SQRT(3) feet - roughly 1 foot 9 inches.
When we come to the 4-D basic unit we are confronted with the obvious problem of being unable to visualize it as it exists in reality. We are forced to represent it with our 3-D model of two 3-D cubes in opposition:
|(3-D) cube 1||(3-D) cube 2|
Extending our concept of a maximum-length diagonal into the 4-D cube, we are intuitively led to represent it as a line extending from corner A of cube 1 to opposite corner B of cube 2. Looking back on the methods we used for finding these diagonals on our previous two models, we come to realize that this is indeed the maximum length diagonal for a 4-D cube.
While looking for the maximum distance within a square we realize that it must be from one corner to another corner, since the corners represent the largest possible distances from the center of the figure.
Starting from corner A, we have three possibilities: AB, AC, or AD. But since AB and AC are already defined lengths, namely 1 foot each, the only possibility left to explore is AD. Measuring it, we find it to be SQRT(2) feet long, which, together with BC, is indeed the longest possible distance found within the figure.
We followed the same reasoning while examining the 3-D cube:
Starting from corner A we have seven possibilities: AB, AC, AD, AE, AF, AG, and AH. Three of these, AC, AB, and AE, are automatically eliminated since they represent the pre-defined lengths of 1 foot each. Also, AD, AF, and AG are eliminated, since they each represent lengths of SQRT(2) feet. Is the last candidate, AH, longer than SQRT(2) feet? Measuring it we found it to be SQRT(3) feet long, the longest possible distance found in the cube (along with BG, DE, and CF).
Going back to the figure of the 4-D cube above, we apply the same method. Without laboring through all the fifteen different possibilities extending from corner A, we quickly see that the diagonal AB is the only undefined length present. All other lengths have been previously defined, since they represent either lengths of 1 foot, SQRT(2) feet, or SQRT(3) feet, as previously demonstrated, since AB is the only diagonal which cannot be contained within any of the eight 3-D cubes which comprise the 4-D supercube.
What, then, is the length of this diagonal? So far the lengths have been 1 foot (or SQRT(1) feet), SQRT(2) feet, and SQRT(3) feet. Following this apparent pattern, may we venture to guess that it might be SQRT(4) feet, that is, 2 feet? Such a conjecture would be very presumptuous indeed. Surely there must be a way of determining its length.
There is a method, and it is precisely the same as the one we used before: constructing a right triangle.
Let us construct a triangle within the 4-D cube: one side AB, another side BC, and the third side AC.
AC we know to be SQRT(3) feet long, since it is the maximum diagonal of the 3-D cube defined by the points A, C, D, E, F, G, H, and I.
A trial sketch of this triangle might look like this:
Now if we were able to demonstrate that the angle at C is a right angle, we could apply the Pythagorean theorem and find the length of d3 to be SQRT(SQRT(32) + 12), or 2 feet.
Is angle C a right angle? Let us examine the nature of the line segments AC and BC. We know that AC is contained within the 3-D cube defined by ADFEHICG. Below it is an identical twin, a 3-D cube of the same measurements, located in a 3-D space exactly parallel to the first 3-D cube. The connecting lines BC and all those parallel to BC represent the paths of extension from the first cube to the second. And, since we defined this extension to be one "straight up" into a fourth dimension, that is, perpendicular to the cubes described above, the line BC is by definition perpendicular to AC. In fact, BC is perpendicular to EC, DC, IC, GC, or any other line within that 3-D space which passes through C.
Therefore, the angle at C does measure 90° , and the length of d3 is indeed SQRT(4), or 2 feet.
In a basic unit of 5 dimensions, our method is exactly the same. Without going into all the details, the length of the longest diagonal contained within the 5-D cube is found to be SQRT(5) feet. This method of measuring maximum diagonals in progressively complex basic units might be illustrated as follows:
Connecting the right triangles we had constructed earlier, we create this "spiraling fan" effect, the spokes of which steadily increase in length according to the sequence SQRT(2), SQRT(3), SQRT(4), SQRT(5), SQRT(n), where n represents the number of dimensions contained in the basic unit.
Therefore we can conclude that in a basic unit of 9 dimensions there will be found a maximum diagonal three feet long. This is quite a startling fact, when we consider that we are dealing with cube-like figures, whose edges measure only one foot each. Imagine fitting a yardstick into a box whose edges measure only one foot each! But in space of nine or more dimensions this is a very real possibility. Moving rapidly up the scale we find a 100-D box containing the proverbial 10-foot pole, the equivalent of 3 1/3 yardsticks! Or even more fantastic is a complex 27,878,400-D box containing an object one mile long!
Here it is readily apparent that as dimensions increase, so does the possibility of distance within an enclosed space. It is at this point that we catch a glimpse of the tremendous complexities possible within spaces of higher dimensions, and of the limitations of the primitive 3-D space which surrounds us.
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