Revisiting Cubes - Visual and Data Structure Methods
by
Allen Klinger and Paul F. Stelling
Abstract
Visual proof of a proposition known to the early Greeks: The cube of every positive integer is a sum of consecutive odd integers. This proposition is shown to hold for all compound/non-prime positive integers that are either odd or divisible by 4. For any such number n, the unique smallest such sequence has size equal to the smallest prime factor of n. Finally, a method for deriving the number of such sequences for each n is given.
Introduction
In the first century A.D. one answer to the question "How can the cubes be represented in terms of the natural numbers?" was given as follows:
First statement: |
"The cube of every [natural] number is Cubical numbers are always equal to the sum of successive odd [natural] numbers and can be represented this way." |
This first statement paraphrases text [1] which cites Nicomachus [2]. We show here that structural concepts predominate over notation in enabling understanding why first statement holds. The first two cases expressed in the usual way show its meaning:
2^{3} = 8 = 3 + 5 (1)
3^{3} = 27 = 7 + 9 + 11 (2)
While the first statement strikes most people as unclear, Equations (1) and (2) suffice to support a persuasive visual proof. However, further questions arise with the next larger integer. One representation of it is as follows:
4^{3} = 64 = 13 + 15 + 17 + 19 (3)
In addition there are the following alternative representations:
4^{3} = 64 = 31 + 33 (4a)
= 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 (4b)
These representations suggest questions about the size and number of these sequences. The first visual proof shows the following elaboration on the first statement:
Second statement: |
"There may be shorter or longer lists of consecutive odds that sum to the cube of any positive integer but there always will be such a list the size of the integer itself." |
Although this second statement hints at how to derive a sequence of consecutive odds summing to a cube, it does not provide a method. Nor does it give real insight into the nature of the problem. For example, how would 7^{3 }be decomposed as the sum of consecutive odds? Are there decompositions of size other than 7? Concise mathematical notation enables terse representation of the value 343, but constrains thought. Visual reasoning [3] can be used to resolve the dilemma. One can see what cube means and use it to generate sums of consecutive odd decompositions for the cube of 7, 12, 17, 25, or any other natural number.
That the cube of any positive integer has a value that can be represented as the sum of two or more consecutive odd positive integers is easy to show visually. We begin by sketching how to do so for equations Equations (1) and (2), and then show how to extend the respective approaches for all even and odd positive integers. The approach is fundamental, based on data structures, and applicable to the following more general questions:
The answers to these questions can be found easily using the visual proofs.
This paper consists of five sections. Section 1 presents a visual proof of the original problem. Section 2 simplifies that proof using tables. Extensions and generalizations follow in Section 3, which answers the questions posed above. Section 4 presents an arithmetic proof of the extended problem, while Section 5 summarizes our results.
First Statement - Visual Basis
Formally
For any positive integer n > 1 the arithmetic cube n^{3} can be represented using a cube structure consisting of n´ n´ n blocks, each 1´ 1´ 1 in size, as shown in Figure 1:(1a) and Figure 1:(2a). The cube has n layers of n^{2} blocks. These layers can be numbered 1 through n, beginning with the bottom layer. Clearly, the number of blocks in each layer is even for even n, odd otherwise.
If n is even then the two middle layers are numbered (n/2) and (n/2 + 1). [E.g., if for n = 2 then these layers are numbered (2/2) = 1 and (2/2 + 1) = 2.] Removing one block from layer (n/2 + 1) and adding it to layer (n/2), as shown in Figure 1:(1b)û(1c), results in two adjacent layers with consecutive odd numbers of blocks. For values larger than 2, continue this process, removing (2i û 1) blocks from each layer (n/2 + i) and adding them to layer (n/2 + 1 û i) for each i from 2 to (n/2). Thus for n even, n^{3} is the sum of the n consecutive odd integers from (n^{2} û n + 1) to (n^{2} + n û 1), which can be verified by summing.
Similarly, if n is odd then the middle layer is numbered (n + 1)/2. [E.g., if for n = 3 then the middle layer is numbered (3 + 1)/2 = 2.] Removing two blocks from the layer above it [(n + 1)/2 + 1 = (n + 3)/2] and adding them to the layer below [(n + 1)/2 û 1 = (n û 1)/2] as in Figure 1:(2b)û(2c), gives three adjacent layers holding consecutive odd numbers of blocks. Continue this, taking 2i blocks from each layer [(n + 1)/2 + i] and adding them to layer [(n + 1)/2 û i], for i = 2 to [(n û 1)/2]. Thus for n odd, n^{3} is again the sum of the consecutive odd integers from (n^{2} û n + 1) to (n^{2} + n û 1). This also can be verified by summing.
Figure 1. Base cases for cubes as sums of odd integers: n=2 and n=3.
Visually or Informally
Once the images in Figure 1 have been drawn, the decomposition process has two parts. The key observation is that all positive integers are either odd or even one approach can be used for all the even integers, and another, only slightly different approach, can be used for all the even integers.
For even n, the image based on n = 2 suffices. The horizontal plane between layers (n/2) = 1 and (n/2 + 1) = 2 symmetrically divides the layers of the cube. The layer immediately above the plane is decremented by one to obtain an odd value in the decomposition. The layer immediately below the plane is correspondingly incremented by one. The process continues, decreasing and increasing the adjacent layers by two more until all layers are exhausted.
When n is odd so is the value n^{2}. The value n^{2} is in the sum the number of elements in the middle layer, and as such is an element of the sequence. Clearly it is just the number of elements in the middle layer. The layer immediately above is that value decreased by two., and the, The layer immediately below is incremented increased by two to obtain the next smaller and larger odd values in the decomposition. The process continues in an analogous manner to the situation case when n is even, that is, one proceeds to decrease and increase the adjacent layers That is, the adjacent layers are decreased and increased by two more until all layers are exhausted.
The visual proof is essentially an the observation that the Figure 1 representations of Equations (1) and (2), and similar representations for larger n, correspond to data structures. Further, tThese structures can be used to generate a sequence of consecutive odd integers that sum to any required cube. ThisThe demonstrations in Figure 1 for Equations (1) and (2) isform the base stepcases of an argument that is true for any n by mathematical induction. As a result,for any integer n these data structures provide athe basis for the algorithmic derivation of values. These values satisfy satisfying both the first statement as well asand the second statement for any integer n.
Second Visual Proof
We now note that the proof relies only on three facts:
The proof does not use the fact that each layer is initially configured as a square. Representing each layer as a row is useful, since it replaces the initial (threeûdimensional) cube by a simpler (twoûdimensional) array. The array consists of n rows with n^{2} blocks in each row. The number of blocks in each row is odd if n is odd, and even if n is even. The same approach to moving blocks can be used as when the blocks were arranged in squares, as illustrated in Figure 2.
Figure 2. Simplified representation of the base cases.
Note that this construction works for any positive integer n that can be represented as the product of two integers of the same even/odd parity. I.e., let the integers be r (rows) and c (columns), where r ≤ c and c and r are either both even or both odd. That is, begin with a data structure as in Figure 2:(1a) [Figure 2:(2a)], and end with one like that shown in Figure 2:(1b) [Figure 2:(2b)].
Clearly the construction applies not just for cubes, but for any positive compound integer n that is either odd or divisible by 4. As a result, every positive integer greater than one that can be represented as k^{i}, where k > 1 and i ≥ 2 are integers is either odd or divisible by 4. Thus, for such a k^{i} we can use k rows and k^{(iû1)} columns, since k ≤ k^{(iû1)}, and k^{(iû1)} is even iff k is even. So we have the following answer to a question posed earlier: decomposition into the sum of consecutive odd integers can be done for any integer power ≥ 2, not just for cubes.
Finally, Suppose consider any positive compound integer n that is odd or divisible by 4 as described abovewe have such an n. Then the smallest number of rows that can be used is the smallest prime factor of n. It follows that the smallest number of consecutive odds that sum to n is just the smallest prime factor of n. Similarly, the largest number of rows that can be used is the largest factor f of n that is smaller not larger than and such that f and n/f are either both even or both odd. Thus, the largest number of consecutive odds that sum to n is f as just defined.
It immediately follows that every positive integer greater than one that can be represented as k^{i}, where k > 1 and i ≥ 2 are integers, can be represented as the sum of consecutive odd integers. In this situation, we can clearly use k rows and k^{(iû1)} columns. (This holds since k ≤ k^{(iû1)}, and k^{(iû1)} is even iff k is even.)
There is one remaining question. remains: For any given positive integer n, how many such sequences are there? Let be the number of sequences of consecutive odd integers that sum to n. Based on the visual construction, it is easy to see that is the same as the number of suitable rectangles that can be formed with n blocks. I.e., the number of rectangles where the number of rows and columns are either both even or both odd, and where the number of rows is no greater than the number in each row. Clearly, the number of rows must be a factor of n. A factor of n is proper if it is not 1 or n. Let be the number of factors of n, and let be the number of proper factors of n.
For odd n it must hold that is the number of proper factors of n that are > 1 and . This number is . If n is even but not divisible by 4, then . If n is divisible by 4, then is the number of factors of that are . This number is . Note that is even unless n is the square of an integer. Also, is easily computed as the product of using the exponents of the prime factorization of . SpecificallyE.g., if the prime factorization of is, then . For example, the value 64, the cube of 4 (64=4^{3}), is both even and divisible by 4: . Hence, 64 can be represented as the sums of sequences of odd numbers. These sequences are of length 2, 4, and 8, and appear above in Equations (3) and (4).
An Arithmetic Proof
We assert that any composite positive integer n that is either odd or divisible by 4 can be represented as the sum of consecutive positive odd integers. Further, there is such a sequence of size p, where p is the smallest prime factor of n. This can be demonstrated by summing the consecutive odd integers over an interval from to . That sum is given as:
(5)
By using (5) it is possible to derive the results given in the previous sections, namely that:
However, the derivation is beyond the scope of this paper, Moreover, it would be more complicated and difficult to understand than the simple visual proof presented above.
Conclusions
We have demonstrated simple visual proofs that for any integer n ³ 2, n^{3} can be represented as the sum of consecutive (positive) odd integers. The extended visual proof based on array data structures led to further results. They were visual and arithmetic proofs that this property not only holds for all n^{i} where i ³ 2, but for all composite positive integers that are either odd or divisible by 4.
The value of the visual methods and their data structure representations is evident because they enabled further extensions. We used them to show two things. First, whenever n can be represented as a sum of consecutive (positive) odd integers, the shortest such sequence of odd integers has length equal to the smallest prime factor of n. Second, we showed how to computea method for computing the number of such sequences for any given n. These Those extensions were facilitated and made comprehensible by insights gained using the visual representations.
We believe that data Data structures and visual representations of mathematical concepts are essential useful tools. Our experiences have convinced us that visual proofs lead to facilitate mathematical understanding and. We hope that the concepts we presented here canwill be used elsewhere. We believe the visual and data structures approaches have value both for both mathematicians and lay peoplenon-mathematicians. We welcome further communication or discussion of the points raised here.
References
[1] Naps, Thomas L., Introduction to Data Structures and Algorithm Analysis, 2nd Edition, West Publishing Co., St. Paul, Minnesota, 1992, p. 50, prob. 4.
[2] Nicomachus, Introduction Arithmetica, first century A.D. (see [1]).
[3] Adams, James L., Conceptual Blockbusting: A Guide to Better Ideas, 3rd ed., Addison-Wesley, Reading, Massachussetts, 1986.
[4] Knuth, Donald Erwin, The Art of Computer Programming: Fundamental Algorithms (Vol. I), Addison-Wesley, Reading, Massachusetts, 1968.
[5] Graham, Ronald, Donald Erwin Knuth, and Oren Patashnik, Concrete Mathematics: A Foundation for Computer Science, Addison-Wesley, Reading, Massachusetts, 1988.