## Sunday, December 28, 2008

### A Different 'View' of Sums of Cubes? An Algebraic "Proof Without Words!"

A well-known and intriguing formula usually proved by Mathematical Induction states that
13 + 23 + 33 + ... +n3 = (1+2+3+...+n)2 .

In words:
The sum of the cubes of the first n positive integers equals the square of the sum of the first n positive integers (or the square of the nth triangular number).

Students as early as middle school can investigate numerical patterns of sums of powers of positive integers and can be led to such discoveries. However, in this post we will look at a different kind of "proof." Proofs without words can be fascinating, challenging and can develop a student's spatial reasoning. Just as there have been many visual proofs of the Pythagorean Theorem (dissection type), mathematicians have sought visual arguments for many other numerical patterns and algebraic formulas. The Greeks of antiquity developed many classical arguments of this type, necessitated perhaps by not having our symbolic algebra available.

You will surely find other examples of this on the web (e.g., "Cut-the-Knot") but I thought it might be nice to bring it down to a middle school or Algebra 1 level by having students play with some particular cases of this general formula. I have always been intrigued by this topic, ever since I saw several visual proofs of the Pythagorean Theorem. Later on I was introduced to the genius of Sidney Kung and Roger B. Nelson (Google them!). Prof. Kung's extraordinary visual proofs were (and may still be) a staple of Mathematics Magazine, an MAA publication. You may also recall I have published a couple of other such proofs, one of which came from a student of mine. Look here.

Part I
Let's try to demonstrate that 13 + 23 = (1 + 2)2

Before displaying the visual we will begin with an arithmetic-algebraic approach:

Think of (1+2)(1+2) as a special case of the form (a+b)(a+b):
Thus, (1+2)(1+2) = (1⋅1) + (1⋅2) + (2⋅1) + (2⋅2)
Now for some creativity. Since cubes involve a product of THREE factors, we can introduce an extra factor of "1" in each term:
(1+2)(1+2) = (1⋅1⋅1) + (1⋅1⋅2 )+ (1⋅2⋅1) + (1⋅2⋅2).

Even without a visual, we can see the first term on the right is 13!!
It will take some work to show that the sum of the other three terms is 23. Ok, with this background, here is a
PROOF WITHOUT WORDS Do you think your students will "see" the proof?? My crude attempt at a graphic leaves a lot to be desired! It may be helpful to have manipulatives such as algebra tiles available or have students physically build these models. I would encourage that strongly!

So we are proving a numerical formula using a sum of volumes. You might say we turned squares into cubes!!

Do you think this investigation is through? Of course not -- I did all the work for you. Now here is the real test:

Part II

Show that 13 + 23 + 33 = (1 + 2 + 3)2
using a "Proof Without Words."

Ok, I'll give you a little hint although you don't need to use this:
Rewrite
(1 + 2 + 3)2
as ((1 + 2 )+ 3)2

Have fun! Just think, if we have a sum of 4th powers, we might need hypercubes!

Anonymous said...

I remember a proof without words of this that is purely planar, though I don't remember where I saw it. The case n = 2 goes like this:

First, draw a square with sides (1+2) and (1+2), marking the various subrectangles.

Second, draw a 1 by 1 square in the lower-left of the figure. Then, draw a 2 by 2 square immediately to its right, and another immediately above it. In chess terms, first you have a1, and then b1, c1, b2, c2, and then a2, b2, a3, b3. These three squares almost fill the 3 by 3 square but leave a 1 by 1 hole in the upper right (c3), and then a 1 by 1 square (b2) is the overlap in the middle. Those two regions have the same size, and they cancel, leading to (1+2)^2 = 1 1^2 + 2 2^2,

For the n = 3 case, place 3 3 by 3 square around the figure in an L shape. (Well, an upside-down L.)

For the n = 4 case, place a vertical stack of n/2 (= 2) 4 by 4 squares on the right of the n = 3 figure. Place another horizontal stack of (n/2) 4 by 4 squares above the n = 3 figure. The two stacks overlap in a 2 by 2 square, but they don't quite fill the 10 by 10 square, leaving a 2 by 2 hole.

Dave Marain said...

That's beautiful, Eric. There's something elegant about representing n^3 as n(n^2) visually! This reminds me of the elegant but simpler 'proof' of the formula for the nth triangular number which I should also demonstrate visually. As you know, this also uses the idea of stacking up a triangular number (L-shaped) and interlocking it with a congruent shape but upside down. The two pieces form a rectangle leading to the famous formula. There must be many websites that show this (I'm sure it's in MathWorld and Cut-the-Knot). It would be great if there were a compendium of such Proofs Without Words so that we could look these up just as they have an encyclopedia of number sequences.

The method I used is probably out there already although I haven't yet found it. I was just playing around with it and it seemed to lend itself to an inductive argument both arithmetically and geometrically.

Thank you for this idea and Happy New Year!

Anonymous said...

Dave,

I think (but do not know!) that the diagram would be more effective if you were trying to show that:
(n+1)^3 = n^3 + 3n^2 + 3n + 1,
in other words, if n was relatively large compared to 1, the eye would focus on the "frosted parts" and the single vertex.

I like your use of color.

Jonathan

Dave Marain said...

You may be right Jonathan but I wasn't really considering the 'cube of a sum', rather I was constructing a 'sum of cubes.' I was actually trying to set up a visualization of the inductive step. Thus (1+2+3+4)^2 = ((1+2+3) + 4)^2 =
(1+2+3)^2 +(2)(4)(1+2+3) + 4^2. Assuming we have already represented (1+2+3)^2 as a sum of cubes, we need to show that the remaining terms can be re-formed to represent 4^3. This can be done numerically but the fun is to do this visually by actually assembling the pieces to form another cube. I have no idea what value this has. I was just playing around with it and thought it would be "fun" but then again my idea of fun is...

thaumkid said...

Just an aside, this is called Nicomachus's theorem. I found that at http://mathworld.wolfram.com/PowerSum.html

I thought that site had some pretty and "visual" demonstrations... =) Happy May!

ScaffoldedMath said...

If I posted a comment before, forget it. I made a major mistake. I took this proof to mean that the sum of ANY two consecutive integers is equal to the square of the sum of the integers. I failed to see that starting from 1 is necessary.

Can anyone show a way to prove this algebraically? I'd love to use it in class, but I'm not the best at proofs, especially ones where there is a set number (here, 1).

Dave Marain said...

Thaumkid -- thanks!!

ZeroSum--
One method of proof uses math induction but that may be above your student's level. There is a general strategy for deriving the sum of powers of integers. However, I left it in the margins of my notebook and I misplaced it!!

Seriously, when I have time I will do a video of this cool method which is related to the Taylor expansion of polynomials.