To Simplify, or Not To Simplify? A Lesson from Medieval Iraq: Fraction Simplification versus Clarity

Author(s): 
Valerio De Angelis (Xavier University of Louisiana) and Jeffrey A. Oaks (University of Indianapolis)

Consider the formula for the sum of the first \(n\) integers, \[\sum_{i=1}^n i = \frac{n(n+1)}{2}.\]

This formula is both well-known, and easy to remember: we take the product of the two consecutive integers that start at \(n\), and divide by \(2\).

On the other hand, the formula for the sum of the squares of the first \(n\) integers, \[ \sum_{i=1}^n i^2=\frac{n(n+1)(2n+1)}{6},\]

may also be well-known, but it is not quite as easy to remember. But we can remedy that as follows: \[ \sum_{i=1}^n i^2=\frac{2n(2n+1)(2n+2)}{24}.\]

If we do not simplify this fraction, we can memorize it as the product of the three consecutive integers that start at \(2n\), divided by \(24\) (or \(4!\)).

A similar phenomenon occurs if we look at the sum of the squares of odd integers: the formula

\begin{equation}\sum_{i=0}^n (2i+1)^2 = \frac{(n+1)(2n+1)(2n+3)}{3}       (1) \end{equation}

is more easily memorized as

\begin{equation} \sum_{i=0}^n (2i+1)^2 = \frac{(2n+1)(2n+2)(2n+3)}{6},     (2)\end{equation}

the product of the three consecutive integers starting at \(2n+1\), and divided by \(3!\). Similarly, the formula for the sum of squares of even integers,

\begin{equation} \sum_{i=1}^n (2i)^2 = \frac{2n(n+1)(2n+1)}{3},        (3)\end{equation}

is easier to remember as

\begin{equation} \sum_{i=1}^n (2i)^2 = \frac{2n(2n+1)(2n+2)}{6},        (4)\end{equation}

the product of the three consecutive integers starting at \(2n\), divided by \(3!\).

It should not go unnoticed that the verbal explanations of these last two reformulations are pretty much the same, and that they are more easily grasped than the notational versions. Taking the verbal explanations as a guide, if we call \(n\) the highest number, then the formulas become identical. If \(n\) is odd, then

\begin{equation} 1^2+3^2+5^2+\ldots+n^2=\frac{n(n+1)(n+2)}{6},   (5)\end{equation}

and if \(n\) is even, then1

\begin{equation}2^2+4^2+6^2+\ldots+n^2=\frac{n(n+1)(n+2)}{6}.    (6)\end{equation}

To express the sums this way we had to abandon our handy sigma notation, which students initially resist anyway. And it is here, having distanced ourselves a little from the confines of our notation, that the essential nature of the rules becomes apparent.

Before the advent of modern algebraic notation, rules like these were routinely expressed verbally. Taking a large step back in time to the latter 13th century, the Moroccan polymath Ibn al-Banna̅ ͗ explained the sum of squares of odd integers and the sum of squares of even integers with wording that reflects formulas (5) and (6). He wrote in his Condensed [Book] on the Operations of Arithmetic (Talki̅ṣ a ͑ma̅l al-ḥisa̅b) that adding the odd squares "is given by multiplying a sixth of the upper extreme by the product of the two numbers that come after it," and the rule for the even squares is identical: "multiply a sixth of the upper extreme by the product of the two numbers that come after it."2

It can be argued that a student with limited algebraic skills or notational experience would find the rules as described by Banna̅ ͗ easier to understand and remember than the two rules (1) and (2). The instructions "multiply a sixth of the upper extreme by the product of the two numbers that come after it" is more accessible than our \({1\over 6}n(n+1)(n+2)\) to a person who struggles with basic algebra. In fact, in Ibn al-Banna̅ ͗ 's time, these rules were considered to be part of arithmetic, not al-jabr wa l-muqbala, as algebra was called then. Ibn al-Banna̅ ͗ thus gives the rules in his chapter on addition, not in his later chapter on algebra. We may view the medieval rules as identical with \({1\over 6}n(n+1)(n+2)\), but in the latter we have taken the step of naming the upper extreme \(n\), and forming an algebraic expression to replace Ibn al-Banna̅ ͗ 's sequence of operations. Arabic algebra was a specific technique of arithmetical problem-solving, and was not used to state arithmetical rules like these.


Notes

 

1. This formula, of course, is also the sum of the first \(n\) triangular numbers.  By noting that consecutive triangular numbers add up to a square, one can compose simple proofs for the rules.

2. See [2, 43.4,9] (i.e., lines 4 and 9 of page 43), our translation. These rules are also found in several other medieval Arabic books.