Al-Jabr

The word “algebra” comes to us from the title of a book, Hidab al-jabr wal-muqubala written by Abu Abdallah Mohammed ben Musa al-Khowarizmi (there are variations in the transliterations of both the title and the author) around AD 825. He was not the inventor of algebra; indeed, his book was a compendium and extension of known strategies, mostly from Hindu mathematicians. What made his book relevant was its breadth, comprehensiveness, and simplicity of presentation. Even so, the book is known to the West in large part because of its Latin translation, Liber algebrae et almucabala, by Robert of Chester, some three hundred years after the original book.

Furthermore this book does not contain the first widespread use of symbols for unknown variables; markers for the unknown had been used previously, and al-Khowarizmi does not use them here. Instead, examples in the book are worked entirely with words, numerals, and drawings. This makes it very difficult to read. So, despite its importance as a historical document, reading it requires a rare blend of a mathematician with historical interest and a keen understanding of language.

I am currently working through portions of Frederic Rosen’s 1831 translation of the original Arabic text. If al-Khowarizmi’s prose was not impenetrable enough, Rosen’s translation uses words that are not in common modern use: “Quadrat” for “square” and “quadrangle” for “rectangle” stand out. Rosen has attempted a loyal translation, providing some clarifications in footnotes but otherwise leaving the text to stand on its own. This is of course a valiant goal for a historian of mathematics, but for someone who wants to know what the text actually means to say, it’s problematic.

The al-Jabr begins by identifying three basic mathematical objects: Numbers, roots, and squares. His root is our unknown variable; his square is the unknown squared, and his number is our constant. These are further interrelated in ways that al-Khowarizmi describes at some length (to page 13 of the translation).

I shall leave that aside for the time being, because I’d rather get to the first example of an actual worked problem, which is \(x^2 + 10x = 39\). This problem is translated by Rosen as “a square and ten roots are equal to thirty-nine dirhems“.

benMusa1To solve this problem, al-Khowarizmi first draws a square. He then draws four rectangles around the square, and four smaller squares at each corner. The final image would be rendered in modern geometry as shown.

The square is represented by the red square in the middle, \(ABCD\); each side is the unknown value, and so the area is the square of that value (that is, \(AD = DC = x\) and the area of \(ABCD = x^2\)). He divides the ten roots into four equal parts, shown as the blue rectangles in the diagram. Each of these have side lengths of \(x\) and \(\frac{5}{2}\), for a total area of \(4 \cdot \frac{5}{2} \cdot x = 10x\).

He turns his attention to the four green squares. These are each squares with a side length of \(\frac{5}{2}\), giving an area of \(\frac{5}{2}\cdot\frac{5}{2} = \frac{25}{4}\). Since there are four of them, the total area is \(4\cdot\frac{25}{4} = 25\).

The original problem states that the total area of the blue and red rectangles is \(39\), so the total area of the big square is \(39 + 25 = 64\). Hence each side length of the square is \(8\). Put in geometric terms, \(FI = 8\), while \(FA = BI = \frac{5}{2}\), so \(FA + BI = 5\). Since \(AB = FI – (FA + BI)\), this means that \(AB = 3\), which is the positive solution to the original problem, \(x^2 + 10x = 39\).

benMusa2Al-Khowarizmi then proceeds to solve the problem again using a simpler diagram. In this case, the red square still has side lengths of \(x\), but this time \(FA = AL = 5\) (giving each rectangle an area of \(5x\), for a total area of \(10x\)). So the green square has an area of \(25\), and the total square an area of \(64\).

This is a process we now call “completing the square”. Al-Khowarizmi’s diagrams show fully why it is called that.

What particularly struck me about this solution is that it is entirely geometric; there is no “algebra” to speak of here. When al-Khowarizmi speaks of “squares”, he means literally that: A square with an unknown side length. “Numbers” are squares with known side lengths (the root of the number represented), while “roots” are rectangles with one side length known (a root of a number) and one unknown (a root of a square).

Naturally, this means that quadratics cannot have negative solutions. The solution in this case of \(x = -13\) is out of reach.

This geometric approach also means that problems that today seem obviously related need to be approached differently. al-Khowarizmi’s second example is \(x^2 + 21 = 10x\) (that is, “a square and twenty-one dirhems are equal to ten roots”).

benMusa3For this case, al-Khowarizmi creates a much different diagram.

Once again, \(ABCD\) represents the unknown square. The rectangle \(ADRQ\) has an area of \(21\). He draws \(FE\) as the midsegment of \(BCRQ\), so that the area of \(FERQ\) is half of the area of \(BCRQ\). We know this area to be \(5x\), and its height to be \(x\), so its width is \(5\).

He then creates the square \(FVSQ\), with a height of \(5\). He creates segment \(\overline{UT}\) so that \(UV = VE\), meaning that \(US = EF\). Since \(VE = ED\), that means that the blue rectangles (\(SUTR\) and \(EDAF\)) have the same area. Therefore, since the area of \(ADRQ\) is \(21\), so too is the area of \(FERQ\) and \(SUTR\), together.

We now have the square \(FVSQ\) with a side length of \(5\) and hence an area of \(25\). This means that the purple square, \(VUTE\), must have an area of \(4\) and hence a side length of \(2\). Since \(VE = ED\), \(ED = 2\). Since \(EC = ER = 5\), \(DC =5 – 2 = 3\).

So one of the solutions of  \(x^2 + 21 = 10x\) is \(x = 3\). In this case, al-Khowarizmi notes (without amending the diagram) that it could be that \(EF\) is actually inside of the square \(ABCD\), giving the second solution of \(x = 5 + 2 = 7\).

Because al-Khowarizmi is limited to geometric explanations, rather than what we would today consider algebraic ones, he has to go to greater lengths to prove relationships.

Al-Khowarizmi then explains the third situation, where it is the square that is by itself (in the example, \(x^2 = 3x + 4\)), which I will present in a separate post.

2 Comments

  1. Pingback: Three and a half methods for finding square roots – Curious Cheetah

  2. Pingback: Mathematics without Negatives | Algebra II

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