### Ibn al-Banna's proofs based in the 'squaring' rule

In *Lifting the Veil,* Ibn al-Bannāʾ follows the order of the contents of his *Condensed Book*. The "squaring" rule is given in the chapter on multiplication, and in his commentary he gives two modifications of this rule, together with a third rule to go with them. These are stated in precisely the forms that will be applied in his proofs later in the book:

The first type of the types of multiplication, known as "squaring", deems it necessary that the surface of two numbers with a square of half of the difference between them be equal (*mithl*) to a square of half of their sum. And if one of them is multiplied by the other and is subtracted from a square of half of their sum, it leaves a square of half of the difference between them. And a square of half of the difference between them is clearly a square of the difference between one of them and half of their sum. [Ibn al-Bannāʾ 1994, 260.5]

The "surface" (*sa**ṭ**ḥ*) of two numbers is their product (see **Note 11**, below). In modern notation, Ibn al-Bannāʾ's rules are:

(1) \(pq+\left({\frac{1}{2}}|p-q|\right)^2 = \left({\frac{1}{2}}(p+q)\right)^2\)

(2) \(\left({\frac{1}{2}}(p+q)\right)^2 - pq = \left({\frac{1}{2}}|p-q|\right)^2\)

(3) \(\left({\frac{1}{2}}|p-q|\right)^2 = {\left|{\frac{1}{2}}s-p\right|}^2,\) where \(s\) is \(p+q\).

The "squaring" rule had already been removed from the context of finger-reckoning in the *Condensed Book,* and here it is even further alienated from its origin, both because of the ways it is rearranged, and because *Lifting the Veil* is a book beyond the interests and abilities of the majority of secretaries and merchants learning practical arithmetic.

Rule (1) will be used for the proofs of the type 4 and 6 equations, while rules (2) and (3) will be used for the proof of type 5. Incidentally, rule (3) serves to convert rule (2) into the arithmetical version of Euclid's *Elements* proposition II.5, though Ibn al-Bannāʾ does not mention the connection with Euclid.

Note that these rules do not belong to algebra, but to arithmetic generally. We should already suspect this because they appear in the chapter on multiplication, before Ibn al-Bannāʾ introduces algebra. But there are other critical indications, the most obvious being that no unknown or undetermined number in these rules is given a name. Numbers are instead referred to by words and phrases like "one of them" and "the difference between them". For my notational versions I took the step of introducing algebra by naming the two numbers \(p\) and \(q\) and then constructing compound expressions from them. There is nothing of that in the Arabic. Additionally, none of these rules is written with the algebraic verb *ʿadala* ("equal").

Later, in the chapter on algebra, Ibn al-Bannāʾ gives his two sets of proofs. One feature of these proofs that sets them apart from the proofs in other authors is that instead of explaining one path toward obtaining the root or thing, he shows how the rule and the equation give rise to two simple equations, one of them type 1 (\(ax^2=bx\)) and the other type 3 (\(bx = c\)).

Ibn al-Bannāʾ runs through his arguments in general, without reference to particular equations. His directions are terse, so when Ibn al-Hāʾim reproduced them in his 1387 *Commentary on the Poem of al-Yāsamīn* he also worked them out in terms of sample equations. (See **Note 12**, below.) I will do the same, and with the same examples. Ibn al-Bannāʾ begins his proofs (see **Note 13**, below):

You can understand the cause (*ʿilla*) of the procedures for the composite types from the multiplication by "squares" that we mentioned before. You always make the two multiplied numbers the number and the *māl.*

Unlike the proofs in his algebra book, the arguments here take place in the context of one of the arithmetical rules. In each of the three proofs the two numbers of the rule (my \(p\) and \(q\)) will be the number (the constant term) and the *māl,* and the equation serves only to make substitutions. The operations called for in the rules are worked out, and the resulting expressions are then "confronted" to form new equations.

For the type 4 proof, Ibn al-Hāʾim and I follow Ibn al-Bannāʾ's steps for the standard example dating back to al-Khwārizmī: "a *māl* and ten roots equal thirty-nine", or in modern notation, \(x^2+10x=39\). Unlike the rules, the equation is stated in terms of a named unknown (the "thing") and with the verb *ʿadala*. Ibn al-Bannāʾ will work in the context of rule (1), and in our example the two numbers are 39 and \(x^2\). It does not matter which is which.

Then the difference between them is the things in the fourth type.

Their difference is \(39-x^2,\) which according our type 4 equation is "ten things", or \(10x\).

You multiply one of them by the other, giving *māl*s \(\left[39x^2\right],\) and you add to them a square of half of their difference, giving *māl*s [adding \(25x^2\) gives \(64x^2\)]. This is a square of half of their sum. You take its root, giving half of their sum, which are things \(\left[8x\right].\) So keep it in mind.

We have now calculated \(pq+\left({\frac{1}{2}}|p-q|\right)^2\) as \(64x^2\), whose square root is \(8x\). By rule (1) this should be equal to \({\frac{1}{2}}(p+q)\), which he now calculates:

Then you direct your attention to half of their sum to get a *māl* and half of the things which are with it, since the number is equal to the *māl* and the things. Adding the *māl* gives two *māl*s and the things, and half of that is a *māl* and half of the things.

Adding an \(x^2\) to both sides of \(x^2+10x=39\) gives \(2x^2+10x=x^2+39\), which is their sum. So half of the sum is also half of the left side, which is \(x^2+5x\). He has now computed the two parts of the rule as algebraic expressions, and with them he sets up equations:

Confront it with the remembered amount, leaving things equal a *māl,* which is the first type.

A common way to express the setting-up of an equation from two algebraic expressions was to "confront" (*qabila*) them. By this act the equation \(x^2+5x=8x\) is established, which simplifies to the type 1 equation \(x^2=3x\). Solutions to simple equations are transparent, so there is no need to explain them.

And if you want, add half of the things to the remembered amount, to get things equal the number, and that is the third type.

Adding \(5x\) to both \(x^2+5x\) and \(8x\) results in \(x^2+10x\) and \(13x\), and from the original equation we know that the first of these is 39. So one confronts them to set up the type 3 equation \(39=13x\).

Ibn al-Bannāʾ next proves the rule for the type 6 equation, also using rule (1). I explain it in terms of Ibn al-Hāʾim's example \(x^2=3x+4\). Here the two numbers of the rule are 4 and \(x^2\). Again, it does not matter which is which.

Work out the sixth type similarly, except that half of their sum is all of the number and half of the things which are with it.

Adding 4 to both sides of the equation gives \(x^2+4=3x+8\), so \({\frac{1}{2}}(x^2+4)\) is \(4+{1\over 2}(3x)\).

This is equal (*mithl*) to the things resulting from the remembered root.

Since the product of the numbers is \(4x^2\) and the square of half their difference is \(\left({1\over 2}(3x)\right)^2\) or \(2{1\over 4}x^2\), their sum is \(6{1\over 4}x^2\). The "remembered root'' is then \(\sqrt{6{1\over 4}x^2}\), or \(2{1\over 2}x\). So \(4+{1\over 2}(3x)\) is equal to \(2{1\over 2}x\). By the particular word used for "equal" (*mithl*) we know that this is not intended as an algebraic equation, but is an equality still in the realm of the arithmetical rule.

So if you add half of the things to what you remembered from the things, your result is the first type.

Adding \(1{1\over 2}x\) to both, substituting \(x^2\) for the \(4+3x\), and confronting the two parts gives the type 1 equation \(x^2=4x\).

And if you subtract half of the things from the remembered [root], you get the third type.

Subtracting \(1{1\over 2}x\) from both gives the type 3 equation \(x=4\).

The type 5 equation is solved a little differently. I follow Ibn al-Hāʾim by explaining it in terms of the equation \(x^2+21=10x\). Now the two numbers are 21 and \(x^2\), though this time it will make a difference which is which.

And in the fifth type a square of half of the things is a square of half of the sum of the numbers.

In \(x^2+21=10x\) the square of half of \(10x\) is the square of \({1\over 2}(x^2+21)\). This is \(25x^2\).

You subtract from it the product of one of them by the other \(\left[21x^2\right],\) leaving a square of the difference between one of them and half of their sum.

Subtracting \(21x^2\) from \(25x^2\) leaves \(4x^2\), which by rule (2) is \(\left({1\over 2}|p-q|\right)^2\). But by rule (3) this is equal to \(\left|{1\over 2}s-p\right|^2\). So \(4x^2\) is \(\left|{1\over 2}s-p\right|^2\). Because \(p\) can be either \(x^2\) or 21, \(4x^2\) is either \(|5x-x^2|^2\) or \(|5x-21|^2\). And because we do not know which of \({1\over 2}s\) (\(5x\)) or \(p\) (21 or \(x^2\)) is greater, \(\sqrt{4x^2}\) might be \(x^2-5x\), \(21-5x\), \(5x-x^2\), or \(5x-21\). He considers the four cases in order:

If you take its root to get things \(\left[2x\right],\) and you add it to half of their sum, which is half of the things which are in the equation \(\left[5x\right],\) and you equate that with the *māl,* it results in the first type \(\left[7x=x^2\right].\) And if you equate it with the number [21] it gives the third type \(\left[7x=21\right].\)

And if you subtract [from the \(5x\)] a root of half of their remembered sum \(\left[2x\right]\) and you equate it with the *māl,* it likewise results in the first type \(\left[3x=x^2\right].\) And if you equate it with the number it results in the third type \(\left[3x=21\right].\)

In all four cases the word "equate" is conjugated from *ʿadala,* the particular verb used in algebraic equations. This act sets up the type 1 and 3 equations (written above in square brackets).

One might wonder how it is that these arguments were supposed to prove the rules for solving the three composite equations. The proofs read more like derivations of alternative methods, and no attempt is made by Ibn al-Bannāʾ to link his calculations with the different steps of the rules. We do know that he intended these to be proofs from his use of the word "cause" (*ʿilla*) in the beginning. This is the same word al-Khwārizmī and others used in their proofs. Based on other proofs in works of Thābit ibn Qurra and al-Karajī, my guess is that our author was just being exceptionally concise. The rules *can* be derived from these arguments—it just takes some work, just as it is not easy to decipher his line of reasoning without following along with sample equations.

**Note 11**. Use of the term "surface" (*sa**ṭ**ḥ*) for the product of two numbers is borrowed from Greek number theory, specifically from definition 16 from Book VII of Euclid's *Elements*. Heath translates the Greek term as "plane".

**Note 12**. Ibn al-Hāʾim's book is titled *Sharḥ al-urjūza al-Yāsmīnīyya,* and is published in [Ibn al-Hāʾim 2003]. He explains Ibn al-Bannāʾ's proofs beginning on page 83.

**Note 13**. Ibn al-Bannāʾ's Arabic text begins at [Ibn al-Bannāʾ 1994, 309.14]. In what follows for types 4 and 5, I copy from [Oaks 2018], with some minor adjustments.