This document refers to marbles and pompoms . Black rocks can be used for marbles and white rocks for pompoms.

The empty set or zero: no marbles, no pompoms.

**Addition:** plus equals

**Closure under addition:** Marbles added to marbles can only ever produce marbles. Addition is closed.

**Identity for addition:** That which when added does not change the number of marbles.

plus _____ equals The identity for addition is the empty set, zero marbles, or zero.

**Commutative property:** The order in which plus is done does not matter: swap the marbles around. No change in the result. Addition obeys the commutative property.

**Associative property:** Even with parentheses forcing a changed order, the result is the same. Addition obeys the associative property.

**Domain:** The domain is marbles.

**Marble multiplication:**
plus plus equals . Let this be written as 3 × equals . Note that multiplication is a number of repeats and in marble math is not necessarily a number of marbles.

**Closure:** Marbles repeated can only ever produce marbles. Multiplication is closed.

**Identity for multiplication:** That number of repeats which does not change the number of marbles. × _____ equals The identity for multiplication is a single repeat, or one.

**Commutative property:** The order in which multiplication is done does not matter: marbles and repeats are interchangeable. No change in the result. Multiplication obeys the commutative property. 3 × equals 5 ×

**Associative property:** With marbles the application of the associative property is less clear. Repeats of repeats would have to be illustrated. The commutative property suggests that operations involving only repeats is associative.

**Domain:** The domain is marbles, multiplication does not produce anthing other that marbles.

**Subtraction:**
minus equals

The above can also be read: five remove two is three. Here too is a subtlety: where plus is merely a simple union of all the marbles, minus involves counting the second set and removing the count number from the first set. This is a radically different and much more complex operation.

**Identity for subtraction:** The number of marbles one can take away without changing the number of marbles is zero, empty set. Zero is the identity for subtraction.

**Closure under subtraction:** Is subraction closed? This is best explored by examining systems such as:

minus is ?? In other words, three marbles take away five marbles. Three of the five can be removed, and then there are no more marbles to be removed. One option is to not allow this operation. Roman numerals have no way to express a solution to this problem, effectively forbidding the operation. Think, however, of borrowing where one owes money. One could denote three marbles minus five marbles as leaving a debt of two marbles owed. Pompoms or rocks of another color can be used to signify these debts. Each pompom is a single marble owed.

minus equals
where the pompoms signify that the "table" is owed two marbles. Pure marble math is NOT closed under subtraction, with the inclusion of markers for "marbles owed" (pompoms or other markers), subtraction gains closure. The desire for closure - important to solving equations - forces the creation of negative numbers.

Now Al'Mat can be explored with pompoms, where pompoms are marbles that are owed.
Pompoms can be added: plus equals . Put in terms of marbles owed, wwo owed marbles plus one owed marble results in the owing of three marbles. In number theory, a negative plus a negative is a positive.

plus equals (no marbles). Pompoms effectively "wipe out" equal numbers of marbles. Thus pompoms are anti-marbles and marbles are anti-pompoms. When they meet they destroy each other one for one.

This leads to a whole new set of rules. Pompoms plus marbles can be pompoms, marbles, or the empty set (no marbles). If the marbles outnumber the pompoms, then plus
produces marbles. If the pompoms outnumber the marbles, then plus produces pompoms.

**Replacing minus with plus:**

Note that the pompoms provide a way to eliminate minus altogether:

minus is can be rewritten as a plus statement with the same result:

plus is

Note too that the first of the two equations CANNOT be order reversed and yield the same result:

minus equals

But the second of the two equations can be order reversed and attain the same result:

plus is

plus can be order reversed, minus cannot. In math this means plus can commute and
minus cannot.

**Commutative property for subtraction:** Subtraction is NOT commutative.

**Multiplication with pompoms: ** Pompoms can be also be repeated, in other word, multiplied:

2 × equals

**Distribution:** 2 × [ plus ]
equals
[ plus ][ plus ]

which is the same number of marbles as:

This same result could have been attained by "dealing" the multiplication operation into the brackets:

[2 ×
plus 2 × ]
equals
[ plus ]
which is the same number of marbles as:

This "dealing" into the parentheses of the repeat count is called **distribution**.

Distribution can be used to get at a fundamental mystery: what happens if multiplying is done with pompoms as the repeat element? The outline of the argument goes as follows: plus is (no marbles)

If marbles are asserted to be equivalent to positive integers, then:

× [ plus ] (is still no marbles)

Distributing the marble results in the following: [ × plus ×
] (must still be no marbles for
consistency)

bear in mind that marble × marble is marble and pompom × marble (pompom
repeated once) is pompom (in order to generate the consistent result of no marbles).
This is still obvious but important: repeating pompoms yields pompoms. Bear
in the back of one's mind that what we are saying is that a negative times a positive is
negative. This now provides a way to do: [ plus ] ×
is no marbles which is

[ × plus
× ] must still be no marbles

but the front is just:

[ plus × ]
must somehow result in no marbles

therefore we had better choose to let
× be equal to !

Weird but true, pompoms × pompoms are marbles.

Those are the rule of Al'Mat. In modern terms:

- marbles plus marbles are marbles: pos + pos= pos
- marbles plus pompom are whichever is the greater numerically: pos + neg = pos or neg
- pompoms plus pompoms are pompoms: neg + neg = neg
- marbles × marbles are marbles: pos * pos = pos
- pompoms × marbles are pompoms: neg * pos = neg
- pompoms × pompoms are marbles: neg * neg = pos

**Division:** Just a few notes on division. Marbles and pompoms that are multiples of a number can be divided into piles of that number. More generally, however, division is not closed. Five marbles cannot be divided into three equal piles, not without busting up marbles. If one allows for marbles to be broken, then fractional marbles become possible. The desire for closure leads to division forcing the creation of fractions. Numbers that are ratios or otherwise known as rational numbers return closure under division.

Red yarn circles are variables containing an unknown, unspecified number of marbles or pompoms. Each red circle must contain the same number of marbles or pompoms.

If equals

To find the answer to an Al'Jabr problem with red yarn circles on one side and marbles or pompoms on the other, "deal" the marbles or pompoms into the red yarn circles. In Al'Jabr, since all red yarn circles contain the same number of marbles or pompoms, the answer can be abbreviated by giving the number of marbles in one red yarn circle.

Another type of Al'Jabr involves having marbles or pompoms on the same side as the red
yarn circles.

plus equals

By looking, one should deduce that one marble should be placed in each red circle. One
way to work towards this would be to find a way to "wipe out" the marbles on the
left side. But what are pompoms for anyway? Toss five pompoms into the left side [boom!],
toss five into the right side [boom!] and when the smoke clears one is left with:

is

Deal out the marbles and the result is the answer: a red yarn circle has one marble. Now we have a system for solving these Al'Jabr problems. Note that the system demands that what we do on one side of the "equals" we must also do on the other side of the "equals." We must be fair. If the left side gets five pompoms, then the right side must get five pompoms. Again, bear in mind that this is being done with real yarn and marbles in the classroom. A pen, pencil, or other object "separates" the two sides.

At this point various Al'Jabr problems can be done, including ones with pompoms:

plus is

The solution, and let the students work on this, can be arrived at by either taking three pompoms from both sides or by "tossing in" (plusing three marbles on both sides and then dealing the pompoms out. As you create problems, be sure they have an integer solution. Fractional solutions introduce complexities that are probably better avoided at this time. The solution above is each circle winds up with two pompoms.

Consider the Al'Jabr:

plus equals

Tossing in four pompoms to both sides and dealing marbles is one way to find the
answer. There is another for this problems: breaking apart. [Do this by physically sliding
the elements around on a table - anyway in which stuff can moved except across the
"equals" is legal because plus was commutative!]. The problem can be broken into
three problems:

plus is

plus is

plus is

Now just solve one of the Al'Jabrs.

Note that in the break apart, the red yarn circles, the left side marbles, and the right side marbles are all broken apart. This "break apart" process is why, when we multiply both sides by 1/3, we must "distribute" the one third, but when we add three to both sides we do not need to "distribute." Adding is simply tossing in three marbles.

Al Jabrs with red yarn circles on both sides:

plus
equals
plus

Just as marbles can be removed from both sides, yarn circles can also be removed from both sides. This leads to "yarn circles are removed from yarn circle and marbles/pompoms are removed from marbles/pompoms." Without actually saying so, the concept of "like terms" has arisen naturally from the system. Without realizing it, students are solving 3x + 5 = x + 8.

Negative" yarn circles could be deployed to handle negative x values. A negative yarn circle with marbles would be equivalent to the positive yarn circle with pompoms, a positive yarn circles with pompoms would be a negative circle with marbles. Solve the problem using the negative yarn circles, then make the change to a positive yarn circle. This might potentially prove useful for solving quadratics such as x²-8x+15=0.

Al'Jabr can be, in a limited way, extended to quadratic equations, but it is probably not productive. Here algebra tiles are probably a better option. The key is to use a new type of variable for the square term: a square grid. Perhaps an egg carton rearranged. The filling order of this new variable is very special. See the example below.

. | . | . | plus plus | is | |

. | . | . | |||

. | . | . |

Remove two marbles from both sides

. | . | . | plus | is | |

. | . | . | |||

. | . | . |

Start dealing the marbles in the following way. Note that I cannot place the marbles physically inside the yarn circles on this page.

. | . | . | plus | is | |

. | . | . | |||

. | . |

Next deal:

. | . | . | plus | is | |

. | |||||

. |

Note that the filling of the square variable must be done such that one always forms a square. The variable must also be considered an infinitely expandable grid. In a classroom one might lash together three egg cartons to create a six by six square variable.

plus | |||

The answer is the number in each yarn circle or the number of marbles along an edge of the grid. Note that this system is limited to marbular answers.

Students at the College may be able to solve 3x +5 = 17 correctly, but when faced with 5 + 3x = 17 will subtract the 3x from the 17 as their first step. I call this "positional memorization." The student has memorized, "subtract the second element from whatever is on the left side." Al'Jabr will let me say, "Can you remove yarn circles from marbles?" Al'Jabr makes concrete like terms.

Another common difficulty is distribution. In solving 3x + 6 = 18 it is perfectly legal to subtract six from both sides, subtracting that six from only the constant term on the left side of the equation. Once the students learn this, they then become confused when you suddenly insist that multiplying both sides by 1/3 must be applied to both terms on the left side. Why? Adding didn't require adding to both terms, so why should multiplication be different? Bear in mind, some of our students would accept 3x +6 = 9x as a perfectly legal statement. Some would add six to both sides of 3x+ 6 = 18 and get 9x + 12 = 27. Subtracting six from both sides could lead to -3x = 12 in some student's view.

The above adding of six can be explained as tossing in six marbles to both sides. Obviously marbles combine only with marbles. Now multiplying both sides by 1/3 can be equated to "breaking apart." The students now have a concrete, in-their-mind's-eye, view of what the operations are doing.

When I began the journey into Al'Jabr I had no idea whether it would work. The previous term a "transition to algebra" (~40% arithmetic/~60% linear algebra) class performing standard drill and practice from the textbook had experienced a failure rate of 74%. The department chair asked that this not occur again - but course repeat rates were hovering around 50%. Neither number was acceptable to me. I asked for wide latitude to try anything I chose to improve success rates.

While visiting Cordoba, Spain on vacation in 1994, I came to learn of Averröes, Ibn Al-Arabi, Maimonides, King Alfonso X, and Muhammed Al Riquti. The later founded a school at Murcia. It was through these people and in these places, according to what I learned in Cordoba, that the knowledge held in the middle East in the Muslim cultures passed into Europe. I was sitting in the Calahorra Tower when the concept of Al'Mat and Al'Jabr crystallized in my mind.

I used Al'Jabr to begin the first few weeks of PreAlgebra the next term. Al'Jabr acted as an advance organizer, if you will, for the students. Then the class moved into algebra and regular drill, practice, and textbook plus College standard chalk and talk. 60% of the class moved on to the next class, only 40% had to repeat the course. More importantly, I had a whole new language I could use when students made common distributional errors - errors that I saw less frequently than in previous terms.

I also did not see the "rapid extinction" of memorized knowledge so often seen in students in the math courses. The students appeared to be moving from pure memorization of unimportant features such as positionality and moving towards seeing mathematics as a language with a few simple rules.

A year later Al'Jabr was utilized by me in our recently minted lowest level course: MS 095 PreAlgebra. Here Al'Jabr came late in the term, after set theory and other explorations. In this course topics were not organized by mathematical complexity. Thus the world of golf ball algebra had already been done before Al'Jabr.

Notebook material originally created in 1994 by Dana Lee Ling. Edited in 2000 and 2017.