NUMBER THEORY AND QUADRATICS ON THE PYTHABACUS

THE PYTHAGOREAN TRIANGLE

REAL NUMBER AXIOMS

THE PYTHAGOREAN THEOREM

FACTORING QUADRATICS

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THE PYTHAGOREAN TRIANGLE AND THEPYTHAGOREAN ABACUS

The numerical array to the left may betermed the Pythagorean Triangle. Ineach row the sequence of numbersascend and descend about the numbersin the central column, these are thecounting numbers starting at the topwith one.The mid-sequence number of each row isthe first occurrence of that number in thearray. All subsequent occurrences of thatnumber will fall in diagonal columnsdescending left and right from this mid-point.Think of the sequence of numbers ineach row as a special instance of a group

of sequences. These all begin with one and ascend to a given counting number,which is repeated (n) times, then the sequence descends to one. A diagonal columnconnects the array rows with ascending and descending numbers of sequences with(n) greater than one, the number of times the mid-sequence number is repeated. Thesum of the numbers included in these sequences will equal the product of the mid-sequence numbers of the connected rows. In the special case where (n) equals onethere is no diagonal displacement of the sequence, it is included completely in agiven row, and the sum of the sequence equals the mid-sequence number times itself, it's square.

PYTHABACUS

It so happens that the sequencesrevealed by the array of thePythagorean Triangle map therectangular arrays of the PythagoreanAbacus. As seen here to the right, therows of beads on the rods of therectangular array revealed betweentwo triangular arrays, with bases ofthree beads, corresponds to the row ofnumbers in the Pythagorean Trianglewith the mid-sequence number three.

To the left it is shown on the abacusthat the rows of beads of therectangular array correspond to thesequence formed by connectingascending and descending numbers ofrows with mid-sequence numbers fourand six with a column offours.

The numbers of these sequences can beconfigured and added together to revealodd and even sequences. The arrows inthe figure to the right indicate thatnumbers should be configured in rollsbeginning in the top roll placingnumbers left to right, from one untilreaching the mid-sequence number, thencontinuing in subsequent rolls right to left, through the sequence back to one.Adding the columns contracts the sequences to odd or even sequences. Suchsequences will begin with a number equal the number of times the mid-sequencenumber is repeated. For example in the first sequence the three occurs only onetime, so the contracted sequence begins with one. In the next sequence, the fouroccurs three times so the contracted sequence begins with three. In the last sequencethe three occurs two times so the contracted sequence begins with two. Theextension of the contracted sequence will equal the mid-sequence number. Thus thecontracted sequence can be determined with out the above configuration.

The contracted sequences of the Pythagorean triangle can be revealed by pushing each bead in turn along the indicated rod in the indicated direction. Angled arrays are revealed with numbers of beads corresponding to the numbers in the contracted sequences.

The arrays above demonstrate that every product may be expressed as a sequence of odd or even numbers that will begin with a number equal the number of times the mid-number of the Pythagorean Triangle sequence is repeated and with an extension equal the mid-number.

AXIOMS OF THE PYTHAGOREAN ABACUS

CLOSURE: If (a=4) is the number of beadsin the blue section, on the base rod of theabacus pictured, the apex of an associatedtriangular array of beads (brown),designated (A), will be on rod (a=4) from thebase. If (b=3) is the number of beads in thered section the apex of an associatedtriangle of beads, designated (B), will be onrod (b=3) from the base. Let (ai) be theordinal number a bead in section (a) liesfrom section (b) and (bi) the ordinal numbera bead in section (b) lies from section (a).Then a section including adjacentsubsections of (a) and (b) bounded by (ai) and (bi) will be associated with atriangular array with apex on rod (ai+bi) from the base. For example, if we choosebead (a3) and bead (b2) as boundaries for adjacent subsections (a3+a2=5) beads willbe included in the subsections, and the apex bead of an associated triangle will be onrod (5). The total number of such apexes will equal (a*b) and form a rectangulararray between triangles (A) and (B). The separation of triangles (A) and (B) andisolation of this rectangle corresponds to the operation of multiplication on theabacus.

a b

C

A×B

AB

A

The orientation of the rectangular product area between the reversed triangles is also reversed but the number of beads included in the area is unchanged.

4 × 3 3 × 4

b=3 a

B

3 3A

2 2

DISTRIBUTIVE: A given section ofbase beads may be separated intonon-equal as well as equalsubsections. (Bi)=(Bi2)+(Bi3), thered plus purple base beads, in thepicture on the left. The number of beads in the rectangle between (A)and (B) equals [(a) x (b)]=[3 x 5=15]or [(a) x (Bi2+Bi3)] which for eachgroup of columns is(a)x(Bi2)+(a)x(Bi3)]=[(3x2)+(3x3)].The number of beads remainsunchanged no matter what thearrangement of non-equalsubsections of a section, or the distribution of columns in the rectangle between (A)and (B).

B

2 3A

9

6

Bibi

Ai

a 1

a n B

1

A

THE PYTHAGOREAN THEOREM AND THE PYTHAGOREAN

ABACUS

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The Pythagorean Theorem states that given a right triangle with sides (a, b, and c), as shown, (a² + b²) will equal (c²). In other words the area of the yellow and red squares will fit into or equal the area of the orange square. A two dimensional proof of this follows that may be taught to fourth graders.

First construct on each side of square (c) right triangles such that a larger square is formed with sides equal (a+b)

Next construct right triangles on the inside of square (c) in the manner shown to the right: make the upper left red, the lower left yellow and the remaining two orange.This will demonstrate that the area of (c²) equals four times the area of the original right triangle plus the area of a square in its' middle.

ab

c

a + b

ab

c

1

24

3

cc

?

14

32

a + b

Now outline a square with sides equal (b) in the upper right corner of the larger square and a square with sides equal (a) in the lower right corner of the larger square. This outlines the area (a²+b²). The Pythagorean Theorem states that this area should fit inside the (c) square. Well if we take a look at the orange area it is all inside of the (c) square and equals most of the area of (a²+b²) except for two blue right triangles, one in the upper right corner and one in the the lower right corner. It is easily seen that the area remaining inside the (c) square, the red and the yellow triangles, is exactly equal to the two blue right triangles. Therefore, the Pythagorean theorem is proved. 

a

b

DEMONSTRATING THE

PYTHAGOREAN THEOREM ON THE ABACUS

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Let a=2 and b=3 then the area of the the large square equals (2+3)²=5² . So multiply 5x5 on the

abacus.

It can be seen that the combined area of each two adjacent triangles constructed on the sides of (c²) equals the area of a rectangle with sides or factors equal (a and b). The checkered rectangle shows the area of these rectangles given a=2 and b=3). The square of beads on the abacus above can be arrayed to show these

rectangles in a similar manner.

The color coding shows which beads represent which area, three beads for each triangle, and the one bead remaining, in the middle, represents the area of the orange square. That the two rectangles on the right plus the middle square equals a²+b² is dramatically shown, when the middle bead, representing the middle square, is pushed to the right into the two rectangles.              

QUADRATICS ON THE PYTHAGOREAN ABACUS

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One of the most elegant math activities on the abacus is factoring quadratics. Lets begin with the equation X²+2X+1=25. First we multiply the square root of 25 times itself on the abacus, forming a rectangle with 25 beads between the triangles of the factors.

The last term of the polynomial is (1), so find the middle rod of the rectangle, it will have a number of beads equal the square root, and push the right most bead of the middle rod to the right to represent the last or right most term of the polynomial.

There are (4) beads remaining on the middle rod. If we push, to the left, the bead adjacent to the (1) bead we pushed to the right we will also push out a rectangle of beads, that will have a number of beads equal the square of the beads remaining on the middle rod, (4²=16). It happens that in this case (X) must equal (4) and (2X) must equal the beads remaining in the columns between the arrays we formed by pushing apart the beads on the middle rod, in this case 2X=8.

The three arrays will correspond one-to-one with the three terms of the polynomial. Therefore X²+2X+1=4²+2x4+1=16+8+1=25. In general for a polynomial of the form X²+b(X)+c, on the middle rod, simply push a number of beads equal (c) to the right and the beads remaining, on the middle rod, to the left. The beads remaining on the middle rod will equal X and the rectangle formed around the remaining beads will equal X².

Since we know that X=4 we can partition the root factors of twenty-five into X+Y. So in this case X=4 and Y=1. We can now see that the red rectangle, equal X² is the product of the red sections of base beads, equal X, and the single blue bead rectangle, equal Y², is the product of the blue sections of base beads, equal Y.  

The purple columns are each a product of one red section, and one blue section of base beads so equal 2(XxY). Therefore, (X+Y)x(X+Y)= X²+2(YxX)+Y². Thus we have the formula for the square of a binomial sum.

FINDING THE FORMULA FOR THE SQUARE OF A BINOMIAL SUM