1. NUMBER THEORY AND QUADRATICS ON THE PYTHABACUS THE PYTHAGOREAN TRIANGLE REAL NUMBER AXIOMS THE PYTHAGOREAN THEOREM FACTORING QUADRATICS MENU

2. PYTHABACUS

6. 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.

7. 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.

8. a b C A×B A B

9. 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

10. b=3 a B 3 A 2

11. B 2 3 A 9 6

12. Bi bi Ai a 1

13. an B 1 A

14. THE PYTHAGOREAN THEOREM AND THE PYTHAGOREAN ABACUS Click or scroll down to begin

15. 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. a b c a + b a b c 1 2 4 3 c c ? 1 4 3 2 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

16. DEMONSTRATING THE PYTHAGOREAN THEOREM ON THE ABACUS Click screen or scroll down to begin

17. 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.

18. QUADRATICS ON THE PYTHAGOREAN ABACUS Click or scroll down to begin

19. 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.

20. 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