### A Double Find Day

**Post-Script:**In typing this piece out on my front patio in the warm autumn sun, the bottom of the first page came upon me faster than I anticipated, but the Quiet-Riter's rubber rollers didn't slip badly at all, permitting me to type within one line of the bottom edge of paper, further evidence as to the pristine condition of this machine.

For those unaware of abacus operation, each rod represents a column or place value for a number to reside. The machine is cleared when all the beads are moved away from the dividing bar, as is shown in the photos. Each bead in the upper portion is worth five of that place value if moved down to the bar, while each bead in the bottom portion moved up to the bar are worth one apiece. The Japanese termed these upper and lower regions "heaven" and "earth." Thus, numbers from zero to nine can be represented on each rod by the proper combination of upper and lower beads.

It should be noted that you can use the thumb and fingers of your hands to represent numbers using the abacus method, with each hand capable of registering a value up to nine, with the thumb worth five and each finger worth one. Thus, you can represent any number between zero and 99 using both hands, with the right hand the one's column and the left hand the ten's column.

Addition on the abacus is performed by first entering a number, then the second number is entered on top of the first, with the provision that, if insufficient beads remain to complete the calculation, complementary arithmetic is used, as a form of borrowing. Thus, to perform the problem of "4+3" the number four is entered by pushing up four of the one-value beads. Next, it can be seen that there are insufficient beads below the bar to add three more. Instead, a five bead is pushed down, and two one-value beads are removed, to yield the answer of 7. The way this is thought of is that the five's compliment of three is two, so instead of adding three, you would add five and subtract the complement of three, that being two.

Similar complementary techniques are used for the tens, as in the problem of "3+8" where three one-value beads are moved up to the bar, and then instead of moving toward the bar a value of eight (because there are insufficient beads left in the one's column to form eight), a single bead is moved up in the ten's column to the left and two one's beads are removed from the one's column (since the ten's complement of eight is two).

Problems can get further complicated when they involve a combination of both five's and ten's complements, as in the problem of "6+7" where first the number six is formed by moving toward the bar a five bead and a one bead. Then, to add seven, there are insufficient beads residing on the one's column to enter seven, and so a ten's complement operation needs to be performed by entering a one bead on the ten's column to the left, then subtract the three complement on the one's column. But you cannot directly subtract a value of three, since there resides only a five and a one bead. Instead, the five is subtracted and the five's complement of three - which is two - is added in that column to form the answer of 13. A simpler way of thinking about this was taught to me by a Taiwanese abacus instructor, who informed me that instead of thinking about the fives complement problem in that way, instead to think of "pushing up the number." Thus, to add the seven you would go ahead with the ten's complement step (adding a single bead onto the ten's column to the left) and then simply push up the five bead and two one's beads in one single operation, which amounts to pushing up a value of seven.

This is much harder to describe in words than to show visually, and so perhaps a future blog article will include either step-by-step photos, or a video could be produced to show the operation more clearly. In the meantime, check out the Soroban Abacus Yahoo discussion group for more information, or do an online search for more information.

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