The Bead Unbaffled - An Abacus Manual

Multiplying Negative Numbers (appendix) - Contributed by Nanami kamimura

I've examined Totton's notes on multiplying negative numbers. On reading it, I've discovered some tips, which I will show below.

The Positive Factor

Point 1: In the notes, on step 4, one must subtract the multiplier from the first digit(s) of the product. I've tried the example of -7 (3) x 9 = -63.

a. (3) x 9 = (27), the negative number at this point is -73.
b. Subtract 90 from 27

i. A B C
2 7
- 9 0 ===> ??? (can't be done)

ii. A B C
1 2 7
- 9 0
3 7

With the numbers of the soroban showing (37), the resulting complementary number will read -63, which is exactly the answer.

But what if we turn things around: 9 x -7 (3) = -63.

(Step a is done as above.)

ii. A B C
2 7
+ 3?
5 7 ==> doesn't work because it will not show -63

ii. A B C
1 2 7
- 7?
5 7 ==> this doesn't work either.

So I think, it is not the multiplier that is subtracted from the first digit(s) of the resulting product as step 4 mentioned. Instead, it is the positive factor to be subtracted from the resulting product for the problem to work.

Working with 9s

Point 2: What if the negative number starts with a 9? Unless it is a single digit or has zeros beside it (such as 90, 900, 9,000), chances are that the negative number's complementary digit is at least one digit short. For example, -93 is displayed as just (7) or -956 is displayed as (44). In multiplying, it will create problems when we go to step 4. Here's an example: -916(84) x 5 = -4,580

Step one: Place -916(84) on the soroban and 5 somewhere to the left.

Step two: Using Kojima's method for multiplication, the soroban shows (420) while the negative complementary number is -580. This is now where the entire setup gets confusing: where to subtract the 5, the positive factor. Although I would subtract five on zero, in which place I would subtract becomes a problem, especially if the complementary number is, for example -19993.

The Better Way

Here's a less confusing way. Instead of setting (84) to repsent -916, set (9084) instead; (9084) is also equal to -916 anyway, since 9's complementary negative number is zero unless it is the last digit.

Back to example from the beginning:

Step 1: Set -916(9084) on the soroban and 5 somewhere to the left.

Step 2: Using Kojima's method of multiplication, the soroban shows (45420), with the negative complementary answer would be -54580.

Step 3: Subtract 5, the positive number, from the *first digit* (in this case) of the product.

A B C D E F
4 5 4 2 0
-5         ===> ??? (can't be done)

A B C D E F
1 4 5 4 2 0
-5
9 5 4 2 0 ==> Its negative complementary number is -4,580, which is the answer.

Example 2 to further prove my point: -995(5/9005) x 6 = -5970

Step one: Place -995(5) on the soroban and 6 somewhere to the left.

Step two: Using Kojima's method for multiplication, the soroban shows (30) while the negative complementary number is -70. This is now where the entire setup gets confusing: where to subtract the 6, the positive factor. In fact, I can't locate which rod I'll subtract 6 from.

Instead of setting (5) to repsent -995, set (9005) instead, since it has the same negative complement.

Back to example from the beginning:

Step 1: Set -995(9005) on the soroban and 6 somewhere to the left.

Step 2: Using Kojima's method of multiplication, the soroban shows (54030), with the negative complementary answer would be -45970.

Step 3: Subtract 6, the positive number, from the *first digit* (in this case) of the product.

A B C D E F
5 4 0 3 0
-6         ===> ??? (can't be done)

A B C D E F
1 5 4 0 3 0
-6
9 4 0 3 0 ==> Its negative complementary number is -5,970, which is the answer.

The point here is: To make multiplying a negative number starting with nine (but not ending in zeroes) less confusing, add to the beginning of the positive complement a 9 and the same number of zeroes as the nines in the original negative number since doing so won't affect the original value. Examples -9678=(322/90322); -9983=(17/90017)

Two Negative Numbers

Point 3: What if two *negative* numbers are multiplied? How would one produce a positive answer?

Since two negative factors result in a positive product, we can just convert the negative numbers into positive numbers and multiply them. But if one wants to multiply two negative numbers without converting them into positive ones, here are the steps.

Example: -78(22) x -43(57) = 3,354

Step 1: Set -78(22) on the soroban and the multiplier -43(57) somewhere to the left.

Step 2: Instead of multiplying -78 x -43, multiply (22) x (57) instead. Using Kojima's multiplication method, we get 1254. That's less our positive answer.

Step 3: This step normally tells us to subtract the positive factor. But we don't have a positive factor. So instead, subtract the two complementary numbers taken from the multiplicand and multiplier. In this case, (22) and (57) are subtracted from the first two digits of the product. It doesn't matter which complementary number goes first; the result is the same.

i. A B C D E                             A B C D E
1 2 5 4                             1 1 2 5 4
-2 2      ==> ???(Can't be done)==>   -2 2
9 0 5 4

ii. A B C D E
9 0 5 4
-5 7
3 3 5 4   ==> This is the answer!

The disadvantage of this method is to remember both complements. So as insurance, after setting the multiplier on the soroban, place the multiplicand to its left as a memory aid.

- NANAMI KAMIMURA

August, 2007
Nanami Kamimura