Abacus Math Program – Lesson 9 – Five Pair Complement Part 1

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We are now half way through the Rightlobemath.com skill training program. So far we have learned to represent base 10 numbers on the abacus, add and subtract with 10 pair complements when the operation involves more than 1 rod. If you are not familiar with these concepts please go back through the online abacus math program lessons starting with Lesson 1. 

So far we have learned 3 out of the 6 total skill training abacus methods. Lessons 5, 6, 7 and 8 showed us how to manage operations that involve 2 rods by using the 10 pair complements for both addition and subtraction. Now we will introduce another complement method to handle the situation when we need to use complements on a single rod. The abacus, specifically the soroban, was designed for number representation efficiency and calculation speed. This is the reason the upper or heavenly bead has a value of 5. Due to the design choice of introducing the 5 bead for efficiency and speed the tradeoff was to introduce the need for another complement pair, the 5 pair. The 5 pair complements will help us handle operations involving the 5 bead but remain on a single rod, e.g. 2+3.

There are only two 5 pair complements to consider: 4-1 and 3-2. These are the only two pairs of numbers that add to 5.

Let’s first look at the 5 pair addition rule in examples such as 4+4, 4+3, 3+2, 4+1. In all of these addition examples there are not enough lower earthly beads to add the numbers directly so we have to add the numbers using the upper 5 bead. In the example 3+4, there is only 1 lower earthly bead available to add the 4. We don’t need to use the 10 pair complement addition rule because we have the upper 5 bead available to add the 4. So instead we add 4 in the form of a 5 pair complement by adding 5 and subtracting 1, the 5 pair complement of 4. In effect we are saying that adding 4 is equivalent to adding the combination of (5 – 1).

In the first example we start by adding 12 to the abacus placing 1 on rod A and placing 2 on rod B. Next we add 2 to 12 by adding 2 to 2 on rod B for an interim sum of 14. Then we add 4 to 14 by adding 4 to 4 on rod B. But here we notice there are no available lower beads to add 4. But there is the upper 5 bead available. So to add 4 to 4, we first add 5 pulling the upper bead to the bar and then subtract 1, the 5 pair complement of 4, from the lower 4 beads. So in effect we have added 5 but subtracted 1 so the net addition is 4. The interim sum is now 18. We show the 5 pair finger movements as two separate finger movements but they can be combined and both beads moved at the same time. Lastly we subtract 7 from 18 by subtracting 7 from 8 on rod B for a final answer of 11.

In this next example we start by adding 14 to the abacus first adding 1 to rod A and then 4 to rod B. Next we add 3 to 14 by adding 3 to 4 on rod B. Since there are no available lower beads we will have to use the 5 pair complement addition. So first add 5 on rod B and then subtract 2, the 5 pair complement of 3, from the 4 lower beads. The interim sum is now 17. Next subtract 8 from 17 by subtracting 8 from 7 on rod B. Here we will need the 10 pair subtraction rule. So first subtract 1 from 1 on rod A and then add 2, the 10 pair complement of 8, to 2 on rod B for an interim sum of 9. Lastly add 7 to 9 on rod B using the 10 pair addition rule. First add 1 to 0 on rod A and then subtract 3, the 10 pair complement of 7, from 9 on rod B. The final answer is 16.

In the last example we start by adding 4 to the abacus on rod B. Next add 9 to 4 on rod B by using the 10 pair addition rule. First add 1 to 0 on rod A then subtract 1, the 10 pair complement of 9, from 4 on rod B for an interim sum of 13. Next add 2 to 13 by adding 2 to 3 on rod B. Here we will need the 5 pair addition rule. So first add 5 and then subtract 3, the 5 pair complement of 2, from 3 on rod B. The interim sum is now 15. Lastly add 5 to 15 by adding 5 to 5 on rod B using the 10 pair addition rule. So first add 1 to 1 on rod A and then subtract 5, the 10 pair complement of 5, from 5 on rod B. The final answer is 20.

Next up Lesson 10 the Five Pair Part 2.