Abacus Math Program – Lesson 10 – Five Pair Complement Part 2
In Lesson 9 Five Pair Part 1 we examined the 5 pair addition rule. Here in Lesson 10 Part 2 we are going to examine the 5 pair subtraction rule. In review there are only two 5 pair complements to consider: 4-1 and 3-2. Let’s take a look at the 5 pair subtraction rule in some examples such as 7-4, 6-3, 6-2, 5-1. In all of these subtraction examples there are not enough lower earthly beads touching the reckoning bar to subtract the numbers directly so we have to subtract using the 5 bead and add the 5 pair complement.
For example, consider 7-3. In this example there are only 2 lower earthly beads touching the reckoning bar so we can’t subtract 3 directly. Therefore, we will subtract 3 in the form of a 5 pair by adding 2, the 5 pair complement of 3, and then subtract 5. So in affect we are substituting the combination of (2-5) for -3.
1. Add the 5 pair complement of the number to be subtracted
2. Subtract 5 by moving the 5 bead away from the reckoning bar
Here we would like to note a bead movement efficiency consideration. In the 5 pair addition rule we added the 5 bead and then subtracted the 5 pair complement. By doing the bead movements in this order our finger is moving in one direction toward the bottom of the abacus. If we did the bead movement in the opposite order by subtracting the 5 pair complement with our index finger moving down, we would then have to change direction with our index finger to move up and then pull the 5 bead down. So instead of one movement we have 3 directional movements. The same is true for the order of the 5 pair subtraction rule. By adding the 5 bead first and then subtracting the 5 pair complement our finger movement is in one direction.
In this first example we start by adding 9 to the abacus on rod B. Next we add 7 to 9 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 for an interim sum of 16. Next subtract 4 from 16 by subtracting 4 from 6 on rod B. Since there is only 1 lower bead touching the bar we will need to use the 5 pair subtraction rule. First add 1, the 5 pair complement of 4, to 6 and subtract the 5 bead. The interim sum is now 12. Lastly add 6 to 12 by adding 6 to 2 on rod B for a final answer of 18.
In this example we start by adding 6 to the abacus on rod B. Next we subtract 3 from 6 on rod B using the 5 pair subtraction rule. So first add 2, the 5 pair complement of 3, and then subtract 5 by removing the 5 bead from the bar. The interim sum is now 3. Next add 7 to 3 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 3 on rod B for an interim sum of 10. Lastly add 4 to 10 by adding 4 to 0 on rod B for a final answer of 14.
In this last example we start by adding 9 to the abacus on rod B. Next subtract 8 from 9 on rod B for an interim sum of 1. Then add 4 to 1 on rod B using the 5 pair addition rule. Add the 5 bead and subtract 1, the 5 pair of 4, from 1 for an interim sum of 5. Lastly subtract 1 from 5 on rod B using the 5 pair subtraction rule. First add 4, the 5 pair of 1, on rod B and then subtract 5 by moving the upper 5 bead away from the bar. The final answer is 4.
