{"id":362,"date":"2024-02-04T13:04:13","date_gmt":"2024-02-04T21:04:13","guid":{"rendered":"https:\/\/www.rightlobemath.com\/blog\/?p=362"},"modified":"2020-02-04T13:27:46","modified_gmt":"2020-02-04T21:27:46","slug":"abacus-math-program-lesson-7-ten-pair-complement-subtraction-part-1","status":"publish","type":"post","link":"https:\/\/sumomath.com\/content\/abacus-math-program-lesson-7-ten-pair-complement-subtraction-part-1\/","title":{"rendered":"Abacus Math Program – Lesson 7 – Ten Pair Complement Subtraction Part 1"},"content":{"rendered":"

In\u00a0Lesson 5<\/u><\/a>\u00a0and\u00a0Lesson 6<\/u><\/a>\u00a0we introduced the idea of 10 pair complement addition on the abacus. Here in Lesson 7 we\u00a0will introduce the concept of subtraction with 10 pair complements. In review, there are only 5 possible 10 pair complements: 9-1, 8-2, 7-3, 6-4, 5-5. In the case where we want to subtract a number from a target\u00a0rod that would result in a number less than zero, we solve this problem using the 10 pair complement subtraction rule. In other words, if there are not enough beads touching the reckoning bar from which we want to subtract a larger number, we use the 10 pair complement subtraction rule. Notice the test\u00a0for the subtraction rule depends on how many beads are touching the reckoning bar whereas the test for the addition rule depends on the how many beads are not touching the reckoning bar.<\/p>\n

The 10 pair subtraction rule states, if there are not enough beads touching the bar, subtract 1 from the rod on the left and add the 10 pair complement of the number to be subtracted\u00a0to the target rod. Notice how the subtraction rule is the inverse of the addition rule.<\/p>\n

10 Pair Complement Subtraction Rule<\/strong>
1. Subtract 1 from the next rod left
2. Add the 10 pair complement of the number to be subtracted to the target rod<\/p>\n

As always pay attention to efficient finger movements and consistently practice these bead movements. With simple and efficient\u00a0bead mechanics and finger movements\u00a0we are building the foundation for fast and accurate arithmetic calculation.<\/p>\n

Please note that we will continuously build upon everything we have learned in previous lessons. So the following examples may require the use of the 10 pair complement addition rule shown in\u00a0lesson 5<\/u><\/a>\u00a0and\u00a0lesson 6<\/u><\/a>.\u00a0<\/p>\n

In this first example we start by adding 2 to the abacus on rod B. Next we add 8 to 2 on rod B but notice there are not enough available beads so we must use the 10 pair addition rule. So first add 1 to 0 on rod A and then subtract the 8, 10 pair complement 2 from 2 on rod B. The interim sum is 10. Next subtract 9 from 10 by subtracting 9 from 0 on rod B. But since there are no beads touching the reckoning bar on rod B, we will use the 10 pair subtraction rule. First subtract 1 from the next rod left\u00a0by subtracting 1 from 1 on rod A. Then add 1, the 10 pair complement 9, to 0 on the target rod, rod B. The interim sum is now 1. Finally add 6 to 1 on rod B for a final answer of 7.<\/p>\n

\t\t\t\t\t\t\t\t\t\t\"\"\t\t\t\t\t\t\t\t\t\t\t<\/p>\n

In the next example we start by adding 3 to the abacus on rod B. Next we add 5 to 3 on rod B for an interim sum of 8. Next we add 2 to 8 but must use the 10 pair addition rule. So first add 1 to the next rod left by adding 1 to 0 on rod A. Then subtract 8, the 10 pair complement of 2, from 8 on rod B for an interim sum of 10. Lastly subtract 6 from 10 by subtracting 6 from 0 on rod B but notice we will need to use the 10 pair subtraction rule. First subtract 1 from 1 on rod A. Then add 4, the 10 pair complement 6 to 0 on the target rod, rod B for a final answer of 4.<\/p>\n

\t\t\t\t\t\t\t\t\t\t\"\"\t\t\t\t\t\t\t\t\t\t\t<\/p>\n

In the final example we start by adding 1 to the abacus on rod B. Next add 9 to 1 using the 10 pair addition rule. First add 1 to 0 on rod A and then subtract 1, the 10 pair complement of 9, from 1 on rod B. The interim sum is 10. Next subtract 5 from 10 by subtracting 5 from 0 on rod B. We will need to use the subtraction rule. So first subtract 1 from 1 on rod A and then add 5, the 10 pair complement of 5, to 0 on rod B. The interim sum is now 5. Lastly add 4 to 5 on rod B for a final answer of 9.<\/p>\n

\t\t\t\t\t\t\t\t\t\t\"\"\t\t\t\t\t\t\t\t\t\t\t<\/p>\n

Next up Lesson 8 Subtraction Part 2<\/u><\/a>.<\/p>\n","protected":false},"excerpt":{"rendered":"

In\u00a0Lesson 5\u00a0and\u00a0Lesson 6\u00a0we introduced the idea of 10 pair complement addition on the abacus. Here in Lesson 7 we\u00a0will introduce the concept of subtraction with 10 pair complements. In review, there are only 5 possible 10 pair complements: 9-1, 8-2, 7-3, 6-4, 5-5. In the case where we want to subtract a number from a<\/p>\n","protected":false},"author":2,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[],"tags":[],"class_list":["post-362","post","type-post","status-publish","format-standard","hentry"],"_links":{"self":[{"href":"https:\/\/sumomath.com\/content\/wp-json\/wp\/v2\/posts\/362"}],"collection":[{"href":"https:\/\/sumomath.com\/content\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/sumomath.com\/content\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/sumomath.com\/content\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/sumomath.com\/content\/wp-json\/wp\/v2\/comments?post=362"}],"version-history":[{"count":0,"href":"https:\/\/sumomath.com\/content\/wp-json\/wp\/v2\/posts\/362\/revisions"}],"wp:attachment":[{"href":"https:\/\/sumomath.com\/content\/wp-json\/wp\/v2\/media?parent=362"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/sumomath.com\/content\/wp-json\/wp\/v2\/categories?post=362"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/sumomath.com\/content\/wp-json\/wp\/v2\/tags?post=362"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}