Subtraction By Addition

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Subtraction By Addition The link to a 5-video lecture on Subtraction by Addition: http://www.youtube.com/watch?v=g7adGJMUBPk&playnext=1&list=PL0ECD3A8166DA66AB

Definitions of terms used throughout this article: Minuend is the number that is to be subtracted from: 8 – 3 = 5, 8 is the Minuend Subtrahend is the number that is to be subtracted: 8 – 3 = 5, 3 is the Subtrahend Difference is the result of subtracting two numbers: 8 – 3 = 5, 5 is the Difference Algorithm is a step-by-step procedure to obtain the correct result of an operation in arithmetic. The 9’s Complement (9C) Table: basic numbers: 0 to 9, shown in brackets: [0 + 9]

[1 + 8]

[2 + 7]

[3 + 6]

[4 + 5]

The 5 number-pairs = 9 and Define the 9’s Complement of each other.

Special Case of Subtraction by Addition: Special Case Algorithm (– by +): Subtract 7 from 15: Difference = 15 – 7 1 + 5 = +6 << Add the Minuend digits: 15 ≠ 1 + 5 +2 << Add 2; 2 is the 9’s Complement of Subtrahend: 7, because 7 + 2 = 9 +8 << Difference obtained by Addition: 6 + 2 Algorithm (– by +): 15 – 7 = 1 + 5 + 2 >>> 15 – 7 = +6 + 2 >>> 15 – 7 = 8 / +6– 9 / + 2 >>> 15 – 7 = 8 Algorithm (Verified): 15 – 7 = (9 + 6) – (9 – 2) >>> 15 – 7 = + 9 Two questions can be asked in the subtraction: 15 – 7; what number added to 7 will equal 15, or what number added to 7 will equal 9? The second question is the key to the Special Case: what number added to the subtrahend will equal 9? In applying the Special Case, a student thinking may go something like this: 15 + 6 <<< Add the minuend digits 1 + 5 –7 + 2 <<< 2 is the number that when added to 7 will equal 9, or use the 9C Table ? + 8 <<< Difference obtained by Addition: 2 + 6 The Special Case is valid in the ranges: Minuend, 10 to 19 and Subtrahend, 0 to 9; a range that includes the facts in the Subtraction Table. Students that are initially learning addition, when they acquire the skill necessary to add 2-basic numbers that sum to 9, they can be taught to do subtraction by addition using the Special Case. Repetitive use of the Special Case will instill in the students the facts in the Subtraction Table. Five examples using the Special Case Algorithm: 18 –9 ?

+ 9 <<< Add the minuend digits 1 + 8 + 0 <<< 0 is the number that when added to subtrahend 9 equals 9, or use the 9C Table + 9 <<< Difference by +; Special Case verified by: 18 – 9 = 9 + 9 – (9 – 0) = 9 + 0 = 9

16 –8 ?

+ 7 <<< Add the minuend digits 1 + 6 + 1 <<< 1 is the number that when added to subtrahend 8 equals 9, or use the 9C Table + 8 <<< Difference by +; Special Case verified by: 16 – 8 = 9 + 7 – (9 – 1) = 7 + 1 = 8

2 13 –7 ?

+ 4 <<< Add the minuend digits 1 + 3 + 2 <<< 2 is the number that when added to subtrahend 7 equals 9, or use the 9C Table + 6 <<< Difference by +; Special Case verified by: 13 – 7 = 9 + 4 – (9 – 2) = 4 + 2 = 6

11 –6 ?

+ 2 <<< Add the minuend digits 1 + 1 + 3 <<< 3 is the number that when added to subtrahend 6 equals 9, or use the 9C Table + 5 <<< Difference by +; Special Case verified by: 11 – 6 = 9 + 2 – (9 – 3) = 2 + 3 = 5

12 –5 ?

+ 3 <<< Add the minuend digits 1 + 2 + 4 <<< 4 is the number that when added to subtrahend 5 equals 9, or use the 9C Table + 7 <<< Difference by +; Special Case verified by: 12 – 5 = 9 + 3 – (9 – 4) = 3 + 4 = 7 Subtraction Table:

Students initially learning addition are also learning subtraction by default; the Special Case Algorithm enables students to construct or to verify the facts in a subtraction table. The red subscripts are the sum of the minuend digits or the 9’s complement of the subtrahend. Differences are found by addition of the red subscripts. For example, 112 – 36 = 2 + 6 = 8

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101 112 123 134 145 156 167 178 189 –18 9 –27 8

9

–36 7

8

9

–45 6

7

8

9

–54 5

6

7

8

9

–63 4

5

6

7

8

9

–72 3

4

5

6

7

8

9

–81 2

3

4

5

6

7

8

9

–90 1

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7

8

S U B T R A H E N

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4 General Case of Subtraction by Addition valid for all numbers: The Special Case transforms into the General Case by adding a zero to the Basic Equation. Subtract 7 from 15: Difference = Minuend – Subtrahend + (99 – 100 + 1)= Zero +15 – 7 +92 107 8

+ 99 – 100 + 1 <<< Adding a zero does not change the Difference – 07 <<< Subtract 07 from + 99 + 92 <<< Difference 92 is the 9’s Complement of 07 + 15 <<< Add the Minuend: 15 +107 <<< Intermediate Result –100 <<< Subtract 100 +7 <<< Intermediate Result These 9-steps are reduced to 3 by adding the +1 <<< Add +1 9’s Complement of the Subtrahend +8 <<< Difference: directly to the Minuend. (See Note 1)

Final Step: transpose and add the left digit 1 to the right digit 7; = – 100 + 1 Note 1: The General Case can be reduced to 3-steps: Step (1) Transform the subtrahend to its 9’s complement or use the 9C Table page 1 Step (2) Add the 9’s complement of the subtrahend to the minuend. Step (3) Transpose the left-most digit, always equals 1, and add to the right-most digit. Examples: 1 to 5, Algorithm (SA) is the final form of the General Case valid for all numbers. Example (1) Subtract 789 from 4567 4567 4567 <<< Subtrahend and Minuend digits must be equal; subtract 0789 (add 1 leading zero) –0789 +9210 <<< The 9’s complement of 0789 is 9210: (0 + 9), (7 + 2), (8 + 1), (9 + 0) ? 13777 <<< Final step: transpose and add the left digit 1 to the right digit 7; (see Note 1) 3778 <<< Difference obtained by (4) Additions Example (2) Subtract 2365 from 4051 4051 4051 <<< Subtrahend digits are equal to the Minuend digits; subtract 2365 − 2365 7634 <<< The 9’s complement of 2365 is 7634: (2 + 7), (3 + 6), (6 + 3), (5 + 4) ? 11685 <<< Final step: transpose and add the left digit 1 to the right digit 5; (see Note 1) 1686 <<< Difference obtained by (4) Additions Example (3) Subtract 56 from 4051 4051 4051 <<< Subtrahend and Minuend digits must be equal; subtract 0056 (add 2 leading zeros) −56 9943 <<< The 9’s complement of 0056 is 9943: (0 + 9), (0 + 9), (5 + 4), (6 + 3) ? 13994 <<< Final step: transpose and add the left digit 1 to the right digit 4; (see Note 1) 3995 <<< Difference obtained by (4) Additions Example (4) Algorithm (SA) can be used to subtract decimal fractions: Subtract 5.375 from 7.250 7.250 7.250 − 5.375 4.624 ? 11.874 1.875

<<< Subtrahend digits are equal to the Minuend digits; subtract 5.375 <<< The 9’s complement of 5.375 is 4.624: (5 + 4), (3 + 6), (7 + 2), (5 + 4) <<< Final step: transpose and add the left digit 1 to the right digit 4; (see Note 1) <<< Difference obtained by (4) Additions

5 Example (5) Algorithm (SA) can be used to subtract mixed numbers: Subtract 5

The 9's Compliment of 5

3 1 from 7 8 4

3 5 5 3 8 is 3 because, 3 + 5 = 8 = 9 8 8 8 8 8

1 5 1 2 5 7 7 + 3 = 7 ( ) + 3 = 10 << Final step: add the left digit 1,to the right digit 0 (see Note 1) 4 8 4 2 8 8 1 3 7 7 7 - 5 =1 << Difference by Addition;1 =1.875 4 8 8 8 1 3 Note: Example (5) is Example (4),transformed to a mixed number : 7 = 7.250 and 5 = 5.375 4 8 Example (6) Algorithm (SA) can be used in long division and whenever subtraction is required. Divide: 9876 by 567, using multiples of ten as a first approximation of the quotient and Algorithm (SA)

10+ 7 << Multiply by a factor of 10 such that the product is less than or equal to the dividend. 567) 9876 4329 << 10 ´ 567 = 5670 < 9876, the 9's complement of 5670 is 4329 , i.e., 5670+ 4329= 9999 14205 << Addition of the 9's complement 4206 << Transpose & add the left digit 1 to the right digit 5,(see Note 1) 6030 << 7 ´ 567 = 3969 < 4206, the 9's complement of 3969 is 6030,i.e., 3969+ 6030 = 9999 10236 << Addition of the 9's complement 237 << Transpose and add the left digit 1 to the right digit 6, (see Note 1) Ê Quotient is 17, Remainder 237 To develop number sense in the students, two other methods of subtraction are shown: Method A: Subtract 789 from 4567 using Minuend Reformulation (see Note 2) Rule: Minuend = Minuend + zero (= +1 – 1); 4567 = 567 + 1 – 1 + 4000 = 568 + 3999 4567 3999 + 568 <<< Reformulate the Minuend to maximize the nine’s: 4567 = 3999 + 568 − 789 − 789 <<< Subtract the subtrahend from the partial minuend ? 3210 <<< Intermediate result: Subtraction without borrowings and carries +568 <<< Final step: add 568 to reconstruct the original Minuend: 4567 = 568 + 3999 3778 <<< Difference obtained by (3) “easy” Subtractions and (3) Additions Note 2: This may be the easiest method to understand that avoids borrowings and carries. Method B: Subtract 789 from 4567 using Minuend and Subtrahend Reduction Rule of Signs: Minuend – Subtrahend digit is (+); Subtrahend – Minuend digit is (−) 4 5 6 7 <<< Subtract the smaller digit from the larger digit and use the Sign rule −7 8 9 <<< This is the original subtrahend 4(−2)(−2)(−2) <<< This is the reduced Minuend: 4000 and the reduced Subtrahend: – 222 4000 = 3999 + 1 <<< Reformulate the reduced Minuend, add zero; 4000 – 1 + 1 − 222 <<< Subtract the reduced subtrahend from the reduced partial minuend 3777 <<< Intermediate result: Subtraction without borrowings and carries + 1 <<< Final step: add 1; to reconstruct the reduced Minuend: 4000 = 1 + 3999

6 3778 <<< Difference obtained by (6) “easy” Subtractions and (1) Addition

7 Conclusion: Algorithm (SA), Examples (1 to 5) is the only method that does subtraction entirely by addition and is valid for all numbers. Digital computers and hand-held calculators, since their inception, are programmed to transform the subtrahend into its 2’s complement and add that to the minuend to compute differences. Algorithm (SA), the 9’s complement procedure, is almost an exact copy of the 2’s complement procedure used by computers; consequently, subtraction by addition is not a new method to compute differences. Here is an example of the method used by digital computers to calculate differences: Subtraction by addition by a Computer or hand-held Calculator using a 4-bit word: 8421 <<< Base-2 place values start at 1 and double from right to left 15 = 1111 <<< 1111 = 8+4+2+1 = 15 – 7 ≠ 1001 <<< This is the 2’s complement of 0111 = 0+4+2+1 = 7 (see Note*) 8 = 1000 <<< Addition yields 8+0+0+0 because computers are programmed to discard the last carry Note*: Method (a): to compute the 2’s complement of 0111, from right to left, hold the first set bit and complement all the bits thereafter, that is, 0 to 1 and 1 to 0. For example: 0111 >>> 2’s Comp. is >>> 1001 or Method (b) start with: 0111 and complement all the digits: 1000, then add 1 = 0001 to obtain 1001

Algorithm (SA), Subtraction by Addition is a superior method compared to the textbook algorithm for subtraction, because addition is not only a much easier operation to perform than subtraction but also negates the need for extensive modification of the minuend and the need for borrowings and carries. Which method should be taught to elementary school children? – All known methods and those that will be invented in the future. The students themselves should be permitted to choose the method most conducive to their understanding.