Understanding Long Division

 

Introduction.

 

The standard written algorithm for long division, either of whole numbers with remainder, or for decimals, is one of many algorithms for division that were used in the past. We do not know for sure why this particular algorithm became standard instead of any of its competitors, but we may list some plausible reasons.

 

1. This algorithm does not require crossing out numbers or erasing. Therefore it can by carried out with ink on paper, providing a permanent record of the computation.

2. Its correctness can be checked by another person just by carefully reading the written record.

3. It uses space on paper efficiently. (In the past, paper was not a cheap commodity.)

4. After years of practice people can achieve a high computing speed with a minimal rate of errors.

 

But while there are good points of this algorithm, it also carries a heavy cost.

 

1. The logical clarity of the procedure is sacrificed for efficiency.

2. Saving space was achieved at the cost of high mental effort. Students must perfect the mental multiplication of a multi- digit number by a one digit number, before they can achieve a rate of successfully executing long division that is even barely acceptable.

 

Below we will modify this algorithm. We will sacrifice its efficiency in order to restore its logical clarity and to make executing it less difficult. We will explain long division by using a generic example.

 

Remark. You may of course use your own examples, but do not make them simpler. Work one long example slowly, and make it easier by allowing students to use calculators for executing the individual steps. Many simple examples are good for drill, but they do not promote understanding.

 

A generic example of division of a 5 digit number by a 3 digit number.

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Task: Divide 137 into 26010.

 

(1) Show an easier version of the algorithm.

Precompute the multiples of 137 from 2 to 9.

       1           1 3 7
  2    2 7 4
  3    4 1 1
  4    5 4 8
  5    6 8 5
  6    8 2 2
  7    9 5 9
  8 1 0 9 6
  9 1 2 3 3

 

The only operations we use from now on are subtraction and addition.

         2 6 0 1 0         
  - 1 3 7 0 0 100
    1 2 3 1 0  
  - 1 0 9 6 0   80
       1 3 5 0  
     - 1 2 3 3 +  9
          1 1 7 189

Thus the answer is 26010÷137 = 189 + 117/137.

 

Remarks.

In this and other examples given to students, they may use 4-operation calculators (e.g., the TI-108) to precompute multiples using

      [ON/C][ON/C][137][+][=][=] ... .

The additions and subtractions should be done as shown, with regrouping done mentally (without any crossing out or erasing).

 

(2) Show the standard display of this algorithm.

 

      

 

Thus the answer is 26010÷137 = 189 + 117/137, or 189 with remainder 117.

Remarks.

Point out the similarities and differences. The main difference is that the student has to estimate which multiple to use (here, the 1st, 8th and 9th), and that he/she has to compute them by multiplication (137*8 and 137*9). Allow students to use the calculator for multiplication, but they should mentally estimate which multiple fits (i. e., which is the biggest multiple still smaller than the number in the current line).

 

(3) Explain the basic principle of the algorithm.

 

This is the principle:

* Represent the number to be divided as the sum of multiples of the divisor, together with the remainder (namely a number that is smaller than the divisor).

* Divide each multiple by the divisor, and add the results. This sum is the required quotient.

 

Different algorithms may construct different decompositions of the number into multiples of the divisor, but the resulting quotient and remainder are always the same.

 

The standard algorithm uses "decimal multiples" of the divisor,

1, 2, ... , 9, 10, 20, ... , 90, 100, 200, ... , 900, ... .

This choice of divisors is reasonable because in our positional notation, multiplication by powers of 10 is easy.

 

The ancient Egyptians, who used base ten, but not positional notation, designed an algorithm which multiplied the divisor by powers of two, 1, 2, 4, 8, 16, 32, ... . They constructed the next multiple by doubling the previous one. They precomputed all the multiples.

 

(4) Show the Egyptian method using modern notation.

 

Again, we are dividing 26010 by 137.

 

Precompute the multiples of 137.

 

       1           1 3 7
  2    2 7 4
  4    5 4 8
  8 1 0 9 6
  16 2 1 9 2
  32 4 3 8 4
  64 8 7 6 8
  128 1 7 5 3 6

 

You do not need to go any further, because the next multiple is bigger than 26010, the number to be divided.

 

         2 6 0 1 0         
  - 1 7 5 3 6   128
       8 4 7 4  
     - 4 3 8 4     32
       4 0 9 0  
     - 2 1 9 2 + 16
       1 8 9 8  
     - 1 0 9 6 +   8
          8 0 2  
        - 5 4 8 +   4
          2 5 4  
        - 1 3 7 +   1
          1 1 7   189

 

Thus the answer is again the same, 26010/137 = 189 + 117/137

 

Remark.

Computing multiples can be done by the following keystrokes,

      [ON/C][ON/C][2][*][137][=][=] ... .

 

(5) Division of decimals can be reduced to division of whole numbers, as shown by the following example.

            260.11.37 = (26010137) = 189 + 117/137.

 

 So there are 189 whole copies of 137 in 26010;

            189*137=25893

 

And 26010-25893=117, so

            260.1/1.37=189+117/137

 

 If we want to get 2 more decimal places, we compute

            11700/137 = 85 + 55/137.

 

So we have

            117/137 = .85 + 55/13700.

 

And finally,

            189 + 117/137 = 189.85 + 55/13700 189.85.

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