i know how to divide but i dont quit understand why we use those steps told in schools. like for example

  ____
3/450    150  quotient
  3
 -----
  15 
  15
------
  000

could someone please tell me why we do these steps is there any other method or can someone explain why?

  • 1
    Are you asking who invented the operation of division and why, or are you asking who invented the American notation for long division? Note in Russia, e.g., a different notation for long division is used. – gt6989b Feb 20 '14 at 17:32
  • 1
    We do those steps because it yields the right answer. $\ddot\smile$ – MJD Feb 20 '14 at 17:32
  • 1
    As seen here different cultures use different notation (though - of course - essentially the same steps. This diversification already suggests that the algorithm was invented long ago: Almost at the same time the positional system got introduced in the western world, the long division algorithm was published (e.g. by Adam Ries). But I assume the Bybylonians could do it also (in base 60). – Hagen von Eitzen Feb 20 '14 at 17:36
  • 2
    There is an extensive discussion of the history in Smith's History of Mathematics, I think volume 2. – André Nicolas Feb 20 '14 at 17:50
  • 1
    @HagenvonEitzen The Babylonians did not use long division. In fact, it would have been rather cumbersome in base 60. In computing $a/b$, they would use a table of inverses to find the sexagesimal expansion of $1\over b$, or at least a decent approximation of it, and then they would multiply $a\cdot{1\over b}$. For this they had rather extensive multiplication tables. – Per Manne Feb 20 '14 at 19:28
up vote 10 down vote accepted

To solve $450\div 3$ we are asking how many groups of 3 there are in 450. The obvious way to solve this is to subtract groups of 3 from 450, one at a time, and count each group as it is subtracted. Thus:

    450
   -  3   
   ----
    447
   -  3   
   ----
    444
   -  3
   ----
    441
   -  3
    .
    .      (do this many many times)
    .
   ----
      6
   -  3
   ----
      3
   -  3
   ----
      0

We have to repeat this 150 times, after which the 450 has been reduced to 0. So the answer is that 450 contains 150 groups of 3.

It should be clear that there is an easy way to speed up this process: instead of removing a single group of 3 each time, we could remove 10 or even 100 groups of 3 at once. So for example:

    450
  - 300     (100 groups)
  -----
    150
  -  30     (10 groups)
  -----
    120
  -  30     (10 groups)
  -----
     90
  -  30     (10 groups)
  -----
     60
  -  30     (10 groups)
  -----
     30
  -  30     (10 groups)
  -----
      0

Here we have removed 150 groups of 3 as before, but instead of doing them one at a time, we removed 100 right away, and then five sets of 10 groups each, for a total of 150 groups.

We can abbreviate this further:

    -------
  3 ) 450

We begin by removing 100 groups of 3, as before:

      1
    -------
  3 ) 450
      300      (100 groups)
      ---
      150

Now we want to remove sets of 10 groups from 150. Removing a set of 10 groups of 3 will remove 30. Instead of removing the sets one at a time, we apply our knowledge of multiplication to see that 150 is big enough to remove 5 sets of 10 groups of 3, for a total of 150:

      15
    -------
  3 ) 450
      300      (100 groups)
      ---
      150
      150      (5 sets of 10 groups = 50 groups)
      ---
        0

Since the total has reached 0, we do not need to remove any single groups, so we fill in the final 0 in the answer:

      150
    -------
  3 ) 450
      300      (100 groups)
      ---
      150
      150      (5 sets of 10 groups = 50 groups)
      ---
        0

Added 2014-04-23: I learned today that my daughter is being explicitly taught this method in the fourth grade. The example she showed me was:

      --------
    7 )   182

Here, she said, you might happen to know that $7\times 20=140$, so you remove the 140 from the dividend:

      --------
    7 )   182 
        - 140     20
        -----
           42

Then perhaps you remember that $7\times 5 = 35$, so you remove the 35:

      --------
    7 )   182 
        - 140     20
        -----
           42
        -  35      5
        -----
            7

Then you remove the remaining 7:

      --------
    7 )   182 
        - 140     20
        -----
           42
        -  35      5
        -----
            7
         -  7      1
        -----
            0

The remainder is now 0, so you add the right-hand column, $20+5+1$, to obtain the quotient 26.

  • i guess now i know ! – abdul raziq Feb 20 '14 at 17:55
  • I don't know the history, and I hope someone else does. But I would be surprised if this method of division were not prehistoric. – MJD Feb 20 '14 at 17:57
  • 1
    Smith says “We have, however a case described in the 4th century by Theon of Alexandria (c. 390) in which the literal numeral system of the Greeks is used and the work is not unlike our own, except that sexageismal [base-60] fractions are employed.” (p.133) Smith also says that the form of the long division algorithm we use today “is often attributed to Gerbert (c. 980), although it is uncertain whether he originated it”. – MJD Feb 20 '14 at 18:20

You asked if there was any other method, and, looking in David Eugene Smith History of Mathematics 2 (1925) as per André Nicolas's suggestion, I saw this method (p.132), which Smith attributes to the Egyptians.

We want to divide 450 by 3. We write a table as follows:

    1    3
    2    6
    4   12
    8   24
   16   48
   32   96
   64  192
  128  384

The first row has 1 on the left and the divisor 3 on the right; each following row is double the previous one.

Now we find items in the right-hand column that add up to 450, and mark those rows:

    1    3
    2    6  *
    4   12  *
    8   24
   16   48  *
   32   96
   64  192
  128  384  *

The rows with $6,12,48,384$ are marked because $6+12+48+384 = 450$. (It is easiest to find these from bottom to top: $384 < 450$, so we mark $384$ and deduct $384$ from $450$, leaving $66$. Then we move up the column until we find a number less than $66$; in this case $48$. We deduct $48$ from $66$, leaving $18$, and continue as before; $24$ is too big, but $12+6 = 18$ and we are done.)

Then we add up the left-hand numbers in the marked rows, $2+4+16+128 = 150$, which is the answer.

It is quite possible that for this problem, where the dividend is a multiple of 10, the Egyptians would have used the same method this way instead:

    10   30  *
    20   60  *
    40  120  *
    80  240  *

A method that is essentially the same as the modern one, but organized differently on the page, is given by David Eugene Smith History of Mathematics 2 (1925) on pages 136–137. He says:

By far the most common plan in use before 1600 is known as the galley, batello, or scratch method and seems to be of Hindu origin.

Here we compute $65284\div 594$:

Steps 1–6 of the scratch method

Steps 1–4 here are subtracting 59400 from 65284. The difference 5884 is visible in the unscratched digits on the left side of the vertical line in display 4. The ‘1’ on the right of the vertical line is the partial quotient.

Step 5 determines that the remainder, 5884, is too small to contain 10 groups of 594, which would be 5940, so the next digit of the quotient is 0.

Step 6 is preparing to divide 594 into 5884. Smith presents the completed work:

Final disposition of the scratch method

The quotient is 109, and the remainder, 538, is visible along the top of the digits on the left of the bar.

  • This example appears to have been taken from the Treviso Arithmetic of 1478. In the Treviso Arithmetic is is preceded by : 825÷2, 9065÷8, and 9825÷94. Only steps 1, 5, and 6 appear in the Arithmetic; steps 2–4 appear to be interpolations by Smith based on the more detailed explanations from the previous examples. – MJD Apr 15 at 11:46

The recursive step of decimal/radix division works as follows. Write the dividend as $\, n = j \,10^i\! +k \,$ where $\, j> d = $ divisor. Divide $\,j\,$ by $\,d\,$ to get $\,j = q\,d + r.\,$ Then

$$\dfrac{n}d\ =\ \dfrac{j\,10^i + k}d\ =\ \dfrac{(qd+r)10^i + k}{d}\ =\ q\, 10^i +\!\!\! \underbrace{\dfrac{ r10^i+k}d}_{\large\rm recurse\ on\ this} $$

Then recursively apply the algorithm to the indicated fraction. Normally one chooses $j$ minimal, but one may also choose any value of $\,j>d\,$ that is convenient to divide by $\,d.$

Taking your example, write $$ \frac{450}{3} = \frac{4 \times 100 + 5 \times 10 + 0 \times 1}{3} = \frac{4}{3} \times 100 + \frac{5}{3} \times 10 + \frac{0}{3} \times 1. $$ We will arrange the terms so that the standard method is suggested.

First compute $\frac{4}{3} \times 100 = \left(1 + \frac{1}{3} \right) \times 100$ first; next, take the piece $\frac{1}{3} \times 100$ and add it to the piece $\frac{5}{3} \times 10$ to get $\left(\frac{10}{3} + \frac{5}{3}\right)\times 10 = \frac{15}{3} \times 10 = 5 \times 10$. The last piece is just zero. Your conclusion is $$ \frac{450}{3} = 1 \times 100 + 5 \times 10 + 0 \times 1 = 150. $$

I hope this helps!

  • That does not seem to me to resemble the way division is actually done. – MJD Feb 20 '14 at 18:07
  • I suppose I could have given a bit more explanation, and perhaps an example with 4 nonzero digits would have been better. But the first group shows where the $1$ comes from ($3$ goes into $400$ $100$ times), then grouping the $\frac{10}{3}$ and $\frac{5}{3}$ shows where the $5$ comes from ($3$ goes into $150$ $50$ times). These are the exact same determinations one makes when doing long division, including the suppression of the powers of $10$ which I achieved here via factoring. Not sure if that clears anything up! – Rookatu Feb 20 '14 at 18:18

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