In arithmetic, long division is a standard division algorithm suitable for dividing multi-digit numbers that is simple enough to perform by hand. It breaks down a division problem into a series of easier steps.
As in all division problems, one number, called the dividend, is divided by another, called the divisor, producing a result called the quotient. It enables computations involving arbitrarily large numbers to be performed by following a series of simple steps. The abbreviated form of long division is called short division, which is almost always used instead of long division when the divisor has only one digit. Chunking (also known as the partial quotients method or the hangman method) is a less mechanical form of long division prominent in the UK which contributes to a more holistic understanding about the division process.
Place in education
Inexpensive calculators and computers have become the most common way to solve division problems, eliminating a traditional mathematical exercise, and decreasing the educational opportunity to show how to do so by paper and pencil techniques. (Internally, those devices use one of a variety of division algorithms, the faster ones among which rely on approximations and multiplications to achieve the tasks). In the United States, long division has been especially targeted for de-emphasis, or even elimination from the school curriculum, by reform mathematics, though traditionally introduced in the 4th or 5th grades.
In English-speaking countries, long division does not use the division slash ⟨∕⟩ or division sign ⟨÷⟩ symbols but instead constructs a tableau. The divisor is separated from the dividend by a right parenthesis ⟨)⟩ or vertical bar ⟨|⟩; the dividend is separated from the quotient by a vinculum (i.e., an overbar). The combination of these two symbols is sometimes known as a long division symbol or division bracket. It developed in the 18th century from an earlier single-line notation separating the dividend from the quotient by a left parenthesis.
The process is begun by dividing the left-most digit of the dividend by the divisor. The quotient (rounded down to an integer) becomes the first digit of the result, and the remainder is calculated (this step is notated as a subtraction). This remainder carries forward when the process is repeated on the following digit of the dividend (notated as 'bringing down' the next digit to the remainder). When all digits have been processed and no remainder is left, the process is complete.
An example is shown below, representing the division of 500 by 4 (with a result of 125).
125 (Explanations) 4)500 4 ( 4 × 1 = 4) 10 ( 5 - 4 = 1) 8 ( 4 × 2 = 8) 20 (10 - 8 = 2) 20 ( 4 × 5 = 20) 0 (20 - 20 = 0)
A more detailed breakdown of the steps goes as follows:
- Find the shortest sequence of digits starting from the left end of the dividend, 500, that the divisor 4 goes into at least once. In this case, this is simply the first digit, 5. The largest number that the divisor 4 can be multiplied by without exceeding 5 is 1, so the digit 1 is put above the 5 to start constructing the quotient.
- Next, the 1 is multiplied by the divisor 4, to obtain the largest whole number that is a multiple of the divisor 4 without exceeding the 5 (4 in this case). This 4 is then placed under and subtracted from the 5 to get the remainder, 1, which is placed under the 4 under the 5.
- Afterwards, the first as-yet unused digit in the dividend, in this case the first digit 0 after the 5, is copied directly underneath itself and next to the remainder 1, to form the number 10.
- At this point the process is repeated enough times to reach a stopping point: The largest number by which the divisor 4 can be multiplied without exceeding 10 is 2, so 2 is written above as the second leftmost quotient digit. This 2 is then multiplied by the divisor 4 to get 8, which is the largest multiple of 4 that does not exceed 10; so 8 is written below 10, and the subtraction 10 minus 8 is performed to get the remainder 2, which is placed below the 8.
- The next digit of the dividend (the last 0 in 500) is copied directly below itself and next to the remainder 2 to form 20. Then the largest number by which the divisor 4 can be multiplied without exceeding 20, which is 5, is placed above as the third leftmost quotient digit. This 5 is multiplied by the divisor 4 to get 20, which is written below and subtracted from the existing 20 to yield the remainder 0, which is then written below the second 20.
- At this point, since there are no more digits to bring down from the dividend and the last subtraction result was 0, we can be assured that the process finished.
If the last remainder when we ran out of dividend digits had been something other than 0, there would have been two possible courses of action:
- We could just stop there and say that the dividend divided by the divisor is the quotient written at the top with the remainder written at the bottom, and write the answer as the quotient followed by a fraction that is the remainder divided by the divisor.
- We could extend the dividend by writing it as, say, 500.000... and continue the process (using a decimal point in the quotient directly above the decimal point in the dividend), in order to get a decimal answer, as in the following example.
31.75 4)127.00 12 (12 ÷ 4 = 3) 07 (0 remainder, bring down next figure) 4 (7 ÷ 4 = 1 r 3) 3.0 (bring down 0 and the decimal point) 2.8 (7 × 4 = 28, 30 ÷ 4 = 7 r 2) 20 (an additional zero is brought down) 20 (5 × 4 = 20) 0
In this example, the decimal part of the result is calculated by continuing the process beyond the units digit, "bringing down" zeros as being the decimal part of the dividend.
This example also illustrates that, at the beginning of the process, a step that produces a zero can be omitted. Since the first digit 1 is less than the divisor 4, the first step is instead performed on the first two digits 12. Similarly, if the divisor were 13, one would perform the first step on 127 rather than 12 or 1.
Basic procedure for long division of n ÷ m
- Find the location of all decimal points in the dividend n and divisor m.
- If necessary, simplify the long division problem by moving the decimals of the divisor and dividend by the same number of decimal places, to the right (or to the left), so that the decimal of the divisor is to the right of the last digit.
- When doing long division, keep the numbers lined up straight from top to bottom under the tableau.
- After each step, be sure the remainder for that step is less than the divisor. If it is not, there are three possible problems: the multiplication is wrong, the subtraction is wrong, or a greater quotient is needed.
- In the end, the remainder, r, is added to the growing quotient as a fraction, r/m.
Invariant property and correctness
The basic presentation of the steps of the process (above) focus on the what steps are to be performed, rather than the properties of those steps that ensure the result will be correct (specifically, that q × m + r = n, where q is the final quotient and r the final remainder). A slight variation of presentation requires more writing, and requires that we change, rather than just update, digits of the quotient, but can shed more light on why these steps actually produce the right answer by allowing evaluation of q × m + r at intermediate points in the process. This illustrates the key property used in the derivation of the algorithm (below).
Specifically, we amend the above basic procedure so that we fill the space after the digits of the quotient under construction with 0's, to at least the 1's place, and include those 0's in the numbers we write below the division bracket.
This lets us maintain an invariant relation at every step: q × m + r = n, where q is the partially-constructed quotient (above the division bracket) and r the partially-constructed remainder (bottom number below the division bracket). Note that, initially q=0 and r=n, so this property holds initially; the process reduces r and increases q with each step, eventually stopping when r<m if we seek the answer in quotient + integer remainder form.
Revisiting the 500 ÷ 4 example above, we find
125 (q, changes from 000 to 100 to 120 to 125 as per notes below) 4)500 400 ( 4 × 100 = 400) 100 (500 - 400 = 100; now q=100, r=100; note q×4+r = 500.) 80 ( 4 × 20 = 80) 20 (100 - 80 = 20; now q=120, r= 20; note q×4+r = 500.) 20 ( 4 × 5 = 20) 0 ( 20 - 20 = 0; now q=125, r= 0; note q×4+r = 500.)
Example with multi-digit divisor
A divisor of any number of digits can be used. In this example, 1260257 is to be divided by 37. First the problem is set up as follows:
Digits of the number 1260257 are taken until a number greater than or equal to 37 occurs. So 1 and 12 are less than 37, but 126 is greater. Next, the greatest multiple of 37 less than or equal to 126 is computed. So 3 × 37 = 111 < 126, but 4 × 37 > 126. The multiple 111 is written underneath the 126 and the 3 is written on the top where the solution will appear:
3 37)1260257 111
Note carefully which place-value column these digits are written into. The 3 in the quotient goes in the same column (ten-thousands place) as the 6 in the dividend 1260257, which is the same column as the last digit of 111.
The 111 is then subtracted from the line above, ignoring all digits to the right:
3 37)1260257 111 15
Now the digit from the next smaller place value of the dividend is copied down and appended to the result 15:
3 37)1260257 111 150
The process repeats: the greatest multiple of 37 less than or equal to 150 is subtracted. This is 148 = 4 × 37, so a 4 is added to the top as the next quotient digit. Then the result of the subtraction is extended by another digit taken from the dividend:
34 37)1260257 111 150 148 22
The greatest multiple of 37 less than or equal to 22 is 0 × 37 = 0. Subtracting 0 from 22 gives 22, we often don't write the subtraction step. Instead, we simply take another digit from the dividend:
340 37)1260257 111 150 148 225
The process is repeated until 37 divides the last line exactly:
34061 37)1260257 111 150 148 225 222 37
Mixed mode long division
mi - yd - ft - in 1 - 634 1 9 r. 15" 37) 50 - 600 - 0 - 0 37 22880 66 348 13 23480 66 348 1760 222 37 333 22880 128 29 15 ===== 111 348 == 170 === 148 22 66 ==
Each of the four columns is worked in turn. Starting with the miles: 50/37 = 1 remainder 13. No further division is possible, so perform a long multiplication by 1,760 to convert miles to yards, the result is 22,880 yards. Carry this to the top of the yards column and add it to the 600 yards in the dividend giving 23,480. Long division of 23,480 / 37 now proceeds as normal yielding 634 with remainder 22. The remainder is multiplied by 3 to get feet and carried up to the feet column. Long division of the feet gives 1 remainder 29 which is then multiplied by twelve to get 348 inches. Long division continues with the final remainder of 15 inches being shown on the result line.
Interpretation of decimal results
When the quotient is not an integer and the division process is extended beyond the decimal point, one of two things can happen:
- The process can terminate, which means that a remainder of 0 is reached; or
- A remainder could be reached that is identical to a previous remainder that occurred after the decimal points were written. In the latter case, continuing the process would be pointless, because from that point onward the same sequence of digits would appear in the quotient over and over. So a bar is drawn over the repeating sequence to indicate that it repeats forever (i.e., every rational number is either a terminating or repeating decimal).
Notation in non-English-speaking countries
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China, Japan, Korea use the same notation as English-speaking nations including India. Elsewhere, the same general principles are used, but the figures are often arranged differently.
In Latin America (except Argentina, Bolivia, Mexico, Colombia, Paraguay, Venezuela, Uruguay and Brazil), the calculation is almost exactly the same, but is written down differently as shown below with the same two examples used above. Usually the quotient is written under a bar drawn under the divisor. A long vertical line is sometimes drawn to the right of the calculations.
500 ÷ 4 = 125 (Explanations) 4 ( 4 × 1 = 4) 10 ( 5 - 4 = 1) 8 ( 4 × 2 = 8) 20 (10 - 8 = 2) 20 ( 4 × 5 = 20) 0 (20 - 20 = 0)
127 ÷ 4 = 31.75 124 30 (bring down 0; decimal to quotient) 28 (7 × 4 = 28) 20 (an additional zero is added) 20 (5 × 4 = 20) 0
In Mexico, the English-speaking world notation is used, except that only the result of the subtraction is annotated and the calculation is done mentally, as shown below:
125 (Explanations) 4)500 10 ( 5 - 4 = 1) 20 (10 - 8 = 2) 0 (20 - 20 = 0)
127|4 −124 31,75 30 −28 20 −20 0
In Spain, Italy, France, Portugal, Lithuania, Romania, Turkey, Greece, Belgium, Belarus, Ukraine, and Russia, the divisor is to the right of the dividend, and separated by a vertical bar. The division also occurs in the column, but the quotient (result) is written below the divider, and separated by the horizontal line. The same method is used in Iran, Vietnam and Mongolia.
127|4 −124|31,75 30 −28 20 −20 0
In Cyprus, as well as in France, a long vertical bar separates the dividend and subsequent subtractions from the quotient and divisor, as in the example below of 6359 divided by 17, which is 374 with a remainder of 1.
Decimal numbers are not divided directly, the dividend and divisor are multiplied by a power of ten so that the division involves two whole numbers. Therefore, if one were dividing 12,7 by 0,4 (commas being used instead of decimal points), the dividend and divisor would first be changed to 127 and 4, and then the division would proceed as above.
In Austria, Germany and Switzerland, the notational form of a normal equation is used. <dividend> : <divisor> = <quotient>, with the colon ":" denoting a binary infix symbol for the division operator (analogous to "/" or "÷"). In these regions the decimal separator is written as a comma. (cf. first section of Latin American countries above, where it's done virtually the same way):
127 : 4 = 31,75 −12 07 −4 30 −28 20 −20 0
In the Netherlands, the following notation is used:
12 / 135 \ 11,25 12 15 12 30 24 60 60 0
Algorithm for arbitrary base
Every natural number can be uniquely represented in an arbitrary number base as a sequence of digits where for all , where is the number of digits in . The value of in terms of its digits and the base is
Let be the dividend and be the divisor, where is the number of digits in . If , then and . Otherwise, we iterate from , before stopping.
For each iteration , let be the quotient extracted so far, be the intermediate dividend, be the intermediate remainder, be the next digit of the original dividend, and be the next digit of the quotient. By definition of digits in base , . By definition of remainder, . All values are natural numbers. We initiate
the first digits of .
With every iteration, the three equations are true:
There only exists one such such that .
The final quotient is and the final remainder is
In base 10, using the example above with and , the initial values and .
Thus, and .
In base 16, with and , the initial values are and .
Thus, and .
If one doesn't have the addition, subtraction, or multiplication tables for base b memorised, then this algorithm still works if the numbers are converted to decimal and at the end are converted back to base b. For example, with the above example,
with . The initial values are and .
Thus, and .
This algorithm can be done using the same kind of pencil-and-paper notations as shown in above sections.
d8f45 r. 5 12 ) f412df ea a1 90 112 10e 4d 48 5f 5a 5
If the quotient is not constrained to be an integer, then the algorithm does not terminate for . Instead, if then by definition. If the remainder is equal to zero at any iteration, then the quotient is a -adic fraction, and is represented as a finite decimal expansion in base positional notation. Otherwise, it is still a rational number but not a -adic rational, and is instead represented as an infinite repeating decimal expansion in base positional notation.
Calculation within the binary number system is simpler, because each digit in the course can only be 1 or 0 - no multiplication is needed as multiplication by either results in the same number or zero.
If this were on a computer, multiplication by 10 can be represented by a bit shift of 1 to the left, and finding reduces down to the logical operation , where true = 1 and false = 0. With every iteration , the following operations are done:
For example, with and , the initial values are and .
|0||1||1011||0||1011 − 0 = 1011||0|
|1||1||10111||1||10111 − 1101 = 1010||1|
|10||0||10100||1||10100 − 1101 = 111||11|
|11||0||1110||1||1110 − 1101 = 1||111|
|100||1||11||0||11 − 0 = 11||1110|
Thus, and .
On each iteration, the most time-consuming task is to select . We know that there are possible values, so we can find using comparisons. Each comparison will require evaluating . Let be the number of digits in the dividend and be the number of digits in the divisor . The number of digits in . The multiplication of is therefore , and likewise the subtraction of . Thus it takes to select . The remainder of the algorithm are addition and the digit-shifting of and to the left one digit, and so takes time and in base , so each iteration takes , or just . For all digits, the algorithm takes time , or in base .
Long division of integers can easily be extended to include non-integer dividends, as long as they are rational. This is because every rational number has a recurring decimal expansion. The procedure can also be extended to include divisors which have a finite or terminating decimal expansion (i.e. decimal fractions). In this case the procedure involves multiplying the divisor and dividend by the appropriate power of ten so that the new divisor is an integer – taking advantage of the fact that a ÷ b = (ca) ÷ (cb) – and then proceeding as above.
- Arbitrary-precision arithmetic
- Egyptian multiplication and division
- Elementary arithmetic
- Fourier division
- Polynomial long division
- Shifting nth root algorithm – for finding square root or any nth root of a number
- Short division
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