Subtraction Algorithm
The algorithm that is currently taught for subtraction requires "regrouping," which has previously been called "borrowing." This algorithm is usually illustrated with manipulatives, in which 10 is exchanged for 10 ones and 100 is exchanged for 10 tens. It becomes more complex when you have to "borrow across zero" because you have to exchange for 9 tens and 10 ones. Other cases are still more complex. For example, subtract 101 from 10000.
The algorithm which is taught in schools is a paper and pencil simulation of subtraction on an abacus, so probably the best way to explain it is by using an abacus.
Here we show a method of explaining this algorithm without manipulatives. All auxiliary computation should be done on a calculator, so children will not be distracted by the complexity of the individual steps.
1. Numbers can be written as sums in many different ways.
6 = 5 + 1 = 3 + 3, and so on. We will not look at all of them, but only at those ways which help us to subtract. We start with two-digit numbers:
28 = 20 + 8 = 10 + 18
45 = 40 + 5 = 30 + 15
61 = 60 + 1 = 50 + 11
90 = 90 + 0 = 80 + 10
So every two-digit number can be written as something plus a "teen" number (including 10 and 11).
2. This way of writing numbers makes some subtractions easy.
Subtract 29 from 57.
57 = | 50 + 7 = | 40 + 17 | |
29 = | 20 + 9 | ||
57 - 29 = | 20 + 8 | = 28 |
Notice that only 57 was put into something and "teen" form.
Subtract 63 from 74.
74 = | 70 + 4 | |
63 = | 60 + 3 | |
74 - 63 = | 10 + 1 | = 11 |
Here we did not need to put 74 in the "teen" form because 4 is bigger than 3.
Practice with other two-digit numbers. You will see that either the first or the second way shown above will give you the answer, provided that the first number is greater than the second. Check the results using a calculator. Do not forget about examples such as 23 - 21, 87 - 9, and 23 - 13. (You must handle correctly zeroes and one-digit numbers.)
23 = | 20 + 3 | 87 = | 70 + 17 | 23 = | 20 + 3 = | |||
21 = | 20 + 1 | 9 = | 9 | 13 = | 10 + 3 | |||
0 + 2 = 2 | 70 + 8 = 78 | 10 + 0 = 10 |
3. Bigger numbers.
You can also subtract numbers with many digits if you learn that
670 = 600 + 70 = 500 + 170,
4300 = 4000 + 300 = 3000 + 1300,
and so on.
You simply have one or more extra zeroes at the end. (It is very important to count them. You have to keep the right number of zeroes.)
Subtract 129 from 321
321 = | 300 + 20 + 1 | = 300 + 10 + 11 = 200 + 110 + 11 |
321 = | 200 + 110 + 11 | |
129 = | 100 + 20 + 9 | |
321 - 129 = | 100 + 90 + 2 | = 192 |
With bigger numbers you sometimes need to make several changes.
1000 = 900 + 100 = 900 + 90 + 10
3001 = 2000 + 1001 = 2000 + 900 + 101 = 2000 + 900 + 90 + 11,
and so on.
Subtract 999 from 3001
3001 = | 2000 + 900 + 90 + 11 | |
999 = | 900 + 90 + 9 | |
3001 - 999 = | 2000 + 0 + 0 + 2 | = 2002 |
Remark. In the standard algorithm
for subtraction, the rewriting shown above is called regrouping and should be
done mentally.
Webpage Developed by Aous Manshad
Last Modified:
July 17, 2007