# Investigating subtraction.

Subtraction allows us to know how much remains in calculations relating to numerous occupational, financial and social situations. Most students think of subtraction as just a process which can be performed mentally or using a calculator. But an investigation of the details of different subtraction processes provides insight into a deeper understanding of mathematics.In the early years, students learn to subtract a smaller digit from a larger digit using a number line. Soon they realise, for example, that 5 - 2 has the same answer as 7 - 4, which is just (5 + 2) (2 + 2). Thus, adding the same digit to each number (minuend and subtrahend) in the original subtraction still leaves the same answer. This is important to understand before the next challenge arises, where they have to subtract 7 from 12, say. This can be done by rote or extending the number line, but thinking students should be more satisfied by:

12 - 7 = (12 + 3) - (7 + 3) = 15 - 10 = 5

Next comes two-digit subtraction, and there is usually no problem with 96 - 52, since it is a pair of one-digit subtractions. However 61 - 37 is a bit more difficult and three methods have been promoted in the past.

The first 'trading' method is to trade one 'ten' from the 'ten's' part of 61. This 'ten' is traded for 10 'ones' and added to the 1 forming:

61 - 37 = (50 + 11) - (30 + 7) = (50 - 30) + (11 - 7) = 20 + 4 = 24

The second 'borrow and pay back' method is to borrow ten to give to the 1 and then 'pay it back' to the 30 forming:

61 - 37 = (60 + 11) - (40 + 7) = (60 - 40) + (11 - 7) = 20 + 4 = 24

The second method comes up with the correct answer, but I am sure that it confuses the students and directs them unconsciously to just accept the process without understanding. Of course, primary school children do not use either method in the above forms with brackets; instead they have the subtrahend directly under the minuend and use crossing out and pre-indices to denote the trading used in Example 1. However, when they reach high school and are introduced to brackets, mainly through algebra, it is time to reflect on subtraction.

This brings me to the third method of subtraction which unfortunately is not well known. With this method there is never the need to either 'trade' or 'borrow and pay back'.

Consider again:

61 - 37 = (99 - 37) + 61 - 99 = 62 + 61 - 99 = 123 - (100 + 1) = 23 + 1 = 24

The steps are: subtract 37 from 99, then add the answer to 61 yielding 123. Then subtract 100 and add 1 (which looks like remove the one from the hundreds position and add it to the units). Only the simplest subtraction and basic addition have been used. Let us see how this third method handles more complicated subtractions. For example, consider:

6003 - 2987 Step 1: 9999 - 2987 = 7012 Step 2: 7012 + 6003 = 13 015 Step 3: 13 015 - 10 000 = 3015 Step 4: 3015 + 1 = 3016

Finally check, as in all subtraction problems, that the answer plus the subtrahend yields the minuend. Give your students some further examples such as 213 - 85. Should they use 99 or 999?

Discovery exercises involving subtraction

Consider any two-digit number with different digits. Reverse the digits and subtract the smaller number from the larger. Keep doing this to the answer each time until a pattern emerges. For example, consider 71.

71 - 17 = 54 54 - 45 = 09 (keeping the two-digit format) 90 - 09 = 81 81 - 18 = 63 63 - 36 = 27 72 - 27 = 45

Hence, we would just keep cycling round through these numbers ad infinitum.

To see why this is so, write the two-digit number as 10A + B, where A * B are different digits. Reverse subtraction yields either 9(A - B) or 9(B - A), and the result follows.

Next ask your students to consider any three-digit number with at least two different digits. Rewrite the digits in descending order as the minuend, and in ascending order as the subtrahend. Subtract, and keep doing this to the answer each time. Not what was expected? Why does this happen?

Now repeat the exploration for four-digit numbers. If the four-digit number is rewritten as 1000 A + 100 B + 10 C + D the suggested subtraction yields:

999 (A - D) + 90 (B - C).

Now (A - D) and (B - C) can only take integer values from 0 to 9. Ask your students to write out the possible answers. They should discover that there are only twenty that survive and when the process is repeated they all gravitate to one answer.

Anyone challenged to have a go at five-digit numbers?

Happy discoveries!

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Title Annotation: | discovery |
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Author: | de Mestre, Neville |

Publication: | Australian Mathematics Teacher |

Date: | Mar 22, 2018 |

Words: | 777 |

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