# Subtraction

## Elementary subtraction

Subtraction is a basic operation in arithmetic. When subtracting natural numbers, we basically count what remains after "removing" some quantity.

In figure 1 a number is subtracted from an other number: 4 − 2 = 2, verbally: "four minus two equals two" or "two subtracted from four is two" or "four minus two makes two"... This calculation may describe a real-life occurrence like: "I have four apples, then I eat two, and then I have two apples left".

The elements that are subtracted are called the *terms* of the subtraction, and the result is called the *difference*.
The *difference* between 29 and 17 is 12, that is 29 − 17 = 12.

Subtraction is *not* commutative, meaning that changing the order of the *terms* changes the *difference*:

("≠" means "is *not* equal to")

It is not difficult to understand that 4 − 3 equals 1, but how should 3 − 4 be perceived? The result is something less than nothing? In a later chapter about negative numbers more about subtracting a number from a smaller number.

Subtracting any number from the same number always equals zero: 3 − 3 = 0.

Subtracting zero from a number has no effect: 3 − 0 = 3.

### Inverse operation of addition

Subtraction can be seen as the opposite or *inverse* operation to addition. If for instance, we "remove" or
"subtract" the 5 in 2 + 5 = 7, we have to remove it
on both sides of the equal sign, and we actually perform the operation 7 − 5 = 2:

7 | = | 2 + 5 |

7 − 5 | = | 2 + 5 − 5 |

7 − 5 | = | 2 |

In other words:

7 − 5 = 2, because 2 + 5 = 7.

And likewise:

7 − 2 = 5, because 5 + 2 = 7.

If someone would ask you to calculate the value of "?" in "2 + ? = 7",
what calculation would you actually perform? What number must we **add** to 2 to get 7?

Because they are inverse operations, we can use the addition table
for subtraction as well.
For 7 − 5 = 2 we look up 7 *in* the table field with 5 in a corresponding
cell in the top row or left column, and then the answer 2 is in the other
corresponding left column or top row cell.

In the previous chapter on addition, it was mentioned that students start by learning all the additions that result in 10 or less than 10 and the complements to 10. From there, it is a small step to learning subtractions with terms less than or equal to ten. Then, by learning the principle of regrouping, students can start with terms larger then 10.

## Regrouping

Learn to add and subtract numbers under ten fluently in your head. If you master this and understand the principles of the decimal numeral system, you can also mentally subtract larger numbers:

50 − 30 = (5 − 3) × 10 = 2 × 10 = 20.

140 − 40 = (14 − 4) × 10 = 10 × 10 = 100.

140 − 40 = 100 + (40 − 40) = 100 + 0 = 100.

470 − 50 = 400 + (70 − 50) = 400 + 20 = 420.

18 − 3 = 10 + (8 − 3) = 10 + 5 = 15.

55 − 4 = 50 + (5 − 4) = 50 + 1 = 51.

Digits in the ones position are subtracted, digits in the tens position are subtracted, digits in the hundreds position are subtracted etc.
And then the results are all added. This is called *regrouping*:

38 − 23 = (30 + 8) − (20 + 3) = (30 − 20) + (8 − 3) = 10 + 5 = 15

29 − 17 = (20 − 10) + (9 − 7) = 10 + 2 = 12.

What is in brackets belongs together as a whole. Subtracting the tens and subtracting the ones and add both results together.

## Borrowing

23 − 8 = 20 + (3 − 8) = ?

How do we deal with the 3 − 8? Let's try to "borrow" some value from the tens position:

23 − 8 | = | 20 + 3 − 8 | |

= | 10 + 10 + 3 − 8 | ||

= | 10 + 10 − 5 + 5 + 3 − 8 | (− 5 + 5 = 0) | |

= | 10 + 10 − 5 + 8 − 8 | ||

= | 10 + 10 − 5 | ||

= | 10 + 5 = 15. |

We "borrowed" (or better: we "stole") 5 from the 20 and added this to the 3, so we turned 3 − 8 into 8 − 8. This may seem a bit overcomplicated, but this "borrowing" from the next higher position is essential for subtracting. Next some more examples where we need to borrow from the next position.

17 − 9 = 10 + 7 − 9 = 10 − 2 + 2 + 7 − 9 = 10 − 2 + 9 − 9 = 10 − 2 = 8.

46 − 7 = 40 + 6 − 7 = 30 + 10 + 6 − 7 = 30 + 10 − 1 = 30 + 9 = 39.

100 − 7 = 90 + 10 − 7 = 90 + 3 = 93.

57 − 29 = (50 − 20) + (7 − 9) = 30 + 7 − 9 = 20 + 10 − 2 = 28

800 − 304 = (800 − 300) + (0 − 4) = 500 − 4 = 490 + 10 − 4 = 496.

220 − 50 = 200 + 20 − 50 = 200 − 30 = 170.

222 − 53 = 200 + (20 − 50) + (2 − 3) = 200 − 30 − 1 = 170 − 1 = 169.

## Subtraction algorithm

Subtraction is a bit more difficult than addition, especially when you need to borrow. Fortunately, there is also a standard algorithm
for subtraction, sometimes called *long subtraction* or *column subtraction*.
This algorithm organizes the process of regrouping and borrowing systematically and always works for natural numbers:

0 | 12 | |

1 | 2 | 7 |

9 | 5 | |

3 | 2 |

127 − 95 = 32

In the above example the subtraction algorithm is used to calculate 127 − 95 = 32. The subtraction algorithm works pretty much the same as the addition algorithm, only now we can only work from top to bottom and we "borrow" from the next left column instead of "carry" to the next left column. We need to work from top to bottom only, with the largest number on top, because unlike addition subtraction is not commutative.

The recipe is as follows:

From **right to left** we start with the subtraction in the first column: 7 − 5 = 2.
We write the 2 in the bottom row under the horizontal line, as we did in the addition algorithm.
Then the second column: 2 − 9, but 2 is less than 9, so we need to borrow from the next column.
We borrow 100 and add this to the 20, which makes 120, but we are in the tens column thus we write 12 and perform 12 − 9 = 3.
The 2 in the tens column is increased to 12 and the 1 in the hundreds column is lowered to 0.
The hundreds column is now empty, and so the algorithm stops.
If we need to borrow, we only borrow 10 or 100 or 1000 etc. from the next left column.

Some more examples:

6 | 5 | 8 |

5 | 3 | |

6 | 0 | 5 |

658 − 53 = 605

1 | 10 | |

2 | 0 | 7 |

9 | 5 | |

1 | 1 | 2 |

207 − 95 = 122

13 | |||

2 | 3 | 16 | |

3 | 4 | 6 | 7 |

1 | 4 | 7 | 5 |

1 | 9 | 9 | 2 |

3467 − 1475 = 1992

If the next column in the top row contains a zero and we need to borrow from it, we need to borrow double: from the next after the next column and then from the next column. In the next example, the 5 − 7 in the ones column cannot borrow from the 0 in the tens column before this 0 first borrows value from the hundreds column. So, the 1 in the hundreds column becomes 0, the 0 in the tens column becomes 10. Then this 10 becomes 9 because the 5 in the ones position borrows 1 (=10) from it. The 5 becomes 15.

9 | ||

0 | 10 | 15 |

1 | 0 | 5 |

9 | 7 | |

8 |

105 − 97 = 8

9 | 9 | ||

0 | 10 | 10 | 10 |

1 | 0 | 0 | 0 |

5 | 6 | ||

9 | 4 | 4 |

1000 − 56 = 944

In the second example above 0 − 6 in the ones column made the 0 in the hundreds column first borrow 1000 from the thousands, then the 0 in the tens column borrow 100 from this 1000 in the hundreds column and then finally borrows 10 from the tens column.

We can check if we executed a subtraction correctly by adding the result to the subtracted number, as we have seen before: 1000 − 56 = 944 because 944 + 56 = 1000.

### Practice application

With the next app you can practice with the standard subtraction algorithm. Use the algorithm on a piece of paper to subtract the numbers. Click the exercise or click "CHECK" to see if you did it right.

Exercises:

## Non-associative

Suppose we need to calculate 110 − 100 − 10.

We might argue:

110 − 100 = 10 and 10 − 10 = 0.

Thus:

110 − 100 − 10 = (110 − 100) − 10 = 10 − 10 = 0

But we might also argue:

100 − 10 = 90 and 110 − 90 = 20.

Thus:110 − 100 − 10 = 110 − (100 − 10) = 110 − 90 = 20.

The latter does *not* comply with the common agreement.
By convention:

110 − 100 − 10 ≠ 110 − (100 − 10).

Subtraction is *non-associative*. As with the non-commutativity of subtraction, it turns out that subtraction cannot be performed in a random order.
110 − 100 − 10 needs to be interpreted (by convention) as subtracting both 100 and 10 from 110.
Subtracting 10 from 100 and then subtract the result, 90, from 110 is a different calculation. Later this course we will discuss in more detail why this is.
For now we state:

110 − 100 − 10 | = | (110 − 100) − 10 |

= | 110 − (100 + 10) | |

= | 0 |

Thus, with *chain subtraction* we add all the numbers, except the first one, and subtract the result from the first number.
This is also the most effective recipe for using algorithms on chain subtraction:

2000 − 324 − 910 − 567 | = | 2000 − (324 + 910 + 567) |

= | 2000 − 1801 | |

= | 199 |

3 | 2 | 4 | |

9 | 1 | 0 | |

5 | 6 | 7 | |

1 | 8 | 0 | 1 |

2 | 0 | 0 | 0 |

1 | 8 | 0 | 1 |

1 | 9 | 9 |