Codiagnóstico

(to the left) from 7 to 6 (= 7− 1). Finally, the subtraction in the hundreds column is now 6− 3 = 3, and 756 − 389 = 367.

Schematically: (5.3) 6 14 16  7  5 6 − 3 8 9 3 6 7

Activity. Use the subtraction algorithm to compute 2345− 687.

The possibility of trading in a given subtraction m−n as described above depends on having a nonzero digit in m to the immediate left of the column where the trading is to take place. Sometimes there is no such nonzero digit. For example, to compute 50003− 465, the subtraction in the ones column is 3− 5, but the digit to the immediate left of 3 in 50003 is 0. The subtraction algorithm now takes care of this situation as follows (again, in the interest of notational simplicity, we state it explicitly for the subtraction 50003− 465):

Start from the right column as usual, change the subtraction 3−5 to 13− 5 and at the same time change all the zeros in 50003 to the immediate left of the digit 3 to nines and decrease the first nonzero digit to the left of 3 (i.e., 5) from 5 to 4 (= 5− 1), and then carry out the subtraction as before.

Carrying out this algorithm on 50003− 465, we obtain schematically:

(5.4)

4 9 9 9 13

 5  0  0  0 3

− 4 6 5

4 9 5 3 8

Activity. Use the subtraction algorithm to compute 300207− 14629.

5.3. Explanation of the Algorithm

For the explanation of the subtraction algorithm as codified in (5.2)–(5.4), the following subtraction fact is needed. In concrete terms, this fact takes the following form: if you have three piles of oranges, having 7, 8, 9 oranges, respectively, in each pile, then taking 3 + 4 + 5 oranges away from the three piles combined would leave behind the same total number of oranges as taking successively 3 oranges from the pile of 7 oranges, 4 oranges from the pile of 8 oranges, and 5 from the pile of 9 oranges. In other words,

(7 + 8 + 9)− (3 + 4 + 5) = (7 − 3) + (8 − 4) + (9 − 5).

In general, suppose l, m, n, a, b, and c are any whole numbers so that  ≥ a, m ≥ b, and n ≥ c. Then, we claim likewise that the following is valid: (5.5) ( + m + n) − (a + b + c) = ( − a) + (m − b) + (n − c).

A formal mathematical explanation of equation (5.5) can be given using the definition of subtraction; see section 5.6. A complete understanding of equation (5.5), however, must await the introduction of negative numbers in Part 3, and we will revisit this equality on page 392.

It should be remarked that equation (5.5) is valid for any number of pairs of numbers. For example, the analogue of equation (5.5) for five pairs of numbers would read: if l ≥ a, m ≥ b, n ≥ c, p ≥ d, and q ≥ e, then

( + m + n + p + q) − (a + b + c + d + e)

= ( − a) + (m − b) + (n − c) + (p − d) + (q − e).

The proof of this more general version is of course entirely similar to the case of three pairs of numbers.

Now that equation (5.5) is available, we can give the explanation of (5.2), the simplest form of the subtraction algorithm. We first do it schematically (note: the double arrow “⇐⇒” in the following stands for “is equivalent to”): 1 6 5 8 − 2 5 7 ? ? ? ? ⇐⇒ − 1000 + 600 + 50 + 80 + 200 + 50 + 7 ? (5.5) ⇐⇒ − 1000 + 600 + 50 + 80 + 200 + 50 + 7 1000 + 400 + 0 + 1 ⇐⇒ − 1 6 5 82 5 7 1 4 0 1 .

Notice that equation (5.5) was used in the middle “⇐⇒”.

We now rewrite the preceding in a more conventional manner, using the version of (5.5) for four pairs of numbers:

1658− 257 = (1000 + 600 + 50 + 8) − (0 + 200 + 50 + 7)

= (1000− 0) + (600 − 200) + (50 − 50) + (8 − 7) (by (5.5)) = 1000 + 400 + 0 + 1

= 1401.

The second line above corresponds exactly to the column-by-column sub- traction in (5.2).

The explanation for (5.3) retraces the steps, run backwards, of the ex- planation given for the method of carrying in the addition algorithm of

5.3. Explanation of the Algorithm 77

Chapter 4. Again, we first do it schematically: 7 5 6 − 3 8 9 ? ? ? ⇐⇒ − 700 + 50 + 6300 + 80 + 9 ? ⇐⇒ 700 + 40 + 16 − 300 + 80 + 9 ? ⇐⇒ 600 + 140 + 16 − 300 + 80 + 9 ? (5.5) ⇐⇒ − 600 + 140 + 16300 + 80 + 9 300 + 60 + 7 ⇐⇒ − 7 5 63 8 9 3 6 7 . In more conventional notation, this is expressed as follows:

756 = 700 + 50 + 6 = (600 + 100) + 50 + 6 = 600 + (100 + 50) + 6 (Theorem 2.1) = 600 + 150 + 6 = 600 + (140 + 10) + 6 = 600 + 140 + (10 + 6) (Theorem 2.1) = 600 + 140 + 16 so that 756− 389 = (600 + 140 + 16) − (300 + 80 + 9) = (600− 300) + (140 − 80) + (16 − 9) (by (5.5)) = 300 + 60 + 7 = 367.

Note that the second line of the preceding calculation, i.e., (600− 300) + (140− 80) + (16 − 9), corresponds exactly to the column-by-column subtrac- tions in (5.3). Note also that we have avoided using exponents to write the numbers 756 and 389 in their expanded forms but have used the ordinary version instead, because the more complicated notation would have obscured the underlying reasoning.

Although we do not wish to over-emphasize the formalism of the laws of operations discussed in Chapter 2, it is nevertheless worthwhile to point out the critical role played by the associative law of addition in the form of Theorem 2.1 in making trading possible in the subtraction algorithm. Activity. Explain the subtraction algorithm for 315−82, first schematically, and then by the conventional method using (5.5).

Finally, we explain the last part of the subtraction algorithm, specifi- cally, why it works in the case of 50003− 465 (see (5.4)). It is again an extended discourse on the associative law. The crucial observation may be summarized as a collection of addition facts that are a direct consequence of counting: 50000 = 49000 + 1000, 1000 = 900 + 100, 100 = 90 + 10, so that 50000 = 49000 + 900 + 90 + 10. Using Theorem 2.1, we get

50003 = 50000 + 3

= (49000 + 900 + 90 + 10) + 3

= 49000 + 900 + 90 + 13 (Theorem 2.1).

So using the version of (5.5) for four pairs of numbers, we have 50003− 465 = (49000 + 900 + 90 + 13) − (0 + 400 + 60 + 5)

= (49000− 0) + (900 − 400) + (90 − 60) + (13 − 5) = 49000 + 500 + 30 + 8

= 49538.

The second line corresponds to the column-by-column subtractions in (5.4). We note that the subtraction algorithm is again one that works from right to left. Just as in the case of the addition algorithm, one can work from left to right, but the amount of backtracking needed for making corrections would be even greater here than in the case of addition, as you can experience from the following.

Activity. Do the preceding subtraction 50003− 465 from left to right, and compare with the computation from right to left.

It is worth repeating that there is absolutely nothing unnatural about teaching children to do something from right to left.

Finally, as in the case of the addition algorithm, we continue to take note of the leitmotif of Chapter 3, to the effect that the subtraction algorithm reduces subtraction to that between two single-digit numbers. Why it should be mentioned here is that, as we have seen, this is not literally true because when trading takes place, one needs to know not only subtraction between single-digit numbers, but also the subtraction of a single-digit number from numbers ≤ 18. Specifically, the following subtraction facts must be known