Articulación Institucional

whole number by a multidigit number (see (iii) above):

To obtain 852× 73, start with the ones digit 3 of 73: first write down the result of 852× 3, then write down the result of 852 × 7 below the former, but with all the digits shifted one digit to the left of those of 852× 3, and then add column-wise as in the addition algorithm: (6.3) 8 5 2 × 7 3 2 5 5 6 + 5 9 6 4 6 2 1 9 6

In case there is any confusion over this algorithm, here is a further illustrative example. Suppose instead of 852× 73, we have 852 × 473. Then the algorithm calls for writing down the results of the three computations, 852×3, 852×7, 852×4, in successive rows with the columns aligned so that the digits of 852×7 are shifted one digit to the left of those of 852×3, the digits of 852×4 are shifted two digits to the left of those of 852×3, and then add column-wise as in the addition algorithm:

(6.4) 8 5 2 × 4 7 3 2 5 5 6 5 9 6 4 + 3 4 0 8 4 0 2 9 9 6

Activity. Do the following computation using the multiplication algorithm: 5 2 7

× 3 6 4

?

6.2. The Explanation

We have, in the main, avoided the use of exponential notation thus far in writing down the expanded form of a whole number. For the explanation of the multiplication algorithm, however, the use of exponential notation will add immeasurably to the conceptual clarity of the discussion. The slight increase in notational inconvenience is a small price to pay, all things considered.

We first explain (6.1) on 852× 3 by applying the distributive law 852× 3 = (8× 102) + (5× 10) + 2 × 3

= (8× 3) × 102+ (5× 3) × 10+ (2× 3) = (24× 102) + (15× 10) + 6,

where we have made use of Theorem 2.2 on page 44 in the second (boldfaced) row to conclude that (8×102_{)×3 = (8×3)×10}2_{, etc. The underlined items} correspond exactly to the single-digit computations specified in the First Part of the multiplication algorithm, and they also explain the reduction of 852×3 to a sequence of multiplications of single-digit numbers (in this case, 8× 3, 5 × 3, and 2 × 3).

Experience with the addition and subtraction algorithms tells us that we should proceed by working from right to left. Start with the 6 in (24× 102_{) + (15× 10) + 6: this corresponds to the 6 in the ones column of} (6.1). Next, we have 15× 10 = (10 + 5) × 10 = 102+ (5× 10), so that 852× 3 = (24 × 102) + 102+ (5× 10) + 6 = (24× 102) + 102 + (5× 10) + 6 (Theorem 2.1) = {(24 × 102) + (1× 102)} + (5 × 10) + 6.

The last expression accounts for the 5 in the tens column in (6.1) as well as the carrying of the 1 in the hundreds column. Now, the distributive law implies that the sum within the braces{ } in the last row is equal to (24×102)+(1×102) = (24+1)×102= (20+5)×102= (20×102)+(5×102). Since 20× 102 = 2× 103 (we are using Theorem 2.2), we have

852× 3 = (2 × 103) + (5× 102) + (5× 10) + 6.

This then exhibits the carrying of 2 in the thousands column of (6.1) together with the 5 in the hundreds column. The explanation of the procedures in (6.1) is now complete.

Activity. Give an analogous explanation of (6.2).

We now relate (6.1) and (6.2) to the original computation of 852×73. In the process, we will explain the Second Part of the multiplication algorithm as exhibited in (6.3). We apply the distributive law and Theorem 2.2: (6.5)

852× 73 = 852 × (70 + 3)

= (852× 70) + (852 × 3) = (852× 7) × 10 + (852 × 3).

6.2. The Explanation 89

The underlined items clearly exhibit why the multiplication of 852 (or any whole number) by a multidigit number will be known once the multiplica- tions of 852 by single-digit numbers are known. Now, using (6.1), (6.2), and (6.5), we obtain

852× 73 = (5964 × 10) + 2556 = 2556 + 59640.

According to the addition algorithm, we can write schematically:

(6.6) 8 5 2 × 7 3 2 5 5 6 + 5 9 6 4 0 6 2 1 9 6

Because we are used to treating an empty spot as a zero (cf. the discussion of the addition algorithm immediately following (4.1) on page 63), it is customary to omit the 0 at the end of 59640. When this is done, (6.6) becomes exactly (6.3). The shifting of 852× 7 by one digit to the left is therefore explained by the fact that the 7 has place value 70, so that 852× 7 is actually 852× 70.

Activity. Compute 73× 852 in the form 7 3

× 8 5 2

?

and give an explanation. (Note: By the commutativity of multiplication, you know in advance that the answer is 62196 (= 852× 73). So the point is to see that the multiplication algorithm gives the same answer.)

Finally, we give an explanation of the application of the multiplication algorithm to 852× 473 given in (6.4). Omitting some details that have already been supplied in the explanations of (6.2) and (6.3), we have

852× 3 = 2556 , 852× 7 = 5964 , 852× 4 = 3408 , and so by the distributive law

852× 473 = 852 × (4× 102) + (7× 10) + 3

= (852× 4) × 102 + (852× 7) × 10 + (852× 3) = 340800 + 59640 + 2556.

Therefore, according to the addition algorithm, we have schematically: (6.7) 8 5 2 × 4 7 3 2 5 5 6 5 9 6 4 0 + 3 4 0 8 0 0 4 0 2 9 9 6

It is seen that (6.4) is just (6.7) with some zeros omitted.

Activity. Suppose you are presented with the following computation: 5 2 7

× 3 0 0 4

2 1 0 8

+ 1 5 8 1

1 6 0 2 0 8 Is this correct? Why?

We mentioned in the discussion in Chapter 3 that there are many ways of writing the standard algorithms. Here is one alternative version of the multiplication algorithm. 8 5 2 × 7 3 5 9 6 4 + 2 5 5 6 6 2 1 9 6

Procedurally, this means we run the algorithm from left to right, first mul- tiplying by 7 before multiplying by 3. Mathematically, we do not consider such formal differences to be a difference at all. It is worth noting that even the algorithm with a single-digit multiplier can be carried out from left to right. For example, 6718× 5 can be done this way:

6 7 1 8 × 5 3 0 3 5 5 + 4 0 3 3 5 9 0

Activity. Give a precise description as well as explanation of the preceding algorithm.

6.2. The Explanation 91

Pedagogical Comment. In a classroom, the most striking feature of the multiplication algorithm may well be the shifting of 5964 by one digit to the left in (6.3), or the shifting of 3408 by two digits to the left in (6.4). This phenomenon of shifting to the left should be carefully explained to students in terms of place value, in the manner of (6.6), i.e., we are actually looking at 852× 70 and 852 × 400, respectively, rather than 852 × 7 and 852 × 4, so that the shifting of digits is caused by the presence of the zeros. End of Pedagogical Comment.

Exercises

1. Explain to a 4th grader why the multiplication algorithm for 86× 37 is correct.

2. How would you compute 32,897,546,126,349× 87 with a 12-digit calcu- lator? Describe one way you can get it done.

3. Use the multiplication algorithm to compute the following: 1 8

× 5 0 0 0 0 9

?

Now apply the same algorithm to the following: 5 0 0 0 0 9

× 1 8

?

Discuss the pros and cons of these two methods of computing 500009× 18.

4. Use the expanded form of 500009 (and of course also that of 18) to explain both computations in Exercise 3 above. (Note that for a number such as 500,009, the use of exponential notation to write out its expanded form is essentially a necessity.)

5. Compute 4208× 87 by the multiplication algorithm, and explain why it is correct.

6. Use mental math to do each of the following: (a) 12× 45. (b) 43 × 22. (c) Suppose you want to buy thirty-five each of 39-cent stamps, 24-cent stamps, 63-cent stamps, and 84-cent stamps.2 How much money do you need?

7. Ms. Wang took her five classes to visit the museum on successive Mon- days. The train fare for each student was $1.85 (which you may take to be 185 cents if you want to avoid decimals). Her classes have 28, 24, 25, 22, and 26 students. How much train fare did Ms. Wang pay for her students altogether?

8. (a) Which 2-digit number, when multiplied by 89, gives a 4-digit number that begins and ends with a 6? (b) List all the 3-digit numbers which have the following properties: the sum of the digits is 12, and when multiplied by 15 they give a 5-digit number which ends with a 5 (i.e., the ones digit is 5). (Clearly this problem can be done by guess-and- check. You are, however, asked to use reasoning to quickly dispatch it by narrowing down the choices.)

Exercises 93

9. Let a, b, m, n be whole numbers so that a < m and b < n. Explain why ab < mn.

10. The following is the product of a three-digit number by a two-digit number using the multiplication algorithm. Fill in the blanks.

8 7

× 5

2

3 5

Chapter 7

The Long Division