1. Multiplying Whole Numbers © Math As A Second Language All Rights Reserved next #5 Taking the Fear out of Math 9 × 9 81 Extending Single Digit Multiplication

2. next © Math As A Second Language All Rights Reserved Continuing the Evolution of Single Digit Multiplication It is not a particularly noteworthy saving of time to write, 4 × 7 in place of 7 + 7 + 7 + 7. However, with respect to our “boxes of candy” situation that we discussed in single digit multiplication, suppose we wanted to buy 400 boxes at a cost of $7 per box. It would indeed be very tedious to write explicitly the sum of four hundred 7’s; that is… 7 + 7 + 7… + 7 400 next

3. © Math As A Second Language All Rights Reserved The above problem, when stated as a multiplication problem, should be written as 400 × 7. Note However, writing the problem as 7 × 400 gives us the equivalent but simpler addition problem… 400 + 400 + 400 + 400 + 400 + 400 + 400. 1 next note 1However, this obscures the fact that we want the sum of four hundred 7’s; not the sum of seven 400’s. While the answer is the same, the “mental image” is quite different. next

4. © Math As A Second Language All Rights Reserved Using the adjective/noun theme there is another way to visualize the “quick way” of multiplying by 400. Once we know the “number fact” that 4 × 7 = 28, we also know such facts as… 4 × 7 apples = 28 apples next 4 × 7 lawyers = 28 lawyers 4 × 7 hundreds = 28 hundreds

5. © Math As A Second Language All Rights Reserved The latter result, stated in the language of place value (replacing the noun “hundred” is the same as multiplying by 100 which is the same as annexing two 0’s) says that 4 × 700 = 2800 (that is, 2,800). next In other words, once we know that 4 × $7 is $28, we also know that 400 × $7 is $2,800. 2 note 2 Going from 4 ×700 to 400 × 7 might have seemed a bit abrupt. It follows directly from our rules. More specifically, 4 × 700 = 4 × (7 × 100) = 4 × (100 × 7) = (4 × 100) × 7 = 400 × 7.

6. next © Math As A Second Language All Rights Reserved This observation gives us an insight to rapid addition. Using our adjective/noun theme, there is no need to know anything beyond the traditional multiplication table for single digits. next As an example, let’s make a multiplication table for a number such as 13 which isn’t usually included as part of the traditional multiplication tables.

7. © Math As A Second Language All Rights Reserved The idea is that we can think of 13 as being an abbreviation for the sum of 1 ten and 3 ones. Thus, a “quick” way to add thirteen is to add 1 in the tens place and then 3 in the ones place. next For example, starting with 13, 13 + 10 + 3 26 1 × 13 = 2 × 13 = we add 10 to get 23, and then add 3 ones to get 26. 23 next

8. © Math As A Second Language All Rights Reserved next Starting with 26, 26 + 10 + 3 39 1 × 13 = 13 3 × 13 = we add 10 to get 36, and then add 3 ones to get 39. 36 next 2 × 13 =

9. next © Math As A Second Language All Rights Reserved next Continuing with 39, 39 + 10 + 3 52 1 × 13 = 13 3 × 13 = we add 10 to get 49, and then add 3 ones to get 52. 49 next 2 × 13 = 26 4 × 13 =

10. next © Math As A Second Language All Rights Reserved In this way, our multiplication table for 13 would be… next 1 × 13 = 13 2 × 13 = 26 3 × 13 = 39 4 × 13 = 52 5 × 13 = 65 6 × 13 = 78 7 × 13 = 91 8 × 13 = 104 9 × 13 = 117

11. next © Math As A Second Language All Rights Reserved Suppose you now wanted to find the price of purchasing 234 items, each costing $13. You could count by 13’s until you got to the 234 th multiple. next This would be both tedious and unnecessary! However, once we know the “13 table” through 9, we can use the adjective/noun theme to find the answer in a relatively easy way.

12. © Math As A Second Language All Rights Reserved Imagine that the 234 items were stacked into three piles with 200 in the first pile, 30 in the second pile, and 4 in the third pile. next

13. © Math As A Second Language All Rights Reserved Thus, the value of the items in the first pile is $2,600. next 2 × 13 = 26200 × 13 = 2,600 3 × 13 = 3930 × 13 = 390 The value of the items in the second pile is $390. 4 × 13 = 52 Finally, the value of the items in the third pile is $52. next

14. © Math As A Second Language All Rights Reserved Hence, the answer to our question is $2,600 + $390 + $52 = $3,042. next With this technique, it is relatively easy to see how multiplication is really a special format for organizing rapid, repeated addition. However, in the traditional format in which multiplication is presented, this clarity is either lacking or obscured.

15. © Math As A Second Language All Rights Reserved For example, the most traditional method of finding the sum of 234 “thirteen’s” is to write the multiplication problem in vertical form, making sure that the number with the greatest number of digits is written on top. next 7 0 2 2 3 4 × 1 3 2 3 4 3, 0 4 2 next 5 2 1 3 × 2 3 4 3 9 3, 0 4 2 2 6 rather than… Therefore, we often write… next

16. © Math As A Second Language All Rights Reserved Notice that in the this format we are actually finding the cost of 13 items, each of which costs $234. This is not the problem we intended to solve, even though it gives us the same answer. 7 0 2 2 3 4 × 1 3 2 3 4 3, 0 4 2

17. next © Math As A Second Language All Rights Reserved Notice that in this format, we capture what it means to find the sum of 234 thirteen’s next In fact, this is precisely how we solved the problem originally, but it is now written in place value notation. 5 2 1 3 × 2 3 4 3 9 3, 0 4 2 2 6

18. next © Math As A Second Language All Rights Reserved For example, when we wrote “39” we placed the 9 under the 5, thus putting the 9 in the tens place. next 5 2 1 3 × 2 3 4 3 9 3, 0 4 2 2 6 In other words, since the 5 was already holding the tens place there is no need for us to write a 0 next to the 9. However, if we wish to, we can write the zero.

19. next © Math As A Second Language All Rights Reserved This form is a shorter version of… next 5 2 1 3 × 2 3 4 3 9 3, 0 4 2 2 6 which is itself a shorter version of… 0 0 = 4 thirteen’s = 30 thirteen’s = 200 thirteen’s = 234 thirteen’s …and in this form we see immediately the connection between the traditional algorithm and rapid repeated arithmetic. next

20. © Math As A Second Language All Rights Reserved Sometimes there is a tendency to confuse things foreign to us with what we believe is illogical. next Note 5 2 1 3 × 2 3 4 3 9 3, 0 4 2 2 6 7 0 2 2 3 4 × 1 3 2 3 4 3, 0 4 2 traditionalnon- traditional next

21. © Math As A Second Language All Rights Reserved While both forms give the same answer, they represent different questions. The traditional format shows us the sum of thirteen 234’s; while the non-traditional format shows the sum of two hundred thirty-four 13’s. next Note Notice, however, that in both formats each digit in the top number multiplies each digit in the bottom number. This lead us to… next

22. © Math As A Second Language All Rights Reserved The Generalized Distributive Property We learned the distributive property in the form b(c + d) = bc + bd. To multiply a sum of numbers by another sum of numbers, we form the sum of all possible products where one factor comes from the first sum, and the second factor comes from the second sum. For example, to compute the product of 3 + 4 + 5 and 8 + 9, we could form sum… (3 × 8) + (3 × 9) + (4 × 8) + (4 × 9) + (5 × 8) + (5 × 9) next

23. © Math As A Second Language All Rights Reserved In real life, we might have found it more convenient to rewrite… 3 + 4 + 5 as 12 and 8 + 9 as 17 …after which we would simply compute the product 12 × 17. However, while we can simplify 3 + 4 + 5 when we are doing arithmetic, in algebra it is not possible to simplify b + c + d in a similar manner. Note

24. next © Math As A Second Language All Rights Reserved Aside from any other of its practical uses, it is essential to know the generalized distributive property, it if we want to rewrite an expression in which letters are used to represent numbers. next Note For example, to form the product of b + c + d and e + f, we might use the distributive property to write the product in the form… (b × e) + (b × f) + (c × e) + (c × f) + (d × e) + (d × f)

25. next © Math As A Second Language All Rights Reserved The Adjective/Noun Theme And the Multiplication Algorithm To see how the adjective/noun theme was used in the multiplication algorithm, let’s revisit a typical computation such as… next 2 8 4 3 7 × 1 9 6 8 4 1 1 2 1 2, 2 3 6 note 3We have deliberately written the number with the few digits on top to help you internalize the fact that this is just as logical as doing it the traditional way. Students might like to do the problem the traditional way as well to see that the answer is the same in both cases. 3 next

26. © Math As A Second Language All Rights Reserved In this format, the nouns have been omitted. However, if we put them in, it becomes easy to see what is happening. next For example, when we multiplied the 3 in 437 by the 2 in 28 we were really multiplying 3 tens by 2 tens, and according to our adjective/noun theme… 3 tens × 2 tens = 6 “ten tens” or 6 hundred. 2 8 4 3 7 × 1 9 6 8 4 1 1 2 1 2, 2 3 6

27. next © Math As A Second Language All Rights Reserved In this sense, we can view the above algorithm in the form… next thousandshundredstensonesten thousands 28 4 37 56 1 4 2 4 6 3 2 8 11 136 11 12 36 2 36 2 2 36 1 next 7 ones × 8 ones = 56 ones 7 ones × 2 tens = 14 tens 3 tens × 8 ones = 24 tens 3 tens × 2 tens = 6 hundreds 4 hundreds × 8 ones = 32 hundreds 4 hundreds × 2 tens = 8 thousands

28. © Math As A Second Language All Rights Reserved For students who are visual learners, we can explain the above algorithm in terms of an area model. Imagine that there is a rectangle whose dimensions are 28 feet by 437 feet. next Note on the Area Model 437 ft 28 ft

29. next © Math As A Second Language All Rights Reserved On the one hand, the area of the rectangle is 28 feet × 437 feet or 12,236 square feet (that is, 12,236 “feet feet” or 12,236 ft 2 ). On the other hand, we can compute the same area by subdividing the rectangle as shown below. next 400 20 8 307 next

30. © Math As A Second Language All Rights Reserved Computing the area of each smaller rectangle, we obtain… next 400 20 8 307 next 8,000600140 3,20024056 400307

31. next © Math As A Second Language All Rights Reserved Notice how the areas of each piece match the set of partial sums we obtained using the algorithm. next 56 thousandshundredstensonesten thousands 28 4 37 1 40 2 40 6 00 3 2 00 8 000 2 236 1 20 8 307400 8,000 600140 3,20024056 next

32. © Math As A Second Language All Rights Reserved The Ancient Method of Duplation The ancient Egyptians anticipated the binary number system long before the invention of either place value or computers. next They realized that every natural number could be expressed as a sum of powers of 2. The Method of Duplation is a rather elegant way of performing rapid addition by knowing how to multiply by 2 and adding.

33. © Math As A Second Language All Rights Reserved To appreciate the Method of Duplation you might pretend that place value was based on trading in by two’s rather than by ten’s. next To have this seem more relevant, consider a monetary system in which the only denominations are $1, $2, $4, $8, $16, etc. $1$2$4$8$16$32$64$128

34. next © Math As A Second Language All Rights Reserved next For example, to give Rick $19, you could give him a $16 bill, leaving a balance of $3, and then give him a $2 bill and a $1 bill. $1$2$4$8$16$32$64$128 1621++=19 *** next In this form, you would never have to give someone more than one bill of each denomination.

35. © Math As A Second Language All Rights Reserved next So for example, to compute the sum of nineteen 67’s (that is: 19 × 67), the Egyptians would make the following table just by knowing how to double a number (the term, duplation). 1 × 67 = 67 2 × 67 = 134 4 × 67 = 268 8 × 67 = 536 16 × 67 = 1,074 32 × 67 = 2,148

36. next © Math As A Second Language All Rights Reserved next And knowing that 19 sixty-sevens was the sum of 16 sixty-sevens, 2 sixty-sevens, and 1 sixty-seven, they would simply perform the addition as shown below… 4 × 67 = 268 8 × 67 = 536 32 × 67 = 2,148 67 134 1,07416 × 67 = 1,074 2 × 67 = 134 1 × 67 = 67 1,273 next

37. © Math As A Second Language All Rights Reserved Here we have another subtle application of the “adjective/noun” theme. Namely, since 19 = 16 + 2 + 1, next Note 19 sixty-sevens 16 sixty-sevens 2 sixty-sevens + 1 sixty-sevens

38. next In our next presentation, we introduce the concept of unmultiplying ( our name for division). © Math As A Second Language All Rights Reserved 19 × 67 multiplication