Chapter IV: Section III: Multiplication

47. I have said that all questions in arithmetic require nothing but addition and subtraction. I do not mean by this that no rule should ever be used except those given in the last section, but that all other rules only shew shorter ways of finding what might be found, if we pleased, by the methods there deduced. Even the last two rules themselves are only short and convenient ways of doing what may be done with a number of pebbles or counters.

48. I want to know the sum of five seventeens, or I ask the following question: There are five heaps of pebbles, and seventeen pebbles in each heap; how many are there in all? Write five seventeens in a column, and make the addition, which gives 85. In this case 85 is called the _product_ of 5 and 17, and the process of finding the product is called MULTIPLICATION, which gives nothing more than the addition of a number of the same quantities. Here 17 is called the _multiplicand_, and 5 is called the _multiplier_.

17
17
17
17
17
----
85

49. If no question harder than this were ever proposed, there would be no occasion for a shorter way than the one here followed. But if there were 1367 heaps of pebbles, and 429 in each heap, the whole number is then 1367 times 429, or 429 multiplied by 1367. I should have to write 429 1367 times, and then to make an addition of enormous length. To avoid this, a shorter rule is necessary, which I now proceed to explain.

50. The student must first make himself acquainted with the products of all numbers as far as 10 times 10 by means of the following table,[8] which must be committed to memory.

+----+----+----+----+----+----+----+----+----+----+----+----+
| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
+----+----+----+----+----+----+----+----+----+----+----+----+
| 2 | 4 | 6 | 8 | 10 | 12 | 14 | 16 | 18 | 20 | 22 | 24 |
+----+----+----+----+----+----+----+----+----+----+----+----+
| 3 | 6 | 9 | 12 | 15 | 18 | 21 | 24 | 27 | 30 | 33 | 36 |
+----+----+----+----+----+----+----+----+----+----+----+----+
| 4 | 8 | 12 | 16 | 20 | 24 | 28 | 32 | 36 | 40 | 44 | 48 |
+----+----+----+----+----+----+----+----+----+----+----+----+
| 5 | 10 | 15 | 20 | 25 | 30 | 35 | 40 | 45 | 50 | 55 | 60 |
+----+----+----+----+----+----+----+----+----+----+----+----+
| 6 | 12 | 18 | 24 | 30 | 36 | 42 | 48 | 54 | 60 | 66 | 72 |
+----+----+----+----+----+----+----+----+----+----+----+----+
| 7 | 14 | 21 | 28 | 35 | 42 | 49 | 56 | 63 | 70 | 77 | 84 |
+----+----+----+----+----+----+----+----+----+----+----+----+
| 8 | 16 | 24 | 32 | 40 | 48 | 56 | 64 | 72 | 80 | 88 | 96 |
+----+----+----+----+----+----+----+----+----+----+----+----+
| 9 | 18 | 27 | 36 | 45 | 54 | 63 | 72 | 81 | 90 | 99 |108 |
+----+----+----+----+----+----+----+----+----+----+----+----+
| 10 | 20 | 30 | 40 | 50 | 60 | 70 | 80 | 90 |100 |110 |120 |
+----+----+----+----+----+----+----+----+----+----+----+----+
| 11 | 22 | 33 | 44 | 55 | 66 | 77 | 88 | 99 |110 |121 |132 |
+----+----+----+----+----+----+----+----+----+----+----+----+
| 12 | 24 | 36 | 48 | 60 | 72 | 84 | 96 |108 |120 |132 |144 |
+----+----+----+----+----+----+----+----+----+----+----+----+

[8] As it is usual to learn the product of numbers up to 12 times 12, I have extended the table thus far. In my opinion, all pupils who shew a tolerable capacity should slowly commit the products to memory as far as 20 times 20, in the course of their progress through this work.

If from this table you wish to know what is 7 times 6, look in the first upright column on the left for either of them; 6 for example. Proceed to the right until you come into the column marked 7 at the top. You there find 42, which is the product of 6 and 7.

51. You may find, in this way, either 6 times 7, or 7 times 6, and for both you find 42. That is, six sevens is the same number as seven sixes. This may be shewn as follows: Place seven counters in a line, and repeat that line in all six times. The number of counters in the whole is 6 times 7, or six sevens, if I reckon the rows from the top to the bottom; but if I count the rows that stand side by side, I find seven of them, and six in each row, the whole number of which is 7 times 6, or seven sixes. And the whole number is 42, whichever way I count. The same method may be applied to any other two numbers. If the signs of (23) were used, it would be said that 7 × 6 = 6 × 7.

● ● ● ● ● ● ●
● ● ● ● ● ● ●
● ● ● ● ● ● ●
● ● ● ● ● ● ●
● ● ● ● ● ● ●
● ● ● ● ● ● ●

52. To take any quantity a number of times, it will be enough to take every one of its parts the same number of times. Thus, a sack of corn will be increased fifty-fold, if each bushel which it contains be replaced by 50 bushels. A country will be doubled by doubling every acre of land, or every county, which it contains. Simple as this may appear, it is necessary to state it, because it is one of the principles on which the rule of multiplication depends.

53. In order to multiply by any number, you may multiply separately by any parts into which you choose to divide that number, and add the results. For example, 4 and 2 make 6. To multiply 7 by 6 first multiply 7 by 4, and then by 2, and add the products. This will give 42, which is the product of 7 and 6. Again, since 57 is made up of 32 and 25, 57 times 50 is made up of 32 times 50 and 25 times 50, and so on. If the signs were used, these would be written thus:

7 × 6 = 7 × 4 + 7 × 2.
50 × 57 = 50 × 32 + 50 × 25.

54. The principles in the last two articles may be expressed thus: If _a_ be made up of the parts _x_, _y_, and _x_, _ma_ is made up of _mx_, _my_, and _mz_; or,

if _a_ = _x_ + _y_ + _z_.
_ma_ = _mx_ + _my_ + _mz_,
or, _m_(_x_ + _y_ + _z_) = _mx_ + _my_ + _mz_.

A similar result may be obtained if _a_, instead of being made up of _x_, _y_, and _z_, is made by combined additions and subtractions, such as _x_ + _y_-_z_, _x_- _y_ + _z_, _x_-_y_-_z_, &c. To take the first as an instance:

Let _a_ = _x_ + _y_ - _z_,
then _ma_ = _mx_ + _my_ - _mz_.

For, if _a_ had been _x_ + _y_, _ma_ would have been _mx_ + _my_. But since _a_ is less than _x_ + _y_ by _z_, too much by _z_ has been repeated every time that _x_ + _y_ has been repeated;--that is, _mz_ too much has been taken; consequently, _ma_ is not _mx_ + _my_, but _mx_ + _my_-_mz_. Similar reasoning may be applied to other cases, and the following results may be obtained:

_m_(_a_ + _b_ + _c_ - _d_) = _ma_ + _mb_ + _mc_ - _md_.

_a_(_a_ - _b_) = _aa_ - _ab_.
_b_(_a_ - _b_) = _ba_ - _bb_.
3(2_a_ - 4_b_) = 6_a_ - 12_b_.
7_a_(7 + 2_b_) = 49_a_ + 14_ab_.
(_aa_ + _a_ + 1)_a_ = _aaa_ + _aa_ + _a_.
(3_ab_ - 2_c_)4_abc_ = 12_aabbc_ - 8_abcc_.

55. There is another way in which two numbers may be multiplied together. Since 8 is 4 times 2, 7 times 8 may be made by multiplying 7 and 4, and then multiplying that _product_ by 2. To shew this, place 7 counters in a line, and repeat that line in all 8 times, as in figures I. and II.

I.
+---------------+
| ● ● ● ● ● ● ● |
A | ● ● ● ● ● ● ● |
| ● ● ● ● ● ● ● |
| ● ● ● ● ● ● ● |
+---------------+

+---------------+
| ● ● ● ● ● ● ● |
B | ● ● ● ● ● ● ● |
| ● ● ● ● ● ● ● |
| ● ● ● ● ● ● ● |
+---------------+

II.
+---------------+
| ● ● ● ● ● ● ● |
| ● ● ● ● ● ● ● |
+---------------+

+---------------+
| ● ● ● ● ● ● ● |
| ● ● ● ● ● ● ● |
+---------------+

+---------------+
| ● ● ● ● ● ● ● |
| ● ● ● ● ● ● ● |
+---------------+

+---------------+
| ● ● ● ● ● ● ● |
| ● ● ● ● ● ● ● |
+---------------+

The number of counters in all is 8 times 7, or 56. But (as in fig. I.) enclose each four rows in oblong figures, such as A and B. The number in each oblong is 4 times 7, or 28, and there are two of those oblongs; so that in the whole the number of counters is twice 28, or 28 x 2, or 7 first multiplied by 4, and that product multiplied by 2. In figure II. it is shewn that 7 multiplied by 8 is also 7 first multiplied by 2, and that product multiplied by 4. The same method may be applied to other numbers. Thus, since 80 is 8 times 10, 256 times 80 is 256 multiplied by 8, and that product multiplied by 10. If we use the signs, the foregoing assertions are made thus:

7 × 8 = 7 × 4 × 2 = 7 × 2 × 4.
256 × 80 = 256 × 8 × 10 = 256 × 10 × 8.

EXERCISES.

Shew that 2 × 3 × 4 × 5 = 2 × 4 × 3 × 5 = 5 × 4 × 2 × 3, &c.

Shew that 18 × 100 = 18 × 57 + 18 × 43.

56. Articles (51) and (55) may be expressed in the following way, where by _ab_ we mean _a_ taken _b_ times; by _abc_, _a_ taken _b_ times, and the result taken _c_ times.

_ab_ = _ba_.
_abc_ = _acb_ = _bca_ = _bac_, &c.
_abc_ = _a_ × (_bc_) = _b_ × (_ca_) = _c_ × (_ab_).

If we would say that the same results are produced by multiplying by _b_, _c_, and _d_, one after the other, and by the product _bcd_ at once, we write the following:

_a_ × _b_ × _c_ × _d_ = _a_ × _bcd_.

The fact is, that if any numbers are to be multiplied together, the product of any two or more may be formed, and substituted instead of those two or more; thus, the product _abcdef_ may be formed by multiplying

_ab_ _cde_ _f_
_abf_ _de_ _c_
_abc_ _def_ &c.

57. In order to multiply by 10, annex a cipher to the right hand of the multiplicand. Thus, 10 times 2356 is 23560. To shew this, write 2356 at length which is

2 thousands, 3 hundreds, 5 tens, and 6 units.

Take each of these parts ten times, which, by (52), is the same as multiplying the whole number by 10, and it will then become

2 tens of thou. 3 tens of hun. 5 tens of tens, and 6 tens,

which is

2 ten-thou. 3 thous. 5 hun. and 6 tens.

This must be written 23560, because 6 is not to be 6 units, but 6 tens. Therefore 2356 × 10 = 23560.

In the same way you may shew, that in order to multiply by 100 you must affix two ciphers to the right; to multiply by 1000 you must affix three ciphers, and so on. The rule will be best caught from the following table:

13 × 10 = 130
13 × 100 = 1300
13 × 1000 = 13000
13 × 10000 = 130000
142 × 1000 = 142000
23700 × 10 = 237000
3040 × 1000 = 3040000
10000 × 100000 = 1000000000

58. I now shew how to multiply by one of the numbers, 2, 3, 4, 5, 6, 7, 8, or 9. I do not include 1, because multiplying by 1, or taking the number once, is what is meant by simply writing down the number. I want to multiply 1368 by 8. Write the first number at full length, which is

1 thousand, 3 hundreds, 6 tens, and 8 units.

To multiply this by 8, multiply each of these parts by 8 (50) and (52), which will give

8 thousands, 24 hundreds, 48 tens, and 64 units.

Now 64 units are written thus 64
48 tens 480
24 hundreds 2400
8 thousands 8000

Add these together, which gives 10944 as the product of 1368 and 8, or 1368 × 8 = 10944. By working a few examples in this way you will see for following rule.

59. I. Multiply the first figure of the multiplicand by the multiplier, write down the units’ figure, and reserve the tens.

II. Do the same with the second figure of the multiplicand, and add to the product the number of tens from the first; put down the units’ figure of this, and reserve the tens.

III. Proceed in this way till you come to the last figure, and then write down the whole number obtained from that figure.

IV. If there be a cipher in the multiplicand, treat it as if it were a number, observing that 0 × 1 = 0, 0 × 2 = 0, &c.

60. In a similar way a number can be multiplied by a figure which is accompanied by ciphers, as, for example, 8000. For 8000 is 8 × 1000, and therefore (55) you must first multiply by 8 and then by 1000, which last operation (57) is done by placing 3 ciphers on the right. Hence the rule in this case is, multiply by the simple number, and place the number of ciphers which follow it at the right of the product.

EXAMPLE.

Multiply 1679423800872
by 60000
------------------
100765428052320000

61. EXERCISES.

What is 1007360 × 7? _Answer_, 7051520.

123456789 × 9 + 10 and 123 × 9 + 4?--_Ans._ 1111111111 and 1111.

What is 136 × 3 + 129 × 4 + 147 × 8 + 27 × 3000?--_Ans._ 83100.

An army is made up of 33 regiments of infantry, each containing 800 men; 14 of cavalry, each containing 600 men; and 2 of artillery, each containing 300 men. The enemy has 6 more regiments of infantry, each containing 100 more men; 3 more regiments of cavalry, each containing 100 men less; and 4 corps of artillery of the same magnitude as those of the first: two regiments of cavalry and one of infantry desert from the former to the latter. How many men has the second army more than the first?--_Answer_, 13400.

62. Suppose it is required to multiply 23707 by 4567. Since 4567 is made up of 4000, 500, 60, and 7, by (53) we must multiply 23707 by each of these, and add the products.

Now (58) 23707 × 7 is 165949
(60) 23707 × 60 is 1422420
23707 × 500 is 11853500
23707 × 4000 is 94828000
---------
The sum of these is 108269869

which is the product required.

It will do as well if, instead of writing the ciphers at the end of each line, we keep the other figures in their places without them. If we take away the ciphers, the second line is one place to the left of the first, the third one place to the left of the second, and so on. Write the multiplier and the multiplicand over these lines, and the process will stand thus:

23707
4567
------
165949
142242
118535
94828
---------
108269869

63. There is one more case to be noticed; that is, where there is a cipher in the middle of the multiplier. The following example will shew that in this case nothing more is necessary than to keep the first figure of each line in the column under the figure of the multiplier from which that line arises. Suppose it required to multiply 365 by 101001. The multiplier is made up of 100000, 1000 and 1. Proceed as before, and

365 × 1 is 365
(57) 365 × 1000 is 365000
365 × 100000 is 36500000
--------
The sum of which is 36865365

and the whole process with the ciphers struck off is:

365
101001
------
365
365
365
--------
36865365

64. The following is the rule in all cases:

I. Place the multiplier under the multiplicand, so that the units of one may be under those of the other.

II. Multiply the whole multiplicand by each figure of the multiplier (59), and place the unit of each line in the column under the figure of the multiplier from which it came.

III. Add together the lines obtained by II. column by column.

65. When the multiplier or multiplicand, or both, have ciphers on the right hand, multiply the two together without the ciphers, and then place on the right of the product all the ciphers that are on the right both of the multiplier and multiplicand. For example, what is 3200 × 3000? First, 3200 is 32 × 100, or one hundred times as great as 32. Again, 32 × 13000 is 32 × 13, with three ciphers affixed, that is 416, with three ciphers affixed, or 416000. But the product required must be 100 times as great as this, or must have two ciphers affixed. It is therefore 41600000, having as many ciphers as are in both multiplier and multiplicand.

66. When any number is multiplied by itself any number of times, the result is called a _power_ of that number. Thus:

6 is called the first power of 6
6 × 6 second power of 6
6 × 6 × 6 third power of 6
6 × 6 × 6 × 6 fourth power of 6
&c. &c.

The second and third powers are usually called the _square_ and _cube_, which are incorrect names, derived from certain connexions of the second and third power with the square and cube in geometry. As exercises in multiplication, the following powers are to be found.

Number proposed. Square. Cube.
972 944784 918330048
1008 1016064 1024192512
3142 9872164 31018339288
3163 10004569 31644451747
5555 30858025 171416328875
6789 46090521 312908547069

The fifth power of 36 is 60466176
fourth 50 6250000
fourth 108 136048896
fourth 277 5887339441

67. It is required to multiply _a_ + _b_ by _c_ + _d_, that is, to take _a_ + _b_ as many times as there are units in _c_ + _d_. By (53) _a_ + _b_ must be taken _c_ times, and _d_ times, or the product required is (_a_ + _b_)_c_ + (_a_ + _b_)_d_. But (52) (_a_ + _b_)_c_ is _ac_ + _bc_, and (_a_ + _b_)_d_ is _ad_ + _bd_; whence the product required is _ac_ + _bc_ + _ad_ + _bd_; or,

(_a_ + _b_)(_c_ + _d_) = _ac_ + _bc_ + _ad_ + _bd_.

By similar reasoning

(_a_ - _b_)(_c_ + _d_) is (_a_ - _b_)_c_ + (_a_ - _b_)_d_; or,

(_a_ - _b_)(_c_ + _d_) = _ac_ - _bc_ + _ad_ - _bd_.

To multiply _a_-_b_ by _c_-_d_, first take _a_-_b_ _c_ times, which gives _ac_-_bc_. This is not correct; for in taking it _c_ times instead of _c_-_d_ times, we have taken it _d_ times too many; or have made a result which is (_a_-_b_)_d_ too great. The real result is therefore _ac_-_bc_-(_a_ -_b_)_d_. But (_a_-_b_)_d_ is _ad_- _bd_, and therefore

(_a_ - _b_)(_c_ - _d_) = _ac_ - _bc_ - _ad_ - _bd_
= _ac_ - _bc_ - _ad_ + _bd_ (41)

From these three examples may be collected the following rule for the multiplication of algebraic quantities: Multiply each term of the multiplicand by each term of the multiplier; when the two terms have both + or both-before them, put + before their product; when one has + and the other-, put-before their product. In using the first terms, which have no sign, apply the rule as if they had the sign +.

68. For example, (_a_ + _b_)(_a_ + _b_) gives _aa_ + _ab_ + _ab_ + _bb_. But _ab_ + _ab_ is 2_ab_; hence the _square_ of _a_ + _b_ is _aa_ + 2_ab_ + _bb_. Again (_a_- _b_)(_a_-_b_) gives _aa_-_ab_-_ab_ + _bb_. But two subtractions of _ab_ are equivalent to subtracting 2_ab_; hence the _square_ of _a_- _b_ is _aa_-2_ab_ + _bb_. Again, (_a_ + _b_)(_a_-_b_) gives _aa_ + _ab_-_ab_ -_bb_. But the addition and subtraction of _ab_ makes no change; hence the product of _a_ + _b_ and _a_- _b_ is _aa_-_bb_.

Again, the square of _a_ + _b_ + _c_ + _d_ or (_a_ + _b_ + _c_ + _d_)(_a_ + _b_ + _c_ + _d_) will be found to be _aa_ + 2_ab_ + 2_ac_ + 2_ad_ + _bb_ + 2_bc_ + 2_bd_ + _cc_ + 2_cd_ + _dd_; or the rule for squaring such a quantity is: Square the first term, and multiply all that come _after_ by twice that term; do the same with the second, and so on to the end.