What is the value of X 5?

How much will 8 books cost at 3 dollars apiece?

(c.) The number supposed to be taken is called the multiplicand, the number showing how many times the multiplicand is supposed to be taken is called the multiplier, and the result is called the product.

(d.) The multiplier and multiplicand are called factors of the product.

(e.) The product is said to be a multiple of its factors.

Illustrations. In the first of the above examples, 6 is the multiplicand, 7 is the multiplier, and the answer, 42, is the product. 7 and 6 are factors of 42, and 42 is a multiple of 7 and 6.

In the second example, 9 is the multiplicand, 6 is the multiplier, and the answer, 54, is the product. 9 and 6 are factors of 54, and 54 is a multiple of 5 and 6.

(f) The last example would be solved thus:

If 1 book costs 3 dollars, 8 books will cost 8 times 3 dollars, which are 24 dollars.

Here 3 is the multiplicand, 8 is the multiplier, and 24 is the product. 8 and 3 are factors of 24, and 24 is a multiple of 3 and 8.

(g.) In performing the operation, the multiplier must always be regarded as an abstract number.

Illustration. A number can be taken 3 times, 5 times, or 8 times, but it would be absurd to speak of taking it 3 bushels times, 5 houses times, or 8 books times.

(h.) The product must be of the same denomination as the multiplicand.

=56 bushels; 9 times 4 books = 35 tenths, &c.

Illustration. 7 times 8 bushels 36 books; 7 times 5 tenths

NOTE. It must be observed, that there is an apparent exception to the last statement, (h.) when the multiplier is a fraction, for .6 times .04 = .024; .02 times .0003 = .000006, &c. This will be explained in the section on fractions. (See page 173, Note.)

(i.) To examine the nature of the operation on the numbers, let us suppose that a person ignorant of all numerical processes, except that of counting, should be called upon to solve the last question.

(j.) If he had a quantity of dollars, he might lay 3 in one place, 3 more in another, 3 more in another, and so go on laying 3 in a place till he should have 8 piles of 3 dollars each. Since the dollars in each pile would buy 1 book, the dollars in all would buy 8 books; he might then, by counting the dollars in the 8 piles, find how much the books would cost.

(k.) If he should have no dollars, he might still determine the result in a similar way, by using pebbles, sticks, marks, or any thing else of a like character. After learning how to add, he might obtain the result by adding 8 threes together.

(1.) If, after having obtained the result in some way similar to the above, he should remember it, he would ever after be able, without counting or adding, to give the answer to any question requiring the amount of 3 taken 8 times.

(m.) If he should learn in a similar manner the several amounts of 10 and each number below 10, taken as many times as there are units in each successive number from 1 to 10, he would learn the common multiplication table as far as ten. If he should now learn how to apply this knowledge to the decimal system of numbers, he would be master of the process of multiplication.

NOTE. The very common definition, "Multiplication is a short method of addition," is not a good one, any more than would be, “ Multiplication is a short method of counting;" for while it is true that the results obtained by multiplication might be obtained by addition, it is equally true that they might be obtained by counting. It is true that multiplication has a dependence both on addition and counting, but it is equally true that it is as distinct from them as they are from each other, and that when we multiply we neither add nor count.

For instance, when we find the sum of 75798+24687 +39764 + 863284395 + 283 + 86536, by the method explained in article 50, we add them; but when we merely remember, and state that their sum is 317791, we do not add them.

So when we call to mind that 4 and 4 are 8, and 4 are 12, and 4 are 16, and 4 are 20, we add 5 fours together; but when we merely remember that 5 fours, or 5 times 4, are 20, we perform no addition, although as a result, we have in the mind the sum of 5 fours.

75. Product not affected by Change in Order of Factors

(a.) In determining the product of two numbers, it makes no difference which is regarded as the multiplicand, provided the other is regarded as the multiplier.

Thus 6 times 4 4 times 6, or 6 fours

Again: 5 times 33 times 5, or 5 threes

4 sixes = 24. = 3 fives = 15.

(b.) The principle may be proved true for all numbers, by the follow ing arrangement of dots:

:

Considering the dots as being arranged in horizontal rows, there are 3 rows with 5 dots in each row; considering them as being arranged in vertical rows, there are 5 rows with 3 dots in each row; and reckoning in either way we include all the dots.

(c.) Now, if these rows were extended in either direction, always being kept equal to each other, it is evident that the number of rows reckoned in one direction would always be equal to the number of dots that would be in a row were the rows reckoned in the other direction, and that all the dots would be reckoned in both instances. The number that represents the multiplicand when the rows are reckoned in one direction, will represent the multiplier when they are reckoned in the other, while the product, or number of dots, will be unaltered.

(d.) Hence, it must always be true that it makes no difference with the product which of the two factors is taken for a multiplier, provided the other be taken as the multiplicand. It will generally be most convenient to consider the larger factor as the multiplicand, though not always so.

NOTE. In changing the order of factors, the one taken for the multiplier should always be regarded as an abstract number, (see 74, g,) while the other should take the denomination of the original multiplicand. Thus, 4 times 3 apples = 12 apples; or changing the order of the factors, we should have 3 times 4 apples 12 apples. In the first case, 4 is an abstract, and 3 a concrete, number; but in the second, 4 is a concrete, and 3 an abstract, number.

=

So 6 times $8: 8 times $6; 4 times $.09 = 9 times $.04; times $4.6 = 469 times $.05; &c.

76. Simple Multiplication. — When only one Factor is a large Number.

When either factor is a large number, it will be well to consider its denominations separately, and, if we write the results as we obtain them, to begin with the lowest denomination.

What will 7 acres of land cost at $75.69 per acre?

Reasoning Process. · If 1 acre costs $75.69, 7 acres will cost 7 times $75.69, which can be found by multiplying it by 7.

In performing the requisite multiplication, the numbers are usually written in some convenient way, as the following:

Explanation. - Beginning at the right hand, or lowest denomination of the multiplicand, we have 7 times 9 hundredths

6 tenths and 3 hundredths.

=

63 hundredths, or

Writing the 3 in the hundredths' place, and reserving the 6 tenths to add to the product of the tenths by 7, we have 7 times 6 tenths = 42 tenths, and 6 tenths added, = 48 tenths, = 4 units and 8 tenths.

Writing the 8 tenths, and reserving the 4 units to add to the product of the units by 7, we have 7 times 5 units 35 units, and 4 units added,

tens to add to the product 49 tens, and 3 tens added,

Writing the 9 units, and reserving the 3 of the tens by 7, we have 7 times 7 tens = 52 tens, which, being our last product, we write. The result, then, is

529.83.

--

NOTE. As soon as practicable, the explanation should be abbre viated, so as to name only results. Thus, 63 hundredths; 42, 48 tenths, 35, 39 units; 49, 52 tens. Ans. $529.83.

77. Multiplication of Compound Numbers.

What will 8 casks of wine cost at £3 9 s. 7 d. per cask?

Reasoning Process. — If 1 cask costs £3 9 s. 7 d., 8 casks will cost 8 times £3 9 s. 7 d., which can be found by multiplying it by 8.

27 16 8 Product.

Explanation. Beginning with the lowest denomination, we have 8 times 7 d. = 56 d., which, since 12 d. = 1 s., must equal as many shillings as there are times 12 in 56, which are 4 times, with a remainder of 8. Hence, 56 d. = 4 s. 8 d.

Writing the 8 d., and reserving the 4 s. to add to the shillings of the

next product, we have 8 times 9 shillings =72 shillings, and 4 shillings from the former product added, are 76 shillings, which, since 20 s. = £1, must equal as many pounds as there are times 20 in 76, which are 3 times, with a remainder of 16. Hence, 76 s. = £3 16 s.

Writing the 16 s., and reserving the £3 to add with the pounds of the next product, we have 8 times £3 = £24, and £3 added, = £27, which, being the last product, we write.

78. Methods of Proof.

First Method. Go over the work a second time in the same manner as before.

Second Method. Consider the multiplicand as the multiplier, and see if this gives the same result as before.

The figures being in this way presented in a different order, we shall not be liable to repeat any mistake we may have made in the first work.

Third Method. Write out by itself the product of the multiplication of each denomination, beginning either at the left or right, and afterwards add these products together. The sum should equal the former product.

Below is the written work of the examples in 76 and 77, as proved by beginning at the left, and writing each denomination of the product separately.