9. I. The denominator of a finite decimal expressed in fractional form is 1 with as many ciphers annexed as there are places in the decimal. The same is true of the finite part of a mixed circulating decimal.

II. To determine the denominator of a repetend in the fractional form, Art. 193, Supplement, page 391, may be applied, as follows:

3i • •

Proposition I. — Prove that .3 = = -Î = f, and that .54 = ff.

Demonstration.— The ratio of 3J to 10 equals the ratio of ¿ to 1; i.e., 8a:10 = *:l. Hence. -U8^.

Again, .54 = .54A = «£. M* : 100 = A : 1. Hence, 5^ = Z = fi

Proposition' II. — Prove that the terminal fraction of a repetend expressed in the common decimal form (as .6 = .6f) always bears the same ratio to 1 as the entire repetend bears to its denominator when ex

pressed in the fractional form (.Cf = -~); г. с., f : 1 = Of : 10.

III. Any repetend is but another form for a decimal expressed by the figures of the repetend and a terminal fraction, — or for a fraction having for its numerator the mixed fractional number expressed by the figures of the repetend and the terminal fraction, and for its denominator 1 with as many ciphers annexed as there are places in the repetend, — or for any reduction of such fraction; and

The fraction reduced to lower terms by subtracting the terminal fraction from the numerator and 1 from the denominator, has for its numerator the figures of the repetend, and for its denominator as many 9's as there are figures in the repetend. Ilence,

To reduce a repetend to a fraction,—

Rule. Write the repetend for the numerator, and as many O's as there are repetend places for the denominator.

10. A mixed circulating decimal may be expressed in the common decimal form with a terminal fraction,

as .43 = .4| = -& + TJV = цр- (or *«iJ+3, see Art. 259) = fg; also, .2436 = э% + rffo. = *W#A (or iii^^tu, Art. 259) = £Щ

The numerator of each of these results may be obtained by subtracting the finite part from the entire circulating decimal (see page 391, 6); and the denominator, by writing as many 9's as there are repetend places and as many ciphers following as there arc finite places. Hence, To reduce a mixed circulating decimal to a fraction,— II. To reduce shillings and pence to the decimal of a pound.

Rule.—I. For the numerator,—Subtract the finite part from the entire circulating decimal; and

IL For the denominator,—Write as many 9's as there are repetend places with as many ciphers annexed as there are faute places.

Ex. 1. Reduce .456 to a fraction. Ex. 2. Reduce .37468 to a fraction.

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11. Circulating decimals may be changed in form by expanding them. In any such expansion, observe

1. That the order of succession of the repetend figures must remain unchanged.

2. That the number of repetend places must remain unchanged, or be changed to a multiple of that number of places.

e.g., .36 = .303 = .3636 = .36363 = .36363636, and so on.

In the first and second of the expansions above, both the order of succession and the number of places in the repetend remain unchanged. In the third and fourth, the order of succession remains unchanged, but the number of repetend places is changed to a multiple of 2, the original number of repetend places.

Proposition.—Prove,by reducing to fractions in their lowest terms, that the expansions above given produce no change in the value of the decimal.

12. Similar répètent!» begin at tbe same decimal place; Conterminous repetends end at the same decimal place; Like repetends are both similar and conterminous; hence,

Principle. Only like repetends can be added or subtracted.

13. To make repetends similar,—

Rule. Expand the circulating decimals till tJie number of finite places in each equals the greatest number of finite places in any of them.

14. To make similar repetcuds conterminous,—

Rule. Expand each repetend to as many places as are indicated by the least common multiple of the numbers of places in the several repetends.

Ex. Make the circulating dec- Pkocess.

. .... . • . . , Similar and con

imals .80, .8Ï54, .363, .42758 ..... itrminma

similar and conterminous. .86 =.8660 =.866666666

Explanation.— I first examine .8754 =.87645 =.875454545

finite places in any one of them, . 43 758 = .42758 =.427585858 and find it is three in the fourth

number. I therefore expand each repetend to the fourth decimal place, over which I place the repetend dot, and where necessary continue the expansion until I have the original number of repetend places in each. The repctends are now similar.

I find that the numbers of repetend places in the several repetends arc expressed by 1, 2, 3, and 2. I therefore find the least common multiple of 1, 2, 3, 2 — which is в,— and expand each repetend until it contains six repetend places. The repetends are now similar and conterminous.

16. To add or subtract circulating decimals,—

Rule. — I. If necessary, make tlie repctends similar and conterminous. П. Add or subtract as in integers, always carrying from the left-hand repetend column when adding or subtracting the right-hand column.

16. To multiply or divide repetends,—

Rule. — I. lîeduce the repetends to fractions.
II. Multiply or divide as infractions.
III. Reduce the fractional result to a decimal.

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Problems.
15.912
о. 74.237
21.468
54.3Í
2.750
.0018
.81234
«. 27.1235
f. 89.6543
.8184
2.5733
91.25
4.36

14. Add -J- and •$•; then reduce the fractions to decimals, add these results, and reduce the sum to a fraction, making the results agree.

15. Add $ and $ and reduce the result to a decimal; then reduce the fractions to decimals separately, and add the decimals, making the results agree.

SHORT METHODS OF REDUCTION. Page 84; Art. 209.

Pupils should give reasons for all Short Methods not analyzed. I. To reduce hundred-weight and pounds to the decimal of a ton.

Ex. Reduce 3 T. 5 cwt. 24 Ib. to tons Process.

and decimals.

Annex the pounds as hundredths to the number of hundred-weight, and multiply the number thus formed by 5, pointing off two additional decimal places.

Problems.

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Ex. Reduce £18 5s. 9d. to pounds and decimals.

Multiply the number of shillings by 5 and Process.

write the product as hundredths of a pound. . r ли

Multiply the number of pence by 4J and ° x 'J = ~ & '5

write the product as thousandths of a pound. 9 x 4fc 03 75 =z £ .0375

The sum of the two products will be the ¿18.28 75

decimal required.

Problems.

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SHORT METHODS OF MULTIPLICATION. Page 131; Art. 264.

I. To multiply by 29, 39, and so on to 99.

7,4 70 Ex. Multiply 747 x 59. Q

59 = 60-1, and 60 = 0 x 10. 44,820

6 times 10 times a number minus once the number is 59 times 7 J/7

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II. To find the product of two factors, when £, i, \, etc., of one of them is a convenient multiplier. p

Ex. Multiply 42 by 21. 43 x3 = 126

Introducing the factor 3 into the multiplicand, and rejecting я 1 —• 3 7

it from the multiplier does not affect the quotient. S8 2

III. To find the product of two factors each expressed by two figures,—the tens alike and the sum of the ones, 1O.

Ex. Multiply 83 by 87. Process.

80x90 + 21 = 7 231 Multiply the tens of one factor by the nest greater

number of tens, and to this product add the product of the ones of both factors.

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