RULE. Multiply the given Denomr by the Numr of the Fraction, and divide the Product by its Dent; the Quote (if there is no Remainder) is the Numr fought. Examp. To reduce to an Equivalent Fraction, having for its Den 12. It is. Thus, 3x1236, and 3649, the Num fought. a DEMONST. This follows from Corol. to the laft. For let the given Fraction be suppose the Numr fought is c And because by fuppofition, then ad=bc; therefore dividing both by b, it is , if the Dent to which it is to be reduced be d, ad = d = c, according to the Rule. SCHOLIUM. If the Divifion has a Remainder, the Problem is plainly impoffible; yet the given Fraction is equal to the Sum of two Fractions, one of which has the given Dent, and its Numr is the Integral Quote of the Dividend directed, by the preceding Rule; and the other has for its Num" the Remainder of the Divifion, and the Den, is the Product of the given Dent and the Dent of the Fraction reduced. For Examp. if is propofed to be reduced to the Denr 5, I take 4×5=20; then 2072, and 6 remains. Whence I conclude, that + Univerfally, Let it be propofed to reduce is. to the Den' m. And let n) a mq, and r remaining; then the Problem is impoffible. But I fay, that2 + =/ am 71 DEMONST. Since 9+, then dividing both by m, it is==+, by Lemma 2. For expreffing the Sum of 9+ am 2 am the m Part of which is the Sum, is the Sum of the m Parts of q and, i. e. 1 + 2/2. m mn PROBLEM VIII. To reduce a Fraction to an Equivalent one, having a given Num", (if possible.) RULE. Multiply the given Numr by the Den of the given Fraction, and divide the Product by its Numr, the Quote (if there is no Remainder) is the Correfpondent Den fought. Examp. To reduce to a Fraction having 18 for =108, and 108—4—27: So the Fraction fought is Numr " take enam, then is === DEMONST. By reversing the given Fraction, and taking the given Num as a Den, it becomes the fame Cafe with the preceding Problem; and it has been fhewn, that if two Fractions are equal, they are fo.when reverfed. But we may argue for this the fame way as in that Problem: Thus, if then amcn, (Lem. 6.) and m=". SCHOLIUM. If there is a Remainder, the Problem is impoffible; yet we can find two Fractions, the one of which has the given Num, and whofe Difference is equal to the given Fraction. For which, this is the Rule; viz. Having multiplied the given Num into the Denr of the Fraction, and divided the Product by its Numr, take the Integral Quote as a Denx to the given Num. And if from this Fraction you fubtract another, whofe Numr is the Remainder, and the Denr is the Product of the Dents, (viz. of the given Fraction and that last found) this Difference is equal to the given Fraction. Examp. If it's proposed to reduce to a Fraction whofe Numr is 9; work thus, 7x9=63. Then 63512, and 3 remains. And then I fay, being=7x 12.) Univerfally, If it's propofed to reduce to the Num 7, with ✔ remaining, then is not reducible to fuch a Num. But I say, 284 (84 ; and if a = b n DEMONST. Since = q, and r remaining, then is b naqr, (by the Proof of Divifion,) Hence dividing equally. by b, it is n+. And again dividing by 9, it is, 72 agtr aq aqtr. b 49 a q bg big+q. (Lem. 2.) But = (Corol. 2. Lem. 5.) Where fore -= + Hence lastly, by equal Subtraction, According to the Rule. Obferze, The preceding Problems relate all to Abstract Fractions, . e. the Fraction reduced, and that to which it is reduced, are fuppofed to have the fame abfolute Deno-: mination, or all to be applied to the fame Integer; therefore there is none mentioned." The following Problems concern Fractions as they are fpecially Applicate. To reduce a Fraction of an Unit of a higher Value, to an Equivalent Fraction of an Unit of a lower Value, thefe Units having a known Relation to one another, i. c. the leffer being equal to a certain known aliquot Part of the other... RULE. Take the Reciprocal of the Fraction which expreffes what Part or Parts the lower Unit is of the higher, and making that with the given Fraction (of the higher) the two Members of a Compound Fraction, reduce it to a Simple, [by Prob. 4.] i.e. multiply the two Num together and the two Dents, the Products make the Fraction fought. And obferve, if the lower is an Aliquot Part of the higher, we have no more to do but multiply the Num" of the given Fraction of the higher by the Dent of that Part. Examp. I. To reduce of 1 to a Fraction of 1. it is 4 of 1 f for 1 b. is of 17 and 11. is 20h. Therefore of 1 is of 20. by the Rule. Examp. 2. To reduce of 17. to a Fraction of 1 Merk, it is is of 11. Therefore 11, is of 1 Merk, (Lem. 7.) So that of 1. is of 2 of 1 Merk, which, according to the Rule, makes 12, the Fraction fought. DEMONST. In the preceding Examples, I have made the Reafon obvious, But to demonftrate it here Univerfally let it be propofed to reduce of a higher Unit to a Fraction of a lower, which is of the higher. I fay it is of of the lower: for of the higher, this must be of the other, (Lem. 7.) Therefore of () of the lower, according to the Rule. And if >the the lower Unit is an Aliquot Part of the higher, all we have to do, is to multiply the Numr of the given Fraction (of the higher) by the Dent of the Aliquot Part. So it is am SCHOL. If a Compound Fraction, or a Fraction of a Number greater than Unity, is proposed, firft reduce it to a Simple Fraction, and then proceed as above. PROBLEM X. To reduce a Fraction of a lower Unit to a higher, (the lower having a known Relation to the higher.) RULE. Make a Compound Fraction of the given Fraction (of the lower,) and that Fraction which expreffes what Part or Parts the lower is of the higher; and reduce this Compound to a Simple, you have the Fraction fought. Exam. 1. To reduce of 1 b. to the Fraction of 1. it is.) For 1 fb. being therefore of of a 1. which is according to the Rule. Examp. 2. of 1 Merk of 17. For 1 Merk. therefore of 1 Merk is =of = of of 1 .. according to the Rule. I I The Reafon of this Rule is obviously the fame in all Cafes, and needs not be farther. infifted on. SCHOL. In either of the two laft Problems, if there are any intermediate Species be twixt the two given Units; and if inftead of the Relation betwixt the higher and lower, there be given the feveral Relations betwixt the Extremes and the Intermediate Species, then reduce the given Fraction to the first intermediate Species, and from that to the next, till you come to the Species required. Examp. of 17. reduced to the Fraction of I Farthing, is 1920, which is found either all at once by knowing that 1 Farthing is, of 17. or by degrees thus, l=4b-40 d=1920 farthings. By multiplying the Numerators gradually by 20, 12, and 4. PROBLEM XI. To express any Applicate Whole Number, fimple or mixed, by a Fraction of some superiour Integer: CASE I. For a Simple Number, make it the Num3, and for Den" take the Number of the inferiour Species which is equal to 1 of the fuperiour; and that is the Fraction fought. So 8 d. is of 1 fb. or of 1 1. ČASE 2. For a mixed Number, reduce it to the lowest Species expreffed in it, and make that the Num; and the Number of that lower Species which is equal to 1 of the given fuperiour Species make the Den, and that is the Fraction fought. Examp. To exprefs 12 fb. 8d. 3 f. by the Fraction of al. it is for the mixed Number is 611 f. and 1f. is therefore 611f. is 611 times! / PROBLEM. XII. To find the Value of a Fraction of any Unit (or other Number) of a given Name, in Integers of lower Species, (where there are any such.) RULE. The given Fraction being (or made) a Simple Fraction, reduce it to a Fraction of the next lower Species, (by Prob. 9.) which being improper, reduce it (by (by Prob. 1.) and the Integral Quote is the Answer in that Species, if there is no Remainder; but if there is a Remainder, it makes a Fraction of that Species; with which you are to proceed to the next Species, and reduce as before; and fo on to the lowest: then the Integral Number found in each Species, with the Fraction of the lower, if there is a Remainder, make up the complete Answer. [And obferve, if the Fraction of the firft, or any fucceeding lower Species is Proper, it is plain you can have no Integer of that Species; and fo you must proceed, and reduce it to the next continually till you have an Improper Fraction: And if you never find fuch a Fraction, then the given Fraction is not expreffible in Integers.] Examp. 1. of 17. is = 13b. 4 d. which I find thus, != (Prob. 10.) 13 fb. fb = b. (Prob. 1.) and bd. (Prob. 10.) = 4 d. (Prob. 1.) The Reafon of this Rule is evident of itself. SCHOLIUMS. 1. This Problem fuppofes the given Fraction a Proper one; but for an Improper, first reduce it, and the Integral Quote is the first Part of the Value fought. Then proceed with the Remainder according to the Rule. 2. This Rule is accommodated to all Cafes, whether the lower Units be Aliquot or Aliquant Parts of the higher. But because in the Cafes which moft commonly occur, they are Aliquot Parts, therefore the Operation is the more Simple; and the Rule may be expreffed thus, viz. Reduce the Numr of the given Fraction (as an Integer) to the next lower Species, till the Product be equal to, or greater than the Denr; then divide by the Den', the Integral Quote is the Part of the Answer in that Species: Reduce the Remainder to the next Species, and divide as before (by the Den) and fo on to the lowest Species; and you have the Answer either in a Simple Whole Number of one Species, or Mixed of different. And if there is a Remainder upon the laft Species, it makes that Part of the Anfwer belonging to that Species a Mixed Number with a Fraction. And this in effect is the fame as the preceding Rule. Examp. 2. To find the Value of 1. it is 13 b. 8 d. 2 f. Operation. 241. . 20 35) 480b (13b. 35 12 35) 300d (8 d. 280 4. 35) 80 ƒ(2ƒ. 70 This way of ordering the Operation is diftinct and eafy; and it is exactly according to the General Rule of the Problem, which you will readily perceive by comparing. For this Reduction of the Remainder, and then the Divifion of the Product, is exactly the Operation whereby the Fraction made of that Remainder is reduced to the Fraction of the next Species, and that again reduced to a Whole Number. 3. If the Fraction to be valued refers to a Whole Number, greater than 1, as of 57. or to a Mix'd Fraction, as of 6. let the Expreffion be reduced to a fimple Fraction, and then find the Value. If that timple Fraction is improper, reduce it to its equivalent Whole Number, and then find the Value of the Remainders in inferiour Species. For the valuing the Fraction of a Mix'd Whole Number, as of 481. 14b. 8d. it is to be done by multiplying the Mix'd Number by the Num of the Fraction, and dividing the Product by the Derr; for of any kind of Quantity is the fame as of 2 times that Quantity. Or Generally, of any Quantity is of a times that Quantity, which fhews the Reason of this Rule. a B A CHA P. III. ADDITION of FRACTIONS. DEFINITION. DDITION of FRACTIONS, is finding a Fraction equal to all the given PROBLEM. To add two or more Fractions into one Sum. Rule. Reduce all the given Fractions to fimple Fractions, of one Unit, and one Dens, (if they are not fo already;) then the Sum of the Numrs being made a Num to the common Der, makes the fractional Sum fought, (which may be further reduced as the Cafe admits.) SCHOLIUM. In the following Examples, I thought it fuperfluous to write down the Operations; but I have fet down the Effect of every Step in the Work, feparating them from each other by the Mark of Equality, fhewing that what follows is equivalent to what precedes; being only the fame Fractions reduced (where it was neceffary) to a different State, according to the Direction of the Rule: Which therefore being, compared with the Rule, all will be clear and manifest. 40 8 Ex. 4of+of+=+=+84 + 3 + 4 =364 In the preceding Examples the Integer is fuppofed to be the fame in all the given Fractions, therefore I have named none; but in the following we fhall make them dif ferent. Examp. 5. 1+.=°° sb.+zsb. =°7. b. =8b. = 8:10: 1. |