CASE III. r Resolve the given number, or quantity, into two fac. tors, one of which shall be the greatest power contained in it, and set the root of this power before the remaining part, with the proper radical sign between them. (r) ExAMPLES. 1. Let v48 be reduced to its most simple form. JNote 1. When any number, or quantity, is prefixed to the surd, that quantity must be multiplied by the root of the factor above mentioned, and the product be then joined to the other part, as before. exAm PLES. 1. Let 2./32 be reduced to its most simple form. Here 2,732=2/16×2=8/? Ans. 2. Let 5 3/24 be reduced to its most simple form. Here 53/24=53/8×3=103/3 Ans. .Note 2. A fractional surd may also be reduced to a more convenient form, by multiplying both the numerator, and denominator by such a number, or quantity, as will make the denominator a complete power of the kind (r) When the given surd contains no factor that is an exact power of the kind required, it is already in its most simple form. Thus, v 15 cannot be reduced lower, because neither of its *factors, 5, nor 3, is a square. required ; and then joining its root, with 1 put over it, as a numerator, to the other part of the surd. (s) EXAMPLES. 1. Let v; be reduced to its most simple form. 3. Let V125 be reduced to its most simple form. 4, Let V294 be reduced to its most simple form. 5. Let Z56 be reduced to its most simpie form. 6 Let R/192 be reduced to its most simple form. 7. Let 7 v80 be reduced to its most single forn. 8. Let 9%. 81 be reduced to its most simple form. I31 Y ; be reduced to its most simple form. (s). The utility of reducing surds to their most simple forms, in order to have the answer in decimals, will be readily perceived from considering the first question above given, where it is found that V+=#v 14 ; in which case it is only necessary to extract the square root of the whole number 14, (or to find it in some of the tables that have seen calculated for this pupose) and then divide it by 7; whereas, otherwise, we must have first divided the numerator by the denominator, and then have found the root of the quotient, for the surd part; or else have determined the root both of the numerator and denominator, and then divided the one by the other ; which are each of them troublesome Processes when performed by the common rules ; and in the next example, for the cube root, the labour would be much greater. 10. Lett vi. be reduced to its most simple form. / 11. Let V/98a” a be reduced to its most simple form. 12. Let vzo-do be reduced to its most simple form CASE IV. When the surds are of the same kind, reduce them to their simplest forms, as in the last case ; then, if the surd part be the same in them all, annex it to the sum of the rational parts, and it will give the whole sum required. But if the quantities have different indices, or the surd part be not the same in each of them, they can only be added together by the signs + and —. Whence 7A/3 the sum 2. It is required to find the sum of 3/500 and 3/108. 4. It is required to find the sum of 3/; and 2vi ; * 3. It is required to find the sum of v72 and V128. 6. It is required to find the sum of V/180 and v405. 7. It is sequired to find the sum of 3 3/40 and 3/135. 8. It is required to find the sum of 43/54 and 53/128. 9. It is required to find the sum of 9M243 and 10.M.363. 10. It is required to find the sum of 3/; and 7./; O - - 1 11. It is required to find the sum of isv} and */33 `-- CASE V. RULE. When the surds are of the same kind, prepare the quantities as in the last rule ; then the difference of the rational parts annexed to the common surd, will give the whole difference required. But if the quantities have different indices, or the surd part be not the same in each of them, they can only be subtracted by means of the sign —. 2. It is required to find the difference of 3/192 and 3/24. 3. It is required to find the difference of 5/20 and 3V45. 4. It is required to find the difference of #v +. and 11 Whence 60 W6the difference, or answer required. |