.. the denominator is 1, and the fraction is reduced to 3/9+ 3/6+/4. 296. Hence for the sum, or difference, of two cube roots, which is one of the most useful cases, the multiplier will be a trinomial surd consisting of the squares of the two given terms, and their product, with its sign changed.. 2 Ex. 8. Reduce shall be rational. 1/5+/3 a Ans. (V-Vry+Vy). · sty to a fraction whose denomina tor Ans. 1/125/75+1/45—1/27. 297. It may not be improper to take notice here of another transformation which binomial surd quantities may undergo by equal involution, and evolution. Ex. 1. To transform √2+√3 to a universal surd. (27+48)27+2/1296+48=147;.'••/27+ 48=√147=√✓49x3=7√3. Ex. 3. To transform 3/320-3/40 to a general surd. Here (2/320/40)=320-33/4096000+3 3/512000 -40=40; 3/320-3/40-23/5. 298. This transformation is very useful, since, by means of it, we can always reduce the sum or difference of any two surd quantities, if they admit of the same irrational part, to a single surd. This may be proved, in general, thus; if a and /b admit of the same irrational part, they must be of the formam and b'"m; and (a'"m+"/b'"m)"=a'"m+n ~ (a'n(n−1)m2--18'nm) + n(n−1) n⁄ (a'n (n−2m2−2 b'2nm2)+ &c. • 2 b'"m=a'"m—na'n-1×mb'n+ &c. .... b'am... /a+n/b="/ (a'"m+nma'n-'b'n +&c. . . . . b'nm)= the nth root of a rational quantity. Hence the product of a by vb is rational if ✔a and b admit of the same irrational part; also, 3/a3× Vb, or /a/b2, is rational, if 3/a and 3/6 admit of the same irrational part; and, in general, /a-X/b, or wax ✔b"-1, is rational, if / a and b admit of the same irrational part. 299. It is proper to observe, that, for the addition or subtraction of two quadratic surds, the following method is given in the BIJA GANITA, or the Algebra of the HINDOOS, translated by STRACHEY. Thus, to find the sum or difference of two surds, ✔a and b, for instance. RULE. Call ab the greater surd; and, if aXb is rational, (that is, a square), call 2 ab the less surd, the sum will be (a+b +2√ab), (= (√a±√b)2), and the difference (a+b2 Vab). If axb is irrational, the addition and subtraction are impossible; that is, they can only be indicated. Example. Required the sum and difference of 2 and 8. Here 2+8=10=> surd; 2x8=16, ..✔ 16=4, and 2/16 =2X4=8=<surd. Then 10+8= 18, and 10 ..√18= sum, and 2= difference. 8=2; in two places. In the first place add 1, and in the second subtract 1; then we shall have α a √[(√ï+1)3xb]=√a+vb, and √(√% —1). xb]=√a√b. α If is irrational, (that is, not a square), the addition or Ђ subtraction can be only made by connecting the surds by the signs or 1, as they are. STURMIUS, in his Mathesis Enucleata, has also given a method similar to the above. Ex. 4. To transform 2+3 to a general surd. Ans. (5-2/6). Ex. 5. To transform a—2x to a universal surd. Ans. (a+4x-41√√ax). Ans. 33/9. Ex. 6. To transform 33/3+3/72 to a universal surd. SV. METHOD OF EXTRACTING THE SQUARE ROOT OF BINOMIAL SURDS. 300. The square root of a quantity cannot be partly rational and partly a quadratic surd. If possible, let n=a+m; then by squaring both sides, na2+2am+m, and 2a/m=n—a2—m; n- —a2 -m therefore, m' a rational quantity, which is con " 2a trary to the supposition. A quantity of the form a is called a quadratic surd. 301. In any equation x+y=a+b, consisting of rational quantities and quadratic surds, the rational parts on each side are equal, and also the irrational parts. If x be not equal to a, let _x=a+m; then a+m+√y=a +√b, or m+√y=√b; that is, b is partly rational, and partly a quadratic surd, which is impossible, (Art. 300); .'.x=a, and y=b. 302. If two quadratic surds √x and y, cannot be reduced to others which have the same irrational part, their product is irrational. If possible, let ay=rx, where r is a whole number or a fraction. Then ry=r, and y=rx; •'• √y=r√/; that is, ✔y and may be so reduced as to have the same irrational part, which is contrary to the supposition. ง 303. One quadratic surd, ✔✅x, cannot be made up of two others, ✔m and n, which have not the same irrational part. If possible, let √x=√m+√n; then by squaring both sides, x=m+2√✓✅mn+n, and x n=2/mn, a rational - m quantity equal to an irrational, which is absurd. 1 304. Let (a+b)=x+y, where c is an even number, a a rntional quantity, b a quadratic surd, x and y, one or both of them, 1 quadratic surds, then (a-b)=xy. C By involution, a+b=x+cxo°¬1y+c.—1ब3y2+ &c., and 2 since c is even, the odd terms of the series are rational, and c-1 -- the even terms irrational; ...a=x+c. x-2y+ &c., and b=cxc1y+c. c-1c-2 2 3 -xc-3y3+&c., (Art. 301); hence, a—b 305. If c be an odd number, a and b, one or both quadratic surds, and x and y involve the same surds that a and b do respectively, and also (a+b)=x+y, then (a—b)c=x-y. By involution, a + b = x2 + cxo1y+c. C 1 ·.—=— xc-2y2+ &c., where the odd terms involve the same surd that x does, because c is an odd number, and the even terms, the same surd that y does; and since no part of a can consist of y and its x-2y+&c., and b-cx-1y+ parts, (Art. 301), a=x+c. c-1c- 2 C. 2 3 C- 1 c-1 2°¤—31⁄23+&c.; hence, a—b=x—cœ¬1y+c. —1 1 2y- &c. ; .. by evolution, (ab)=x-y. 2 306. The square root of a binomial, one of whose terms is a quadratic surd, and the other rational, may sometimes be expressed by a binomial, one or both of whose terms are quadratic surds. Let a+b be the given binomial, and suppose (a+√b) =x+y; where x and y are one or both quadratic surds; then √(a−√b)=x—y; .. by multiplication, (a2-b)=x2—y3, Ꮖ also, by squaring both sides of the first equation, a+b=x2+2xy+y, and a=x2+y2; .. by addition, a+√(a2-b)=2x2, and by subtraction, a√(a2—b)=2y2; and the root x+y=√[ža+š√(a2—b)]+ √ √(ab)]. From this conclusion it appears, that the square root of a+b can only be expressed by a binomial of the form x+y, one or both of which are quadratic surds, when a2—b is a perfect square. By a similar process it might be shown that the square root of a-b is [a+‡√(a2—b)]−√[‡a—±√(a2—b)], subject to the same limitation. Ex. 1. Required the square root of 3+2√2. Let (3+2/2)=x+y; then (3-2/2)=x-y; by multiplication, (9-8)=x-y; that is, 22-y-1. Also, by squaring both sides of the first equation, 3+2/2 -2xy-y, and 2+y=3; .. by addition, 2x2=4, and x=√2. Again, by subtraction, 2y2=2; .. y=1, and x+y=√2+1 the root required. Or, the root may be found by substituting 3 for a, 2√2= 8 for b, or 8 for b, in the above formula; thus, x+y=√[}++ √(9—8)]+√[÷−√(9——8)=√(3+1)+ √(一支)= 2+1. Ex. 2. Required the square root of 1983. Ans. 4+√3. Ans.7-5. Ans. 23. Ex. 3. What is the square root of 12→✔/140? Ans. 5-2. Ans. 6+5. Ex. 7. Find the square root of 18—10—7. Ans. 5--7. Ex. 8. Find the square root of -1+4—5. Ans. 2+-5. 307. The cth root of a binomial, one or both of whose terms are possible quadratic surds, may sometimes be expressed by a binomial of that description. Let A+B be the given binomial surd, in which both terms are possible; the quantities under the radical signs whole numbers; and A is greater than B. Let [(A+B) X✓Q]=x+y; then/[(A-B) ×√Q]=x-y; .. by multiplication, //[(A2-B2)×Q]=x2-y2; now let Q be so assumed, that (A2-B3) XQ may be a perfect cth power no, then a-y-n. Again, by squaring both sides of the first two equations, we have [(A+B)2×Q]=x2+2xy+y2 ✅[(A—B)3×Q]=x2—2xy+y2 ••• V/[(A+B)3×Q]+V[(A−B)3×Q]=2x2+2y2; which is always a whole number when the root is a binomial surd; take therefore s and t, the nearest integer values of ✓/[(A+B)2 XQ] and [(A-B)3×Q], one of which is greater and the other less than the true value of the corresponding quantity; then since the sum of these surds is an integer, the fractional parts must destroy each other, and 2x2+2y2=s+t, exactly, when the root of the proposed quantity can be obtained. We have therefore these two equations, x2—y2=n, and x2+y2={s |