impossible or imaginary, supposing a, b, and m to be real, and that a and b are positive. 2d. To show how imaginary numbers or quantities may occur in calculation, and to perceive their nature more fully, we will solve the following QUESTION. To find two numbers, whose sum is 8, and product 20. Solution. Because the square of the sum of two numbers equals the sum of their squares increased by twice their product, and the square of their difference equals the sum of their squares diminished by twice their product, it follows that 82-4 × 20 = 64 — 80 =- 16 equals the square of the difference of the numbers; consequently, since - 16 = 16 x1, by extracting the square root we get V-16=4-1 for the difference of the required numbers. Hence, since the half sum of two numbers increased by their half difference equals the greater number, and that the half sum diminished by the half difference equals the 8 4 V 1 less number, it follows that + 8 41-1 2 2 2 4+21-1 and = 4 — 2 √ — 1 will represent the sought num bers, which are imaginary; consequently, no real numbers can be found which will satisfy the conditions of the question. Now, although we can not conceive 4 + 2 - 1 and 4 21 to represent any real numbers, yet they are expressions which, with certain conventions, will be found to satisfy the conditions of the question. Thus, since 4 + 2 - 1 and 4-2-1 have been found by using the rules which we know to be applicable to real numbers and quantities, that is, since 4+2 - 1 and 421 have been found by treating them as being real numbers; reciprocally, when we apply them to see if they satisfy the conditions of the question, we must clearly use the rules which are applicable to real numbers or quantities, that is, we must treat 4 + 2 √ — 1 and 4-2-1 as if they were real numbers. Hence, by adding 4 + 2-1 and 4-2-1 we get 4+2-1+ 4-2-18; since, according to the rules which apply to real numbers or quantities, we must have 21 1-2—1 = 0. 4+2 Also, the product of 4 + 21 and 4-2-1 is ex pressed by (4 + 2 √ − 1) × (4 — 2 √− 1) = 16 + 8 √ — 18-1-4(-1); since, by the rules which are applicable to the multiplication of real numbers and quantities, the product of 21 and 4 is 8 -1, and that of 21-1 and 4 is 81, also the product of 2 - 1 and -2 - 1 must be expressed by - 2 × 2 × √- 1x √-1=4(-1). Because 8-1-8-10, our product is reduced to 164(-1); now because 1 has been obtained from 1 by putting -1 under the sign of the square root, reciprocally we must take the square of -1 by removing the sign of the square root; consequently, for (1) we must put 1, and our product reduces to (4 + 2 √ − 1) × (4 — 2 √ 1) = 16 - 4 x − 1 = 16 + 4 = 20. Hence, 4+2 - 1 and 4-2-1, when used according to the preceding conventions, or when used as if they were real numbers or quantities, have been found to satisfy the conditions of the question; noticing (particularly) that -1 is squared by removing the sign of the root, or putting 1 for (-1). It is clear that observations similar to the preceding will be applicable in all cases when imaginary numbers or quantities are used in calculation. 3d. It follows, from what has been done, that imaginary numbers or quantities must be added by the rules which are applicable to real numbers or quantities. Thus, the sum of √ 1, 3 — 1, and 7 — 1, is √ − 1 + 3 √ − 1 + 7 √ — 1 = 111; observing that the coefficient of 1 is 1. The sum of 1,8 - 1, 14-1, and 6-1 equals 7 √ −1 + 8 = 1; noticing that V-1 and -1 are unlike expressions, because the indices of the roots are different. Similarly, the sum of a b and that of cb is с expressed by (a + c) √ − b = (a + c) × √b × √−1, and that of mya, na, and pvc, is expressed by (m + n) × Va × V=1+pvc × √ — 1. Again, if we subtract 3-1 from 8-1, and b—1 from a 1, the remainders will be 51, and (a - b) -1; and in like manner if we subtract ma from nya, the remainder will be na-ni- a = n√ a × v = 1 − nv a × 1. 4th. If 2n stands for any positive whole number, a for any whole number or quantity, and A for any positive number or quantity, then since - A = A × 1, it is clear that a — A = a VA × 2n v 2n - 1 may be taken as the representa tive of any imaginary monomial. 5th. Multiplication and division of imaginary monomials, which result from the square root. EXAMPLES. 1. To find the square of a and the product of W and - b. Ans. (V-a)2 = (√ a × √— 1)2 = -a, and = √ a √ — 1 × √b × √− 1 = √ab × (√ — 1)2 = — √ ab. 2. To find the product of a and b, and the pro duct ofc and 3. To divide±√ a = ± √a × √−1 by ± √ − b = ± Hence, the product of two imaginary monomials of the square root which have the same sign, equals minus the product of their coefficients multiplied by the product of the numbers or quantities under the surd, regarding them as real; but if the monomials have unlike signs, the product equals plus the product of the coefficients multiplied by the product of the numbers or quantities under the surd, regarding them as real. Hence, also, the quotient of two imaginary monomials of the square root which have the same sign, equals plus the product of the quotient of the coefficients multiplied by the square root of the quotients of the numbers or quantities (under the surd) regarding them as real; but if the imaginaries have unlike signs, the quotient found (as before) must have the negative sign. EXAMPLES. -- 1. Multiply 5-7 by 3-2, and 4-8 by 51—3. Ans. 1514, and 20 1/24 = 40 1/6. 2. Multiply 3-a by 4 3. Multiply-5-3 by 3-6, and — a √ — c -cv-d. v - c by 4. Divide ± 4-10 by 2-5, and 15 - 8 by 5 √ 2. 5. Divide 30 8 by 10 -3. Ans. 31 216. 3 6. Find the second, third, and fourth powers of V 1. 7. Find the product of 2 -1,5 - 1, and -7—1. Ans. 70(-1)=-70-1. 8. Multiply 1-1 by 1--1, and find the square of 1-1. Ans. (1 + √1) × (1 − √ − 1) = 1 + √√ − 1 − √− 1 − (-1)= 2, and (11) 1+2 √1-1 = ± 2 1-1. 9. Multiply a + b by a√-b, and find the cube of c± √ Ans. (a + √ b) × (a — √ — b) = a2 + a √ — b — a v = b — (√ — b)2 = a2 + b, and (c±√ — d)3 = c2 ± 3c2 √-d3cd V-d3. 10. Divide 3 + 2-1 by 3-2√ − 1, a + √ − b by a-v-b. 11. Find the value of (3+4 -1+ √3-4√ 3 — 4 √ — 1) 2. Ans. (3 + 4 √ − 1 + 3-4 √1+10) = 2, a real number; consequently, imaginary expressions are sometimes representatives of real numbers or quantities. 12. Divide a2 + b2 by a + b — 1, and 1 by 1-- 1. Ans. a-by-1, and 1+-1 6th. It is clear, from what has been done, that for (√ — a)2 b we may write a x-b= - Vab. REMARKS. 1 x 1. If we expand 1 and = (1 − a) ̄3by the x binomial theorem, according to the ascending powers of x, Supposing to be positive, it is clear that (1) and (2) are true so long as x is not greater than 1; so that their right members may be taken as representatives of 1 and and vice versa; consequently, if we put 1 for a in |