(as b3) ÷ (a - b) = (as b3) ÷ (a + b) = = a1 — a3b + a2b2 — ab3 + b1. where the latter is evidently not a complete quotient. 265 a+b' These theorems will enable the student to effect important simplifications in the reduction of fractions, and of equations, and must therefore obtain sufficient attention before he proceeds further. PROBLEMS AND THEOREMS ON THE FIRST FOUR RULES OF ALGEBRA. 1. HALF the difference of two quantities added to half the sum gives the greater of them, and subtracted leaves the less. [Let the student select his own symbols, and illustrate it with his own numbers.] 2. If 2s = a + b + c, what are the values of sa, s b, and s c? and what is half their sum equal to? Find also their product, and arrange its terms systematically? 3. The difference between the square of the sum of two numbers and the square of their difference is equal to four times their product; and the sum of these squares of their sum and difference is double the sum of their squares. Prove this. ...... a, 4. The sum of two numbers multiplied by their difference is equal to . 5. If 2s =a+b+c+d, what is the sum, and what the product of s s—b, s — C, S d? and what is the sum of their squares? 6. Let several binomial factors, of which a is the first term, and where a, b, c, d, &c. are the second terms of the several factors, be multiplied together: then describe the manner in which the coefficients of the several powers of x in the product are formed of the quantities a, b, c, d, ... 7. Prove that a2 · a2 + (b + c) 2 = (a+b+c) (−a+b+c); and hence that 4a2b2 — (a2+b2 — c2)2 = (a + b + c) (− a+b+c) (a b+c) (a + b— c). y; and show that if any other odd whole number be substituted for 5 in these expressions, the division will terminate without a remainder. 9. Convert (u + x + y + z)2 into the form (u + x)2 + (u + y)2 + (u + z)2 + (x + y)2 + (x + z)2 + (y + z)2 — 2 (u2 + x2 + y2 + z2); and likewise into the form }{(u+x+y) + (u+x+2)2 + (u+y+z)2 + (x+y+z)2 — (u2+x2+y2+z2)} and again into u2+ (2u+x) x+ {2 (u+x) +y} y + { 2 (u+x+y+z)}z; and show that in this last, u may also change its place with either of the other quantities, x, y, or z. 10. Multiply a + b √−1 by a − b √1, and also by c+d√1; and multiply together four factors, each equal to a+b √−1, and then four others, each of which is a- √-1; and lastly, multiply the factors a+b √−1, c+d √ −1, e+f√−1, g+h√−1, and i + k-1 together. 11. If the term rectangle of two lines, in the first ten propositions of the Second Book of Euclid, be exchanged for the term product of two numbers, and square of a line for the square of a number; show that the propositions thus transformed are also true, whatever those numbers may be. 12. Divide a by 1+ 3, and by (−1 + √√/-3)2: and show that the quotient in the latter case is the same as would be obtained if we divide a by -1--3. Show also that a (— 1 13. Show that the sum of x (x + y is (x + y + z)2. √ −3)3 — a ( − 1 + √/−3)3. + z), y ( y + z + x), and z (z + x + y) 14. Simplify to the utmost degree the expressions x2 + 2xy + y2 − {x2+xy—y2— (2xy—x2—y2)} ; - [a + b — {a + b + c − (a + b + c + d) } ]; a a2 + (-b)2 + (— c)2 + 2a (— b) + 2a (−c) + 2 (− b) (− c). ALGEBRAIC FRACTIONS. ALGEBRAIC FRACTIONS have the same names and rules of operation, as numerical fractions in common arithmetic; as appears in the following Rules and Cases. CASE I. To reduce a Mixed Quantity to an Improper Fraction. MULTIPLY the integer by the denominator of the fraction, and to the product add the numerator, or connect it with its proper sign,+ or - -; then the denominator being set under this sum, will give the improper fraction required. To reduce an Improper Fraction to a Whole or Mixed Quantity. DIVIDE the numerator by the denominator, for the integral part; and set the remainder, if any, over the denominator, for the fractional part: the two joined together will be the mixed quantity required. CASE III. To reduce Fractions to a Common Denominator. MULTIPLY every numerator, separately, by all the denominators except its own, for the new numerators; and all the denominators together, for the common denominator. When the denominators have a common divisor, it will be better, instead of multiplying by the whole denominators, to multiply only by those parts which arise from dividing by the common divisor. Observing also the several rules and directions, as in Fractions in the Arithmetic. 1st fraction by bc, of the 2d by cx, and of the 3d by bx. Here there are no other quantities than a, b, and c, in any of the denominators, and none of the denominators contain more than one term: therefore abc will be the least common denominator. 2c За + 26 2c A similar method to this may often be advantageously employed. To find the greatest common Measure of the Terms of a Fraction. DIVIDE the greater term by the less, and the last divisor by the last remainder, and so on till nothing remains; then the divisor last used will be the common measure required; just the same as in common numbers. But note, that it is proper to range the quantities according to the dimensions of some letters, as is shown in division. Note also, that all the letters or figures which are common to each term of the successive divisors, must be thrown out of them, or must divide them, before they are used in the operation *. * After the student becomes expert in the resolution of expressions into their component factors, the following rule may be advantageously employed. Rule. To find the greatest common measure of two polynomials, one of which contains a letter which the other does not, arrange according to the powers of that letter: the greatest common measure required is that of the coefficients of the said letter and of the other polynomial. Arranging according to the quantity a, the second becomes 8d (z+6) a 14fg (z+6), or (8d — 14fg) (z+6). In like manner the first may be resolved into 6bc (z+6) + 10mx (≈ + 6), or (6bc + 10mx) (≈ + 6). The operation, therefore, is reduced to that upon 8d 14fg and 6bc + 10mx. But these have obviously no common measure but 2: therefore 2 (z+6), or 2z + 12, is the greatest common measure required. By similar processes the pupil will readily find that the greatest common measure of |