* Here a, the first term of the dividend, being divided by Na, the first term of the divisor, gives a for the first term of the quotient. For a=a, and a = a, and the difference of the exponents is 1-, or; therefore a divided by a gives aa =√a, as above. Or it may be considered thus: ask what quantity being multiplied by a will give a, and the answer is va; then the divisor being multiplied by da,

1

the product is a-Nab; but there being no term in the dividend,

Here it is obvious, that the division cannot terminate without a remainder; therefore we write the divisor under the remainder with a line between them, and add the fraction to ac, the other two terms, to complete the quotient.

But when the dividend does not precisely contain the divisor, then we generally express the whole quotient as a fraction, having reduced it to its lowest terms, or rejected the letters and factors, that are found in every term of the dividend and divisor.

5. Thus,

that corresponds to -Nab, the second term of this product, we subtract a- ab from a-b, the dividend, and the sign of the quantity - ab being changed, the remainder is +√ab-b.

Now + √ab, the first term of this remainder, divided by Na,

the first term of the divisor, gives - E for the second term

of the quotient, by which we multiply the divisor, and the prod

uct, viz. +√ab-b, being subtracted from the aforesaid re mainder, nothing remains; and the quotient is Na+ √b.

5. Thus, a'bx+acx'+ax3 divided by adx+an

2

2

6. And a+ab+d* divided by a2-acta*c* gives

a+ab+d2

a-ac+ac

Here the quotient cannot be reduced to lower terms, be cause the factor a is not to be found in the term d2.

But it is to be observed, that though a fraction cannot be reduced to lower terms by a simple divisor, yet it may sometimes be so reduced by a compound one; as will appear in the reduction of fractions.

7. Divide a3+x3 by a+x.

2

2

Ans. a2-ax+x2.

8. Divide a3-3a2y+zay-y3 by a-y.

Ans. a2-2ay+y2.

ALGEBRAIC FRACTIONS have the same names and rules of operation, as fractions in Arithmetic.

PROBLEM PROBLEM I.

To find the greatest common measure of the terms of a fraction.

RULE.

1. Range the quantities according to the dimensions of some letter, as is shewn in division.

2. Divide the greater term by the less, and the last division by the last remainder, and so on till nothing remain; then the divisor last used will be the common measure required.

NOTE. All the letters or figures, which are common to each term of any divisor, must be rejected before such divisor is used in the operation.

Therefore x+b is the greatest common measure.

3. Το

1. Find the greatest common measure, as in the last problem.

2. Divide both the terms of the fraction by the com mon measure thus found, and it will be reduced to its lowest terms.

Therefore c+x is the greatest common measure;

cx+x2

and c+x) catan is the fraction required.

a

2. Having