DIVISION OF FRACTIONS.

RULE.

Divide the numerator of one by the numerator of the other, and the denominator of one by the denominator of the other, the result is the quotient; but if they are not divisible by each other, invert the divisor, and proceed as in Multiplication.

INVOLUTION.

RULE.

This being only the multiplication of any given quantity continually by itself, to any required power, the operation will be manifest by an Example.

Required the second power of a + b, and the third and fourth power of the same.

a + b

a + b

a2 + ab

ab + b2

a2 + 2ab+b2 Second power required. a+b

a3 +2a2b + ab2

a2b + 2ab2 + b3

a3 +3a2b+3ab2b3 Third power required. a b

a1 + 3a3b + 3a2b2 + ab3

a3b+3a2b2 + 3ab3 + b4

a2 + 4a3b + 6a2b2 + 4ab3 + 64 Fourth power.

The labour of multiplication with binomials, or compound quantities converted into binomials, may in a great measure be avoided by the application of the 'Binomial Theorem," discovered by Sir Isaac Newton.

66

RULE.

A quantity containing two terms is generally called a binomial; in a binomial, the first term is raised to the given power, and the last term is also raised to that power; for example, if a + b be required to be raised to the second power, it will stand thus (a+b)2, which indicates the second power of a + b, therefore a will become a2 for the first term of (a + b)2, and b will become b2 for the last term; then the intermediate terms are found as follow:-the first term decrease, by unity from the given power, and the second term increase from its first power; therefore, the first being a2, the second will be ab, and the third will be b2, the number of terms being always one more than the given index.

Now we have a2, ab, b2, to which it is required to prefix numerical co-efficients. The co-efficient of the first term is always one, and if we multiply the co-efficient of a by its index, and divide by the number of terms to that place, we find the co-efficient of the following term. In the case of (a + b)2, the index of the first term is 2, and the co-efficient is 1, therefore 2 x 112 for the co-efficient of the second term; now the index of the second term is 1, and the co-efficient is 2, therefore 2 × 1, and divided by the number of terms to that place, is equal to 1, for the co-efficient of the last. Wherefore, for (a + b)2 we have a2 + 2ab + b2; all other powers are similarly raised.

Find the 6th power of a + b.

Ans...(a+b)6 = a + 6a5b + 15a4b2 + 20a3b3 + 15a2b4 + 6ab5 + bo.

We may similarly find the co-efficients and indices of the terms of (a + b)n

Ex. gr....(a+b)" the first term of which will be an; the second will be nan-1b; the third will be

n(n −1)an-2b2; the fourth will be n(n-1) (n-2)

2

2

3

a” — 363; the fifth will be n(n − 1) (n − 2) (n — 3)

an

2

3

4

an — 4b4, &c.; and the last term will be b”. Therefore

(a + b)n = an + nan−1b + n(n−1) an—2b2 +

2

1.-Raise 3x 2y to the sixth power.

Ans..."
.729.x6 2916x5y+4860x4y2

4320x3y32160x2y1 — 576xy + 64y6.

2.- Find the ninth power of x

Ans...x9

y, or (x y)9. 9x8y36x7y2 — 84x6у3 +

126x1y5 + 84x3y6 — 36x2y7 + 9xy — yo.

3. Find the third power of a + b + c.

Ans...a33a2b+3ab2 + b3 + 3a2c +

6abc3b2c+ 3ac2 + 3bc2 + c3

This last Example being a trinomial quantity, it will be necessary to cast it into a binomial, thus ((a+b)+c)3 and proceed precisely as in the others, considering a +b as a single quantity.

It will be requisite to remember that when the last term of a binomial is negative, the alternate terms of the product will be also negative.

Ex. gr....(a — b)2 = a2

· 2ab + b2.

Required to raise (ab) to that power.

First we have a1, a3b, a2b2, ab3, b4, by a decreasing, and b increasing, according to the Rule.

Next we have a1 — a3b + a2b2 ab3 + b, by prefixing the alternate signs.

for the co-efficients of each, according to the Rule.

Lastly we have la 4a3b+6a2b2 4ab3 + b, by prefixing the co-efficients.

EVOLUTION;

OR, THE

RULE FOR EXTRACTING THE ROOT OF ANY QUANTITY.

RULE FOR SIMPLE QUANTITIES.

The root of a quantity is found by dividing its index by the required root; thus, to find the a2, we divide the index by 2, as is evident, since a2 = a2

= a,

the required root. A surd quantity does not admit of such division thus, 37, 3/a2, &c., are surds, since the index of either is not divisible by the root. The roots of numerical co-efficients are extracted as in Arithmetic.