Sect

fraction common divisor denominator numerator quantity

SECT. III.-FRA.CTIONS.

In any fraction the upper number, or the dividend, is called the numerator, and the lower number or divisor is called the denominator. Thus, in the fractioq, a is the numerator, and S the denominator.

If the numerator be less than the denominator, such a fraction is called a proper fraction ; but if the numerator be either equal to or greater than the denominator, it is called an improper fraction ; and if a quantity be made up of an integer and a fraction, it is called a mixed quantity.

a x a - a a + x Thus, a+ is a proper fraction ; and - a are both improper fractions; and b ax - is a mixed quantity.

The reciprocal of a fraction is another fraction, having its numerator and denominator respectively equal to the denominator and numerator of the former.

a Thus, a –is the reciprocal of the fraction -6.• If the numerator and denominator of a fraction be either both multiplied or both divided by the same quantity, the value of the resulting fraction is the same as before.

To demonstrate this proposition we shall throw the definition of a fraction into a categorical form. We shall a accordingly define the fraction b - as such a magnitude, that when it is multiplied by b. the product is a.

SECT-1 431 From this proposition, it is obvious that a fraction may be very differently expressed without changing its value, and that any integer may be reduced to the form of a fraction, by placing the product arising from its multiplication by any assumed quantity as the numerator, and the assumed quantity as the denominator of the fraction. It also appears that a fraction very complex in its form may often be reduced to another of the same value, but more simple, by finding a quantity which will divide both the numerator and denominator, without leaving a remainder. Such a common divisor, or common measure, may be either simple or compound; if it be simple, it is readily found by inspection, but if it be compound, it may be found as in the following problem.

Rule 1. Range the quantities according to the power of some one of the letters, as in division, leaving out the simple divisors of each quantity.

and the last divisor shall be the common measure required.

Note. - It will sometimes be necessary to multiply the dividends by simple quantities in order to make the divisions succeed.

The demonstration of this proposition depends on the Axtom, that whatever divides a number divides any multiple of the number ; and whatever divides two numbers divides their sum or difference. It was given by Euclid in Prop. 2, Eook vii., very much as follows : - Let a, 1; be the quantities, the smaller of which is b. Let a be divided by 5, with a remainder c, b by c, with a remainder d, c by d, with no remainder, d is the greatest common measure of a and b. We have a – pb= c, b – qc = c =rd.

Now, (1.) d is a common measure of a and ; for d divides c qc gc+d pb pb a; i.e., d divides a and b.

(2.) It is the greatest common divisor. For if not, let e be the greatest; then, since e divides a and 5, it divides a and pb, a – pb c qc qc d; i.e., e is less than d, and not mreater.

Cor. Every other divisor of a and b divides their greatest common measure.

Observe that no fraction is in a form to be interpreted until it is reduced to its lowest terms.

Ex. 1. Required the greatest common measure of the quantities a2x – x2 and a3 – 2a2x + ax2. The simple divisor x being taken out of the former of these quantities, and a out of the latter, they are reduced to a2 – x2 and a2– 2ax + x2 ; and as the quantity a rises to the same dimensions in both, we may take either of them as the first ,divisor : let us take that which consists of fewest terms, and the operation will stand thus : Hence it appears that a – x is the greatest common measure required.

Ex. 2. Required the greatest common measure of Sa2b2– I Oat' + 2b4, and 0a45 – 9a3b2+ 3a252– 3a54.

It is evident, from inspection, that I) is a simple divisor of both quantities ; it will therefore be a factor of the common measure required. Let the simple divisors be now left out of both quantities, and they are reduced to 4a2 – 5ab+ b2, and 3a2– 3a2b+ab2– b3; but as the second of these is to be divided by the first, it must be multiplied by 4 to make the division succeed, and the operation -will stand thus : 431 This remainder is to be divided by 1,, and the new dividend multiplied by 3, to make the division again succeed, and the work will stand thus : This remainder is to be divided by - 19b, which being done, and the last divisor taken as a dividend as before, the rest of the operation will be as follows : - In like manner we find 9a4b-9a3b2+3a2b3-3ab4 9a3+3ab" 8a2b2-10ab3+2b4 8ab-2b2 the common measure being b(a - b), as was shown in Example 2, Problem I.

a2 b2 c2+2b, Ex. 2. Reduce „ . „ . , to its lowest terms.

431 431 from which it appears that the common divisor sought is a - b, and remarking that the quantities proposed have also a simple divisor 7), the greatest common measure which is required will be b(a It will be seen that the examples we have given are not on numbers, but on algebraic quantities. In fact, the axiom and the demonstration founded on it apply, with some restrictions and modifications, to such quantities. The most important of the modifications is this : that the divisor, instead of being a whole number, is an expression of the form x + m, where m is of the nature of a numerical quantity, and does not depend on x.

The application of this modified form of the axiom has a wide range in the higher analysis. We offer two additional examples for advanced students.

Ex. 1. If ax2+bx+c, a'x' + b'x + c' have a common divisor of the form x+m, prove that (a' b - ab) (bre - bc') = (a'c - ad)2 Multiply the first expression by a', and the second by a, and subtract the products, the difference (a'b - a/0x+ dc - ac', is by the axiom divisible by x + 431 Again, multiply the first' expression by d , and the second by c, and subtract them; the difference (a'c - ac')x2 + (b' c - bc')x is divisible_by x+m, x +b',c, is x + m 431 431 the condition required.

Ex. 2. If ax3+ 3bx2+ d, bx3+3dx + e, have a common divisor;•then (4bd - ae)3 + 27 (ad2 + = 0.

Treating this question exactly as the last, viz, multiplying first by b and a, and then by e and d, and subtracting, it appears (if u be written instead of bd - ae for brevity) that the two following expressions have a common divisor, 3b2x2 - 3adx+ u and ux2 - 3bex+ 3d2, whence, by the last example, the condition is (3beu - Dad') (3adu - 9b3e) = (a2 - 62d2)2 from which u divides out as a common factor, and the result reduces to that enunciated.

Ex. 1. Reduce - a2-2a2x+ax2 to its lowest terms.

We have already found in the first example of Prob. I. that the greatest common measure of the numerator and denominator is a - x; and dividing both by this quantity, Ex. 3. To find the value of (x+t) - ;:lx - 3when x = 2.

Here the substitution of 2 in place of x renders the numerator and denominator separately equal to 0. This shows (Art. 20) that x - 2 is a divisor of each of them. We get, therefore, x+1 / which when x= 2 becomes - 2.

- x3-4x2+2x+1 result is , which, when 1 is written in place of x, becomes0 , or infinity.

Rule. Multiply the integer by the denominator of the fraction, and to the product add the numerator; and the denominator being placed under this stun, the result will be the improper fraction required.

Ex. 1. Reduce a - x+ to an improper fraction.

PRon. IV. - To Reduce an Improper Fraction to a Whole or Mixed IST'undier.

Rule. Divide the numerator by the denominator for the integral part, and place the remainder, if any, over the denominator ; it will be the mixed quantity required.

ax+2x2 Ex. 1. Reduce a+x x - y and to whole or mixed quantities.

First ax+2x2 + - x2, the answer.

a+x a+x And = x + y a whole quantity, which is the

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