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common measure is 48. But when these substitutions are made for x and a in 3.c + 4a, the highest common divisor, the result is 16 for the greatest common measure which is different from 48. The reason of this discrepancy arises from the fact that in the algebraical process, factors have been rejected which are common measures of the corresponding arithmetical expressions.

6. As the common divisor is a quadratic factor, the remainder will be of the form px +q, and must be equal to zero. And as x is not zero, there must be p=0 and q=0, from which the required condition will be determined. VOTE.— The second remainder in the process is bcro + (c2 +62 - a2)x + bc, which is

c2 +62 - a? equal to bc'x? +

+ 62 - a? -.x+1). If m be put for

the rest of the pro be

bc cess may be considerably simplified.

7. The common divisor is of the form x+b, and the remainder does not contain

This remainder made equal to zero, when reduced, will give the required condition. 8. Let

and
9

denote the two decimals, and suppose r to be the greatest 10* 10~+a common measure of p and q.

D Then Do.

PO and M=

r. 10m+*

.. 10".
10m
r10m+
M10m

P9 PI 9. The highest conimon divisor is a quadratic factor of the form wa + px+q if a and a' are prime to each other.

10. If 2? + ax +b be divided by 3+c, the quotient is x + (a - c) and the remainder b- (a – c)c=0, and b=c(a – c).

If x2 + x +a be also divided by x+c, the quotient is x + (m – c) and the remainder a - - c(m-c)=0.

The product of w? + ax + b and 4+ (m - c) will be the least common multiple. If ca - c) be substituted for b, the required form will be determined.

11. If each expression be divided by x+2, the remainders will not contain 2. If each be made equal to zero, a value of q may be found for each expression, and by equating these values of 9, the required condition will be determined.

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RESULTS, HINTS, ETC., FOR THE EXERCISES ON THE

ALGEBRAIC FRACTIONS.

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I. 1. See Section V., Art. 2, p. 8. 2. Let be a fraction in which a is prime to b. If possible let the fraction 6

6 be equivalent to the fraction having its terms respectively less than a and b.

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=c, and ad is divisible by b. Since b is prime to a, d must od b be divisible by b. This is impossible, for d is less than b.

bc 3. For since

; hence bc is divisible by d. But as d is prime to c,

d must be divisible by d. Let

cmd -m, then b=md, and a= = mc; or a and bare d

d respectively equimultiples of c and d.

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It may be added here, that the fraction formed of the sums of the numerators and the sums of the denominators of three, four, or more unequal fractions is always intermediate in value between the greatest and least of these fractions. 6. For let ; and

be two fractions in their lowest terms, of which ca factor in

dac the denominator of one is not contained in 6 the denominator of the other.

adx + cb

oda a And if be an integer, then adx+co

must also be an integer, which cannot be the case, unless cb be bdx divisible by bdx, that is, unless cb be divisible by each of the factors of bdx.

But b and c are each prime to X. Hence cb cannot be divisible by ac; 22 de

cannot be an integer. 8. See Art. 3, p. 9.

sa bi 9. Let be the fraction, then

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10. (a? +62)(c? +d2)=(ac+bd)2 + (ad - bc)?.

11. The expression is equivalent to x2 +2x+3, which will be integral for all integral values of x. If x3 + ys +be divided by x+y+z, there is a remainder - 3yz(y+z).

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4 H. C. D. is

2 °(x2 + 4xy + 4ya),
3- y
Lowest terms

**-y*

(5x - 1).2 5 22 - 1

2.3 -1

5ax(2x+a) 6 (3x2 – 2ax+a?)

3x + 2a

1-22 7 1-2x + 3x2

1+&+*?

x+2 8 33 - 3.0? + 3x-1

2? + x -1°

1 (22 +230 +2)

22 - 2x +2

10x2 – 24x+15 10 2-3

5.03 - 21.2? + 24. - 18*

X? +1 11 ... (x+2)(x+3)(2x + 1)

ac2 - 1

5x3+x2 +x+1. 2-1

24-1

z? xy +y? 13 3x – 2y

2? + xy + y2'

8(23 + 2? +*+1)x3 - 21(x2 +2+1) 14 2-1

21662 + x+1)x+ - 80x3 + x2 +3+1)

cbx + 2 15 2 + a

2? + 6.6 +62

ab - ac- 62 16 (6+ c)(a-)

a(ab+ac - 62)

3x – (a +23) 17

2?- (a + b)2 + ab
18
(a-1).C +1

(a? +a+1).x2 + (a +1)x - 1

(a− 1)2x2 – (a +1) - 1 Note to 14. If the numerator be subtracted from the denominator, then the renainder is 21(x2 - 1) - 21(x - 1).x: – 8(x2 - 1) of which each term is divisible by 2-1.

Note to 16. If the factor b+c be removed from the numerator and the denomi. nator, the highest common divisor of the resulting numerator and denominator will be found to be a-6. This will be found to be the shorter process.

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1.

5.

9.

29x

&r 38% -2 5412 ab + ac tlc or x+ 2.

3.

4.
21
21
7
504

abe
ayz + bxz + cxy a:b? + a2c2 + b2c2

1

? + a2
6.

7. 8.
XYZ
abc

22-a?
2x + 4

10.
2(9a? + 4x2)

11.
a? +62

12. 0.
x2 + 4x – 21
9.x2 - 4a?

ab
a? – 2ax + b2 22 - xy + 2y2
14.

15.
(x – a)(- )

22 – 4y?

C + ax
cx+by+az
17. 0. 18.

4xy(x? +??)

19. abc

2* + x^yty x' +x"y-ty

13.

a + bx

16.

2r3

VI.

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6+
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40h (m n)abx (m + n)ac a? – 2nc+63 8 9.

10. (x - 3)(x-4) a’-62

box? ---C2

a?-63
(ab){(a+b)x ab}

11x
(a? + b2)x - (a+b)a)
12.

13.
(a – a)(2 - 0)
(x-a)(x - b)(a --O)

156
23

- xy2 + 2y
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11.

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14.

VII.

The examples under this number offer no difficulties. For the reduction of the continued fractions in Ex. 10 reference may be made to the note, pp. 5, 6 of Sect. IX. of the Elementary Arithmetic, where the reduction of a continued numerical fraction is fully explained.

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10. a* - 3(a - b)a2x – (31a— 3062)x? (167a – 1031),3 234

18

108 3026 8 2626 5 11. 25 2 + x3

a? - ab +62

+2.
1225 5
735

a? + ab +62

12.

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IX. 1. ab. 2. 1-a2 +62-c? 2ab-a? ba+c? c? – (a - b)? _ (c+ab)(c - a+b). 2ab 2ab 2ab

2ab (a-b+c)'a-b-c) 3.

2ab 4. 1a? - (b + c)2

1

(a+b+c)(a 6–c)-1-2-6-c_2(6+c).
(a + b + c)?
(a+b+c)*

atbtc atbtc 5.

2(b+c) a-b-c

(a+b){(a+b)3 –c2} (a +"){(a+b)2 -6.

462c? – (a? 62 – c2)2 = (2bc+a? 62 – c?)(26c a2 +62 +ca) (a+b) {(a+b):

(a + b)(a + b + c)(a+b-c) {a? -(6-c)2}. {(6+c)2 – ar} (+6–c)(a – 6+C)(6+c+a)(6+c-a)

a + b

(a --b+c)(b +c-a) 7. d?. 8. a+b-c-d. 9. (a+b+c-d)(a +b+dc)(a +c+d b)(h+c+d-a)

4(ab+cd)? 10. 1- {a? +.}-{1+

a? + b2 - 72

a? + 5? - ? 2ab

2ab Substitute for a?, and the reduced expression is — (a+b+c+d)(a + b-c-d)(n+c-h-d)(b+c-a-d)

4(ab - cd)

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9.
4(a + x)

(x + y)?
10.
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11. a? - 62

3(C+2) (a + x)x

*?-ya (2? – Y?):

13. 1. 14. (a - b-c)?. 22 + y2

12.

c

2

XI. The following are the respective quotients :1 2113

22

1 1. 2x2 +

373 2.

3.

+ 4. 5x2 +*+9. 3

3 3 4

2 395 5. 2a3

2?
X!

3x 30
6.
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1
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+ 46 562' 863

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