SCHOLIUM.

520. The discovery of the number of impossible roots in an equation has given great trouble to Algebraists; and their researches, hitherto, have not been attended with any great success.

*

two roots are impossible or not, according as

tive or negative (Art. 490).

A biquadratic,

x2 + q x2 + rx + 8 = 0,

has two impossible roots, when two roots of the equation

y3 +2qy2 + (q2 − 4s) y — r2 = 0,

are impossible; and all its roots are impossible, when the roots of this cubic are all possible and two of them negative. (Art. 492).

521. Waring has given a rule for determining the number of impossible roots in an equation of five dimensions, but the investigation cannot properly be introduced into an elementary treatise. See Med. Algebraicæ, p. 82.

522. Newton's rule is general and easily applied, but as it is deduced from the nature of the inferior limits, it will not always detect impossible roots (Art. 516). The proof also is defective, as it does not extend to that part of the rule which respects the number of impossible roots. Thus far however it may be depended upon, that it never shews impossible roots, but when there are some such in the proposed equation.

Many other rules, which will frequently discover the impossible roots in any equation, may be seen in the Med. Alg. C. 2.

[* The truth of this latter observation has been materially affected by the recent labours of Sturm and Fourier. The student is referred to Fourier's Analyse des Equations Déterminées; or Stevenson's Algebraic Equations, Chap. IV; or Hymers' Theory of Equations, Section VI.]

END OF PART II.

[APPENDIX I.

NOTE 1.

THEORY OF INDICES.

ARTS. 62, 64, and 69 relate to the subject of Indices or Exponents. The first Article is strictly so called a Definition; but the others are not; at least they are not arbitrary definitions, but are dependent upon something which does not appear. Thus it is strictly a Definition that

a.a.a shall be represented by a3

and a.a. a...to n factors........by a", if n be any positive integer.

Hence it is easily shewn that, when m and n are positive integers,

am. an = am+n (1),

......

Now, the assumptions in Arts. 64 and 69 are made subject to these two Rules, that is, made so that these two rules may hold true for fractional and negative indices, as well as for positive integers. Thus, if Rule (1) be true for negative integers as indices,

and a. a= a+ = a, if Rule (1) be true for fractional indices.

and a. a. a= a++= a, if Rule (1) be true for fractional indices.

And generally,

since a.a.Va. &c. to n factors = ɑ.

a, if Rule (1) be true for fractional indices,

Hence it appears that the Definitions respecting negative and fractional Indices are not made arbitrarily, but have reference to the RULES already determined for the multiplication and division of powers, when the Indices are positive integers.

Therefore am, an = am+n, whether m and n be whole or fractional, positive or negative.

And similarly am ÷ a" = a"-", whether m and n be whole or fractional, positive, or negative.

NOTE 2.

ADDITION AND SUBTRACTION.

IN actual practice it is seldom that either addition or subtraction of algebraical quantities is presented to us as in the Examples, Arts. 85, and 86. The expressions to be handled are more commonly in one line, with or without brackets.

Thus Ex. 3. in Art. 85. would stand 5a-3b+4a7b,

and Ex. 4. in Art. 86..............6a-12b-(-5a-106). The following rules, therefore, are extensively useful.

RULE I. If an expression in brackets be preceded by the sign +, it will not be altered in value if the brackets be struck out.

RULE II. If an expression in brackets be preceded by the sign, the brackets may be struck out, if the signs within the brackets be changed, namely, + into, and into +.

Rule I. is manifestly true.

Rule II. may be proved thus:

To subtract b+c from a, that is, to shew what results from taking away the brackets in the expression

a - (b+c).

If from a the portion b be first subtracted, the result is clearly ab; but there is not enough subtracted by the quantity c, since b + c was proposed to be subtracted; therefore c also must be subtracted, and the result is

or a

Next to subtract b

a-b-c;

(b+c)=a-b-c.

a,

c from that is, to shew what results from taking away the brackets in the expression

a − (b −c).

If from a the quantity b be subtracted, the result is a b;

but there has been too much subtracted by the quantity c, since