Page:EB1911 - Volume 09.djvu/749

Rh and c = 3, d = 4 or c = 3, d = 4) or else a, b to be 3, 4 and c, d to be 1, 2; and it therefore in effect determines (a + b)3 + (c + d)3 to be = 370, and not any other value; that is, (a + b)3 + (c + d)3, as having a single value, must be determinable rationally. And we can in the same way account for cases of failure as regards particular equations; thus, the roots being 1, 2, 3, 4 as before, a2b = 2 determines a to be = 1 and b to be = 2, but if the roots had been 1, 2, 4, 16 then a2b = 16 does not uniquely determine a, b but only makes them to be 1, 16 or 2, 4 respectively.

As to the a posteriori proof, assume, for instance,

t1 = ab + cd, &emsp; y1 = (a + b)3 + (c + d)3, t2 = ac + bd, &emsp; y2 = (a + c)3 + (b + d)3, t3 = ad + bc, &emsp; y3 = (a + d)3 + (b + c)3;

then y1 + y2 + y3, t1y1 + t2y2 + t3y3, t12y1 + t22y2 + t32y3 will be respectively symmetrical functions of the roots of the quartic, and therefore rational and integral functions of the coefficients; that is, they will be known.

Suppose for a moment that t1, t2, t3 are all known; then the equations being linear in y1, y2, y3 these can be expressed rationally in terms of the coefficients and of t1, t2, t3; that is, y1, y2, y3 will be known. But observe further that y1 is obtained as a function of t1, t2, t3 symmetrical as regards t2, t3; it can therefore be expressed as a rational function of t1 and of t2 + t3, t2t3, and thence as a rational function of t1 and of t1 + t2 + t3, t1t2 + t1t3 + t2t3, t1t2t3; but these last are symmetrical functions of the roots, and as such they are expressible rationally in terms of the coefficients; that is, y1 will be expressed as a rational function of t1 and of the coefficients; or t1 (alone, not t2 or t3) being known, y1 will be rationally determined.

16. We now consider the question of the algebraical solution of equations, or, more accurately, that of the solution of equations by radicals.

In the case of a quadric equation x2 &minus; px + q = 0, we can by the assistance of the sign &radic; or undefined find an expression for x as a 2–valued function of the coefficients p, q such that substituting this value in the equation, the equation is thereby identically satisfied; it has been found that this expression is

x = {p ± &radic;(p2 &minus; 4q) },

and the equation is on this account said to be algebraically solvable, or more accurately solvable by radicals. Or we may by writing x = &minus; p + z reduce the equation to z2 = (p2 &minus; 4q), viz. to an equation of the form x2 = a; and in virtue of its being thus reducible we say that the original equation is solvable by radicals. And the question for an equation of any higher order, say of the order n, is, can we by means of radicals (that is, by aid of the sign m &radic; or 1/m, using as many as we please of such signs and with any values of m) find an n-valued function (or any function) of the coefficients which substituted for x in the equation shall satisfy it identically?

It will be observed that the coefficients p, q ... are not explicitly considered as numbers, but even if they do denote numbers, the question whether a numerical equation admits of solution by radicals is wholly unconnected with the before-mentioned theorem of the existence of the n roots of such an equation. It does not even follow that in the case of a numerical equation solvable by radicals the algebraical solution gives the numerical solution, but this requires explanation. Consider first a numerical quadric equation with imaginary coefficients. In the formula x = {p ± &radic;(p2 &minus; 4q) }, substituting for p, q their given numerical values, we obtain for x an expression of the form x = + i ± &radic;( + i), where, , ,  are real numbers. This expression substituted for x in the quadric equation would satisfy it identically, and it is thus an algebraical solution; but there is no obvious a priori reason why &radic;( + i) should have a value = c + di, where c and d are real numbers calculable by the extraction of a root or roots of real numbers; however the case is (what there was no a priori right to expect) that &radic;( + i) has such a value calculable by means of the radical expressions &radic;$($&radic;(2 + 2) ± $)$; and hence the algebraical solution of a numerical quadric equation does in every case give the numerical solution. The case of a numerical cubic equation will be considered presently.

17. A cubic equation can be solved by radicals.

Taking for greater simplicity the cubic in the reduced form x3 + qx &minus; r = 0, and assuming x = a + b, this will be a solution if only 3ab = q and a3 + b3 = r, equations which give (a3 &minus; b3)2 = r2 &minus; q3, a quadric equation solvable by radicals, and giving a3 &minus; b3 = &radic;(r2 &minus;  q3), a 2-valued function of the coefficients: combining this with a3 + b3 = r, we have a3 = {r + &radic;(r2 &minus;  q3) }, a 2-valued function: we then have a by means of a cube root, viz.

a = &#8731;[ {r + &radic;(r2 &minus; q3) }],

a 6-valued function of the coefficients; but then, writing q = b/3a, we have, as may be shown, a + b a 3-valued function of the coefficients; and x = a + b is the required solution by radicals. It would have been wrong to complete the solution by writing

b = &#8731;[ {r &minus; &radic;(r2 &minus; q3) } ],

for then a + b would have been given as a 9-valued function having only 3 of its values roots, and the other 6 values being irrelevant. Observe that in this last process we make no use of the equation 3ab = q, in its original form, but use only the derived equation 27a3b3 = q3, implied in, but not implying, the original form.

An interesting variation of the solution is to write x = ab(a + b), giving a3b3 (a3 + b3) = r and 3a3b3 = q, or say a3 + b3 = 3r/q, a3b3 = q; and consequently

i.e. here a3, b3 are each of them a 2-valued function, but as the only effect of altering the sign of the quadric radical is to interchange a3, b3, they may be regarded as each of them 1-valued; a and b are each of them 3-valued (for observe that here only a3b3, not ab, is given); and ab(a + b) thus is in appearance a 9-valued function; but it can easily be shown that it is (as it ought to be) only 3-valued.

In the case of a numerical cubic, even when the coefficients are real, substituting their values in the expression

x = &#8731;[ {r + &radic;(r2 &minus; q3) }] +  q ÷ &#8731;[ {r + &radic;(r2 &minus;  q3) }],

this may depend on an expression of the form &#8731;( + i) where and are real numbers (it will do so if r2 &minus;  q3 is a negative number), and then we cannot by the extraction of any root or roots of real positive numbers reduce &#8731;( + i) to the form c + di, c and d real numbers; hence here the algebraical solution does not give the numerical solution, and we have here the so-called “irreducible case” of a cubic equation. By what precedes there is nothing in this that might not have been expected; the algebraical solution makes the solution depend on the extraction of the cube root of a number, and there was no reason for expecting this to be a real number. It is well known that the case in question is that wherein the three roots of the numerical cubic equation are all real; if the roots are two imaginary, one real, then contrariwise the quantity under the cube root is real; and the algebraical solution gives the numerical one.

The irreducible case is solvable by a trigonometrical formula, but this is not a solution by radicals: it consists in effect in reducing the given numerical cubic (not to a cubic of the form z3 = a, solvable by the extraction of a cube root, but) to a cubic of the form 4x3 &minus; 3x = a, corresponding to the equation 4 cos3 &minus; 3 cos  = cos 3 which serves to determine cos when cos 3 is known. The theory is applicable to an algebraical cubic equation; say that such an equation, if it can be reduced to the form 4x3 &minus; 3x = a, is solvable by “trisection”—then the general cubic equation is solvable by trisection.

18. A quartic equation is solvable by radicals, and it is to be remarked that the existence of such a solution depends on the existence of 3-valued functions such as ab + cd of the four roots (a, b, c, d): by what precedes ab + cd is the root of a cubic equation, which equation is solvable by radicals: hence ab + cd can be found by radicals; and since abcd is a given function, ab and cd can then be found by radicals. But by what precedes, if ab be known then any similar function, say a + b, is obtainable rationally; and then from the values of a + b and ab we may by radicals obtain the value of a or b, that is, an expression for the root of the given quartic equation: the expression ultimately obtained is 4-valued, corresponding to the different values of the several radicals which enter therein, and we have thus the expression by radicals of each of the four roots of the quartic equation. But when the quartic is numerical the same thing happens as in the cubic, and the algebraical solution does not in every case give the numerical one.

It will be understood from the foregoing explanation as to the quartic how in the next following case, that of the quintic, the question of the solvability by radicals depends on the existence or non-existence of k-valued functions of the five roots (a, b, c, d, e); the fundamental theorem is the one already stated, a rational function of five letters, if it has less than 5, cannot have more than 2 values, that is, there are no 3-valued or 4-valued functions of 5 letters: and by reasoning depending in part upon this theorem, N. H. Abel (1824) showed that a general quintic equation is not solvable by radicals; and a fortiori the general equation of any order higher than 5 is not solvable by radicals.

19. The general theory of the solvability of an equation by radicals depends fundamentally on A. T. Vandermonde’s remark (1770) that, supposing an equation is solvable by radicals, and that we have therefore an algebraical expression of x in terms of the coefficients, then substituting for the coefficients their values in terms of the roots, the resulting expression must reduce itself to any one at pleasure of the roots a, b, c ...; thus in the case of the quadric equation, in the expression x = {p + &radic;(p2 &minus; 4q) }, substituting for p and q their values, and observing that (a + b)2 &minus; 4ab = (a &minus; b)2, this becomes x = {a + b + &radic;(a &minus; b)2}, the value being a or b according as the radical is taken to be +(a &minus; b) or &minus;(a &minus; b).

So in the cubic equation x3 &minus; px2 + qx &minus; r = 0, if the roots are a, b, c, and if is used to denote an imaginary cube root of unity, 2 +  + 1 = 0, then writing for shortness p = a + b + c, L = a + b + 2c, M = a + 2b + c, it is at once seen that LM, L3 + M3, and therefore also