Elements of Algebra

44 ELEMENTS SECT. I. impossible number, gives always an imaginary product: thus, $\sqrt{-3}$ by $\sqrt{+5}$, gives $\sqrt{-15}$. 149. It is the same with regard to division; for $\sqrt{a}$ divided by $\sqrt{b}$ making $\sqrt{\frac{a}{b}}$, it is evident that $\sqrt{-4}$ divided by $\sqrt{-1}$ will make $\sqrt{+4}$, or 2; that $\sqrt{+3}$ divided by $\sqrt{-3}$ will give $\sqrt{-1}$; and that 1 divided by $\sqrt{-1}$ gives $\sqrt{\frac{+1}{-1}}$, or $\sqrt{-1}$; because 1 is equal to $\sqrt{+1}$. 150. We have before observed, that the square root of any number has always two values, one positive and the other negative; that $\sqrt{4}$, for example, is both $+2$ and $-2$, and that, in general, we may take $-\sqrt{a}$ as well as $+\sqrt{a}$ for the square root of $a$. This remark applies also to imaginary numbers; the square root of $-a$ is both $+\sqrt{-a}$ and $-\sqrt{-a}$; but we must not confound the signs $+$ and $-$, which are before the radical sign $\sqrt{}$, with the sign which comes after it. 151. It remains for us to remove any doubt, which may be entertained concerning the utility of the numbers of which we have been speaking; for those numbers being impossible, it would not be surprising if they were thought entirely useless, and the object only of an unfounded speculation. This, however, would be a mistake; for the calculation of imaginary quantities is of the greatest importance, as questions frequently arise, of which we cannot immediately say whether they include any thing real and possible, or not; but when the solution of such a question leads to imaginary numbers, we are certain that what is required is impossible. In order to illustrate what we have said by an example, suppose it were proposed to divide the number 12 into two such parts, that the product of those parts may be 40. If we resolve this question by the ordinary rules, we find for the parts sought $6+\sqrt{-4}$ and $6-\sqrt{-4}$*; but these numbers being imaginary, we conclude, that it is impossible to resolve the question. The difference will be easily perceived, if we suppose the question had been to divide 12 into two parts which multiplied together would produce 35; for it is evident that those parts must be 7 and 5. * $(12-x) \times x = 40$ $12x - x^2 = 40$ $x^2 - 12x = -40$ and $x^2 = 12x - 40$ $\therefore x = +6 \mp \sqrt{36-40} = +6 \mp \sqrt{-4}$