Elements of Algebra

40 ELEMENTS SECT. I. the square root of 12, for example, may appear, we are not ignorant that it must be a number, which, when multiplied by itself, would exactly produce 12; and this property is sufficient to give us an idea of the number, because it is in our power to approximate towards its value continually. 130. As we are therefore sufficiently acquainted with the nature of irrational numbers, under our present consideration, a particular sign has been agreed on to express the square roots of all numbers that are not perfect squares; which sign is written thus $\sqrt{}$, and is read *square root*. Thus, $\sqrt{12}$ represents the square root of 12, or the number which, multiplied by itself, produces 12; and $\sqrt{2}$ represents the square root of 2; $\sqrt{3}$ the square root of 3; $\sqrt{\frac{2}{3}}$ that of $\frac{2}{3}$; and, in general, $\sqrt{a}$ represents the square root of the number $a$. Whenever, therefore, we would express the square root of a number, which is not a square, we need only make use of the mark $\sqrt{}$ by placing it before the number. 131. The explanation which we have given of irrational numbers will readily enable us to apply to them the known methods of calculation. For knowing that the square root of 2, multiplied by itself, must produce 2; we know also, that the multiplication of $\sqrt{2}$ by $\sqrt{2}$ must necessarily produce 2; that, in the same manner, the multiplication of $\sqrt{3}$ by $\sqrt{3}$ must give 3; that $\sqrt{5}$ by $\sqrt{5}$ makes 5; that $\sqrt{\frac{2}{3}}$ by $\sqrt{\frac{2}{3}}$ makes $\frac{2}{3}$; and, in general, that $\sqrt{a}$ multiplied by $\sqrt{a}$ produces $a$. 132. But when it is required to multiply $\sqrt{a}$ by $\sqrt{b}$, the product is $\sqrt{ab}$; for we have already shewn, that if a square has two or more factors, its root must be composed of the roots of those factors; we therefore find the square root of the product $ab$, which is $\sqrt{ab}$, by multiplying the square root of $a$, or $\sqrt{a}$, by the square root of $b$, or $\sqrt{b}$; &c. It is evident from this, that if $b$ were equal to $a$, we should have $\sqrt{aa}$ for the product of $\sqrt{a}$ by $\sqrt{b}$. But $\sqrt{aa}$ is evidently $a$, since $aa$ is the square of $a$. 133. In division, if it were required, for example, to divide $\sqrt{a}$ by $\sqrt{b}$, we obtain $\sqrt{\frac{a}{b}}$; and, in this instance, the irrationality may vanish in the quotient. Thus, having to divide $\sqrt{18}$ by $\sqrt{8}$, the quotient is $\sqrt{\frac{18}{8}}$, which is reduced to $\sqrt{\frac{9}{4}}$, and consequently to $\frac{3}{2}$, because $\frac{9}{4}$ is the square of $\frac{3}{2}$. 134. When the number before which we have placed the radical sign $\sqrt{}$, is itself a square, its root is expressed in the