# Chapter 17. "Of the Calculation of Powers."

### Part I. Section I. Chapter 17. “Of the Calculation of Powers.”

180 We have nothing particular to observe with regard to the Addition and Subtraction of powers; for we only represent those operations by means of the signs + and -, when the powers are different. For example, \(a^3 + a^2\) is the sum of the second and third powers of \(a\); and \(a^5 - a^4\) is what remains when we subtract the fourth power of \(a\) from the fifth; and neither of these results can be abridged: but when we have powers of the same kind or degree, it is evidently unnecessary to connect them by signs; as \(a^3+a^3\) becomes \(2a^3\), etc.

181 But in the Multiplication of powers, several circumstances require attention.

First, when it is required to multiply any power of \(a\) by \(a\), we obtain the succeeding power; that is to say, the power whose exponent is greater by a unit. Thus, \(a^2\) multiplied by \(a\), produces \(a^3\); and \(a^3\), multiplied by \(a\), produces \(a^4\). In the same manner, when it is required to multiply by a the powers of any number represented by \(a\), having negative exponents, we have only to add 1 to the exponent. Thus, \(a^{-1}\) multiplied by \(a\) produces \(a^0\), or 1; which is made more evident by considering that \(a^{-1}\) is equal to \(\frac{1}{a}\), and that the product of \(\frac{1}{a}\) by \(a\) being \(\frac{a}{a}\), it is consequently equal to 1; likewise \(a^{-2}\) multiplied by \(a\), produces \(a^{-1}\), or \(\frac{1}{a}\); and \(a^{-10}\) multiplied by \(a\), gives \(a^{-9}\), and so on.

182 Next, if it be required to multiply any power of \(a\) by \(a^2\), or the second power, I say that the exponent becomes greater by 2. Thus, the product of \(a^2\) by \(a^2\) is \(a^4\); that of \(a^2\) by \(a^3\) is \(a^5\); that of \(a^4\) by \(a^2\) is \(a^6\); and, more generally, \(a^n\) multiplied by \(a^2\) makes \(a^{n+2}\). With regard to negative exponents, we shall have \(a^1\), or \(a\), for the product of \(a^{-1}\) by \(a^2\); for \(a^{-1}\) being equal to \(\frac{1}{a}\), it is the same as if we had divided \(aa\) by \(a\); consequently, the product required is \(\frac{aa}{a}\), or \(a\); also \(a^{-2}\), multiplied by \(a^2\), produces \(a^0\), or 1; and \(a^{-3}\), multiplied by \(a^2\), produces \(a^{-1}\).

183 It is no less evident, that to multiply any power of \(a\) by \(a^3\), we must increase its exponent by three units; and that, consequently, the product of \(a^n\) by \(a^3\) is \(a^{n+3}\). And whenever it is required to multiply together two powers of \(a\), the product will be also a power of \(a\), and such that its exponent will be the sum of those of the two given powers. For example, \(a^4\) multiplied by \(a^5\) will make \(a^9\), and \(a^{12}\) multiplied by \(a^7\) will produce \(a^{19}\), etc.

184 From these considerations we may easily determine the highest powers. To find, for instance, the twenty-fourth power of 2, I multiply the twelfth power by the twelfth power, because \(2^{24}\) is the same amount as \(2^{12}\) multiplied with \(2^{12}\). Now, we have already seen that \(2^{12}\) is 4096; I say therefore that the number 16777216, or the product of 4096 by 4096, expresses the power required, namely, \(2^{24}\).

185 Let us now proceed to division. We shall remark, in the first place, that to divide a power of \(a\) by \(a\), we must subtract 1 from the exponent, or diminish it by unity; thus, \(a^5\) divided by \(a\) gives \(a^4\); and \(a^0\), or 1, divided by \(a\), is equal to \(a^{-1}\) or \(\frac{1}{a}\); also \(a^{-3}\) divided by \(a\), gives \(a^{-4}\).

186 If we have to divide a given power of \(a\) by \(a^2\) we must diminish the exponent by 2; and if by \(a^3\) we must subtract 3 units from the exponent of the power proposed; and, in general, whatever power of a it is required to divide by any other power of \(a\), the rule is always to subtract the exponent of the second from the exponent of the first of those powers: thus \(a^{15}\) divided by \(a^7\) will give \(a^8\); \(a^6\) divided by \(a^7\) will give \(a^{-1}\); and \(a^{-3}\) divided by \(a^4\) will give \(a^{-7}\).

187 From what has been said, it is easy to understand the method of finding the powers of powers, this being done by multiplication. When we seek, for example, the square, or the second power of \(a^3\), we find \(a^6\); and in the same manner we find \(a^{12}\) for the third power, or the cube, of \(a^4\). To obtain the square of a power, we have only to double its exponent; for its cube, we must triple the exponent; and so on. Thus, the square of \(a^n\) is \(a^{2n}\); the cube of \(a^n\) is \(a^{3n}\); the seventh power of \(a^n\) is \(a^{7n}\), etc.

188 The square of \(a^2\), or the square of the square of \(a\),
being \(a^4\), we see why the fourth power is called the *biquadrate*:
also, the square of \(a^3\) being \(a^6\), the sixth power has
received the name of the square-cubed.

Lastly, the cube of \(a^3\) being \(a^9\), we call the ninth power the cubo-cube: after this, no other denominations of this kind have been introduced for powers; and, indeed, the two last are very little used.

#### Editions

- Leonhard Euler.
*Elements of Algebra*. Translated by Rev. John Hewlett. Third Edition. Longmans, Hurst, Rees, Orme, and Co. London. 1822. - Leonhard Euler.
*Vollständige Anleitung zur Algebra. Mit den Zusätzen von Joseph Louis Lagrange.*Herausgegeben von Heinrich Weber. B. G. Teubner. Leipzig and Berlin. 1911. Leonhardi Euleri Opera omnia. Series prima. Opera mathematica. Volumen primum.