# Chapter 8. "Of the Calculation of Irrational Quantities."

### Part I. Section II. Chapter 8. “Of the Calculation of Irrational Quantities.”

326 When it is required to add together two or more irrational quantities, this is to be done, according to the method before laid down, by writing all the terms in succession, each with its proper sign: and, with regard to abbreviations, we must remark that, instead of \(\surd a+\surd a\), for example, we may write \(2\surd a\); and that \(\surd a-\surd a= 0\), because these two terms destroy one another. Thus, the quantities 3 + √2 and 1 + √2, added together, make 4 + 2√2, or 4 + √8; the sum of 5 + √3 and 4 - √3, is 9; and that of 2√3 + 3√2 and √3 - √2, is 3√3 + 2√2.

327 Subtraction also is very easy, since we have only to add the proposed numbers, after having changed their signs; as will be readily seen in the following example, by subtracting the lower line from the upper.

\[\begin{array}{rrrrr} 4&-\surd 2&+2\surd 3&-3\surd 5&+4\surd 6\\ 1&+2\surd 2&-2\surd 3&-5\surd 5&+6\surd 6\\ \hline 3&-3\surd 2&-4\surd 3&+2\surd 5&-2\surd 6 \end{array}\]328 In multiplication, we must recollect that \(\surd a\) multiplied by \(\surd a\) produces \(a\); and that if the numbers which follow the sign \(\surd\) are different, as \(a\) and \(b\), we have \(\surd ab\) for the product of \(\surd a\) multiplied by \(\surd b\). After this, it will be easy to calculate the following examples:

\[\begin{array}{rrrr} 1&+\surd 2&&\\ 1&+\surd 2&&\\ \hline 1&+\surd 2&&\\ &+\surd 2&+2&\\ \hline 1&+2\surd 2&+2&=3+2\surd 2. \end{array}\] \[\begin{array}{rrr} 4&+2\surd 2&\\ 2&-\surd 2&\\ \hline 8&+4\surd 2&\\ &-4\surd 2&-4\\ \hline 8&-4&=4. \end{array}\]329 What we have said applies also to imaginary quantities; we shall only observe farther, that \(\surd -a\) multiplied by \(\surd -a\) produces \(-a\). If it were required to find the cube of -1 + √-3, we should take the square of that number, and then multiply that square by the same number; as in the following operation:

\[\begin{array}{rrrr} -1&+\surd -3&&\\ -1&+\surd -3&&\\ \hline 1&-\surd -3&&\\ &-\surd -3&-3&\\ \hline 1&-2\surd -3&-3&=-2-2\surd -3 \end{array}\] \[\begin{array}{rrrr} -2&-2\surd -3&&\\ -1&+\surd -3&&\\ \hline 2&+2\surd -3&&\\ &-2\surd -3&+6&\\ \hline 2&+6&&=8 \end{array}\]330 In the division of surds, we have only to express the proposed quantities in the form of a fraction; which may be then changed into another expression having a rational denominator; for if the denominator be \(a+\surd b\), for example, and we multiply both this and the numerator by \(a-\surd b\), the new denominator will be \(a^2 - b\), in which there is no radical sign. Let it be proposed, for example, to divide 3 + 2√2 by 1 + √2: we shall first have \(\frac{3+2\surd 2}{1+\surd 2}\); then multiplying the two terms of the fraction by 1 - √2, we shall have for the numerator:

\[\begin{array}{rrrr} 3&+2\surd 2&&\\ 1&-\surd 2&&\\ \hline 3&+2\surd 2&&\\ &-3\surd 2&-4&\\ \hline 3&-\surd 2&-4&=-\surd 2-1; \end{array}\]and for the denominator:

\[\begin{array}{rrrr} 1&+\surd 2&&\\ 1&-\surd 2&&\\ \hline 1&+\surd 2&&\\ &-\surd 2&-2&\\ \hline 1&&-2&=-1. \end{array}\]Our new fraction therefore is \(\frac{-\surd 2-1}{-1}\); and if we again multiply the two terms by -1, we shall have for the numerator √2 + 1, and for the denominator +1. Now, it is easy to show that √2 + 1 is equal to the proposed fraction \(\frac{3+2\surd 2}{1+\surd 2}\); for √2 + 1 being multiplied by the divisor 1 + √2, thus,

\[\begin{array}{rrrr} 1&+\surd 2&&\\ 1&+\surd 2&&\\ \hline 1&+\surd 2&&\\ &+\surd 2&+2&\\ \hline 1&+2\surd 2&+2&=3+2\surd 2. \end{array}\]Another example. Let 8 - 5√2 be divided by 3 - 2√2. This, in the first instance, is \(\frac{8-5\surd 2}{3-2\surd 2}\); and multiplying the two terms of this fraction by 3 + 2√2, we have for the numerator,

\[\begin{array}{rrrr} 8&-5\surd 2&&\\ 3&+2\surd 2&&\\ \hline 24&-15\surd 2&&\\ &+16\surd 2&-20&\\ \hline 24&+\surd 2&-20&=4+\surd 2; \end{array}\]and for the denominator,

\[\begin{array}{rrrr} 3&-2\surd 2&&\\ 3&+2\surd 2&&\\ \hline 9&-6\surd 2&&\\ &+6\surd 2&-8&\\ \hline 9&&-8&=1. \end{array}\]Consequently, the quotient will be 4 + √2. The truth of this may be proved, as before, by multiplication; thus,

\[\begin{array}{rrrr} 4&+\surd 2&&\\ 3&-2\surd 2&&\\ \hline 12&+3\surd 2&&\\ &-8\surd 2&-4&\\ \hline 12&-5\surd 2&-48&=8-5\surd 2. \end{array}\]331 In the same manner, we may transform irrational fractions into others, that have rational denominators. If we have, for example, the fraction \(\frac{1}{5-2\surd 6}\), and multiply its numerator and denominator by 5 + 2√6, we transform it into this, \(\frac{5+2\surd 6}{1}=5+2\surd 6\); in like manner, the fraction \(\frac{2}{-1+\surd -3}\) assumes this form, \(\frac{2+2\surd -3}{-4}=\frac{1+\surd -3}{-2}\); also

\[\frac{\surd 6 + \surd 5}{\surd 6 - \surd 5}=\frac{11+2\surd 30}{1} =11+2\surd 30.\]332 When the denominator contains several terms, we may, in the same manner, make the radical signs in it vanish one by one. Thus, if the fraction \(\frac{1}{\surd 10 - \surd 2 - \surd 3}\) be proposed, we first multiply these two terms by √10 + √2 + √3, and obtain the fraction \(\frac{\surd 10 + \surd 2 + \surd 3}{5-2\surd 6}\); then multiplying its numerator and denominator by 5 + 2√6, we have 5√10 + 11√2 + 9√3 + 2√60.

#### 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.