# Chapter 9. "Of the Addition and Subtraction of Fractions."

### Part I. Section I. Chapter 9. “Of the Addition and Subtraction of Fractions.”

94 When fractions have equal denominators, there is no difficulty in adding and subtracting them; for ²⁄₇ + ³⁄₇ is equal to ⁵⁄₇, and ⁴⁄₇ - ²⁄₇ is equal to ²⁄₇. In this case, therefore, either for addition or subtraction, we alter only the numerators, and place the common denominator under the line, thus;

⁷⁄₁₀₀ + ⁹⁄₁₀₀ - ¹²⁄₁₀₀ - ¹⁵⁄₁₀₀ + ²⁰⁄₁₀₀ is equal to ⁹⁄₁₀₀;

²⁴⁄₅₀ - ⁷⁄₅₀ - ¹²⁄₅₀ + ³¹⁄₅₀ is equal to ³⁶⁄₅₀, or ¹⁸⁄₂₅;

¹⁶⁄₂₀ - ³⁄₂₀ - ¹¹⁄₂₀ + ¹⁴⁄₂₀ is equal to ¹⁶⁄₂₀, or ⅘;

also ⅓ + ⅔ is equal to ³⁄₃, or 1, that is to say, a whole; and ²⁄₄ - ¾ + ¼ is equal to ⁰⁄₄, that is to say, nothing, or 0.

95 But when fractions have not equal denominators, we can always change them into other fractions that have the same denominator. For example, when it is proposed to add together the fractions ½ and ⅓, we must consider that ½ is the same as ³⁄₆, and that ⅓ is equivalent to ²⁄₆; we have therefore, instead of the two fractions proposed, ³⁄₆ + ²⁄₆, the sum of which is ⅚. And if the two fractions were united by the sign minus as ½ - ⅓, we should have ³⁄₆ - ²⁄₆, or ⅙.

As another example, let the fractions proposed be ¾ + ⅝. Here, since ¾ is the same as ⁶⁄₈, this value may be substituted for ¾, and we may then say ⁶⁄₈ + ⅝ makes ¹¹⁄₈, or 1⅜.

Suppose farther, that the sum of ⅓ and ⅓ were required, I say that it is ⁷⁄₁₂; for ⅓ = ⁴⁄₁₂, and ¼ = ³⁄₁₂: therefore ⁴⁄₁₂ + ³⁄₁₂ = ⁷⁄₁₂.

96 We may have a greater number of fractions to reduce to a common denominator; for example, ½, ⅔, ¾, ⅘, ⅚. In this case, the whole depends on finding a number that shall be divisible by all the denominators of those fractions. In this instance, 60 is the number which has that property, and which consequently becomes the common denominator. We shall therefore have ³⁰⁄₆₀, instead of ½; ⁴⁰⁄₆₀, instead of ⅔; ⁴⁵⁄₆₀, instead of ¾; ⁴⁸⁄₆₀, instead of ⅘; and ⁵⁰⁄₆₀, instead of ⅚. If now it be required to add together all these fractions ³⁰⁄₆₀, ⁴⁰⁄₆₀, ⁴⁵⁄₆₀, ⁴⁸⁄₆₀, ⁵⁰⁄₆₀; we have only to add all the numerators, and under the sum place the common denominator 60; that is to say, we shall have ²¹³⁄₆₀, or 3 wholes, and the fractional remainder ¹¹⁄₂₀.

97 The whole of this operation consists, as we before stated, in changing fractions, whose denominators are unequal, into others whose denominators are equal. In order, therefore, to perform it generally, let \(\frac{a}{b}\) and \(\frac{c}{d}\) be the fractions proposed. First, multiply the two terms of the first fraction by \(d\), and we shall have the fraction \(\frac{ad}{bd}\) equal to \(\frac{a}{b}\); next multiply the two terms of the second fraction by \(b\), and we shall have an equivalent value of it expressed by \(\frac{bc}{bd}\); thus the two denominators are become equal. Now, if the sum of the two proposed fractions be required, we may immediately answer that it is \(\frac{ad+bc}{bd}\); and if their difference be asked, we say that it is \(\frac{ad-bc}{bd}\). If the fractions ⅝ and ⁷⁄₉, for example, were proposed, we should obtain, in their stead, ⁴⁵⁄₇₂ and ⁵⁶⁄₇₂; of which the sum is ¹⁰¹⁄₇₂ and the difference ¹¹⁄₇₂.

98 To this part of the subject belongs also the question, Of two proposed fractions which is the greater or the less? for, to resolve this, we have only to reduce the two fractions to the same denominator. Let us take, for example, the two fractions ⅔ and ⁵⁄₇; when reduced to the same denominator, the first becomes ¹⁴⁄₂₁, and the second ¹⁵⁄₂₁, where it is evident that the second, or ⁵⁄₇, is the greater, and exceeds the former by ¹⁄₂₁.

Again, if the fractions ⅗ and ⅝ be proposed, we shall have to substitute for them ²⁴⁄₄₀ and ³⁵⁄₄₀; whence we may conclude that ⅝ exceeds ⅗, but only by ¹⁄₄₀.

99 When it is required to subtract a fraction from an integer, it is sufficient to change one of the units of that integer into a fraction, which has the same denominator as that which is to be subtracted ; then in the rest of the operation there is no difficulty. If it be required, for example, to subtract ⅔ from 1, we write ³⁄₃ instead of 1, and say that ⅔ taken from ³⁄₃ leaves the remainder ⅓. So ⁵⁄₁₂, subtracted from 1, leaves ⁷⁄₁₂.

If it were required to subtract ¾ from 2 we should write 1 and ⁴⁄₄ instead of 2, and should then immediately see that after the subtraction there must remain 1¼.

100 It happens also sometimes, that having added two or more fractions together, we obtain more than an integer; that is to say, a numerator greater than the denominator: this is a case which has already occurred, and deserves attention.

We found, for example (Art. 96), that the sum of the five fractions ½, ⅔, ¾, ⅘, and ⅚ was ²¹³⁄₆₀, and remarked that the value of this sum was 3³³⁄₆₀ or 3¹¹⁄₂₀. Likewise, ⅔ + ¾, or ⁸⁄₁₂ + ⁹⁄₁₂, makes ¹⁷⁄₁₂, or 1⁵⁄₁₂. We have therefore only to perform the actual division of the numerator by the denominator, to see how many integers there are for the quotient, and to set down the remainder.

Nearly the same must be done to add together numbers compounded of integers and fractions; we first add the fractions, and if the sum produces one or more integers, these are added to the other integers. If it be proposed, for example, to add 3½ and 2⅔; we first take the sum of ½ and ⅔, or of ³⁄₆ and ⁴⁄₆, which is ⁷⁄₆, or 1⅙; and thus we find the total sum to be 6⅙.

#### 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.