ch1.2.8-calculation-irrational-quantities

Chapter 1.2.8 – Of the Calculation of Irrational Quantities

Summary: Euler extends the four arithmetic operations to expressions containing surds and imaginary quantities, and introduces the conjugate technique for rationalizing denominators.

Sources: chapter-1.2.8

Last updated: 2026-04-28

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Addition and Subtraction

Like surds combine just as like terms do; unlike surds are simply listed side by side (§326): (source: chapter-1.2.8)

$$\sqrt{a}+\sqrt{a}=2\sqrt{a},\sqrt{a}-\sqrt{a}=0.$$

Examples:

• $(3+\sqrt{2})+(1+\sqrt{2})=4+2\sqrt{2}$
• $(5+\sqrt{3})+(4-\sqrt{3})=9$
• $(2\sqrt{3}+3\sqrt{2})+(\sqrt{3}-\sqrt{2})=3\sqrt{3}+2\sqrt{2}$

Subtraction follows the same rule after sign change (§327). (source: chapter-1.2.8)

Multiplication

Key rules (§328): (source: chapter-1.2.8)

$$\sqrt{a}\cdot \sqrt{a}=a,\sqrt{a}\cdot \sqrt{b}=\sqrt{ab}.$$

Examples:

$$(1+\sqrt{2}{)}^2=3+2\sqrt{2},(4+2\sqrt{2})(2-\sqrt{2})=4.$$

Imaginary Quantities

The same rules apply with the key identity $\sqrt{-a}\cdot \sqrt{-a}=-a$ (§329): (source: chapter-1.2.8)

$$(-1+\sqrt{-3}{)}^3=8.$$

(The computation passes through the intermediate square $-2-2\sqrt{-3}$, then multiplying again by $-1+\sqrt{-3}$ gives $8$.)

Division — Rationalizing the Denominator

To divide, write the result as a fraction and multiply numerator and denominator by the conjugate of the denominator, eliminating the radical (§330–331): (source: chapter-1.2.8)

$$\frac{3+2\sqrt{2}}{1+\sqrt{2}}\cdot \frac{1-\sqrt{2}}{1-\sqrt{2}}=\frac{-\sqrt{2}-1}{-1}=\sqrt{2}+1.$$

Further examples: (source: chapter-1.2.8)

$$\frac{8-5\sqrt{2}}{3-2\sqrt{2}}=4+\sqrt{2},\frac{1}{5-2\sqrt{6}}=5+2\sqrt{6},\frac{\sqrt{6}+\sqrt{5}}{\sqrt{6}-\sqrt{5}}=11+2\sqrt{30}.$$

The conjugate technique generalizes: multiplying $a+\sqrt{b}$ by $a-\sqrt{b}$ yields the rational $a^2-b$. (source: chapter-1.2.8)

Multi-Radical Denominators

When the denominator has several radical terms, the conjugate is applied one radical at a time (§332): (source: chapter-1.2.8)

$$\frac{1}{\sqrt{10}-\sqrt{2}-\sqrt{3}}\stackrel{\times (\sqrt{10}+\sqrt{2}+\sqrt{3})}{\to }\frac{\sqrt{10}+\sqrt{2}+\sqrt{3}}{5-2\sqrt{6}}\stackrel{\times (5+2\sqrt{6})}{\to }5\sqrt{10}+11\sqrt{2}+9\sqrt{3}+2\sqrt{60}.$$

Related pages

• ch1.2.7-extraction-of-roots-compound
• ch1.2.9-cubes-and-cube-root-extraction
• rationalization
• irrational-numbers
• square-roots-and-irrational-numbers
• imaginary-numbers