Rationalising Denominators

Understanding how to simplify denominators, especially those involving surds, is essential in both solving equations and applying mathematical concepts in fields such as engineering and science.

Understanding Rational and Irrational Numbers

Brief Overview

Understanding both rational and irrational numbers is crucial for simplifying expressions. These types of numbers play a major role across various mathematical applications.

• Rational Number: Numbers that can be expressed as fractions $\frac{p}{q}$, where $p$ and $q$ are integers, and $q \neq 0$.
  • Examples include $\frac{1}{2}$, $5$, and $-6$.

• Irrational Number: Numbers that cannot be expressed as a fraction of two integers.
  • Key Features:
    • Decimal expansions that neither terminate nor repeat.
  • Examples include $\sqrt{2}$, $\pi$, and $e$.

Surds and Their Properties

• Surds: Expressions with roots that are irrational numbers.
  • Not all roots are surds, such as $\sqrt{4} = 2$.

Key Concept: Surds in Real Numbers

• Historical Insight: Ancient Greeks discovered that $\sqrt{2}$ could not be represented as a fraction, which led to the introduction of irrational numbers.
• Real-world Importance: Recognising these distinctions can influence precision in scientific calculations.

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Rationalising Denominators

Importance of Rationalisation

Rationalising the denominator simplifies complex operations and enhances precision in calculations.

• Simplification: Makes calculations clearer and reduces errors, especially during operations like multiplication and division.
• Exactness and Convention: Ensures mathematical precision and compliance with conventions, vital in fields like engineering and finance.

Steps to Rationalise Denominators

1. Identify the Surd Denominator.
2. Multiply Numerator and Denominator by the Surd.
3. Simplify the Result.

Simple Example: Single Surd

Problem: Rationalise $\frac{1}{\sqrt{2}}$.

• Solution:
  • Multiply both numerator and denominator by $\sqrt{2}$: $\frac{1}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2}$
  • This gives us $\frac{\sqrt{2}}{2}$ as our rationalised result.

Binomial Denominator

• Concept of Conjugates:
  • Conjugate Pairs: Formed by changing the sign between the terms in a binomial, e.g., the conjugate of $1 + \sqrt{3}$ is $1 - \sqrt{3}$.
  • Application:
    • Changing the sign eliminates surds based on the difference of squares: $(a+b)(a-b)=a^2-b^2$.

Example: Utilising Conjugates

Problem: Rationalise $\frac{1}{1 + \sqrt{3}}$.

• Solution:
  • Multiply numerator and denominator by the conjugate $1 - \sqrt{3}$:
  • $\frac{1}{1 + \sqrt{3}} \times \frac{1 - \sqrt{3}}{1 - \sqrt{3}} = \frac{1 - \sqrt{3}}{(1)^2 - (\sqrt{3})^2}$
  • This gives $\frac{1 - \sqrt{3}}{1 - 3} = \frac{1 - \sqrt{3}}{-2}$
  • Simplifying further: $\frac{\sqrt{3} - 1}{2}$

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Solutions to Practice Exercises

1. Rationalise: $\frac{2}{\sqrt{3}}$. Solution: $\frac{2}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{2\sqrt{3}}{3}$

2. Rationalise: $\frac{1}{2 - \sqrt{5}}$. Solution:

   • Multiply by conjugate: $\frac{1}{2 - \sqrt{5}} \times \frac{2 + \sqrt{5}}{2 + \sqrt{5}}$
   • $= \frac{2 + \sqrt{5}}{(2)^2 - (\sqrt{5})^2} = \frac{2 + \sqrt{5}}{4 - 5} = \frac{2 + \sqrt{5}}{-1} = \frac{\sqrt{5} + 2}{1}$

3. Complex Challenge: $\frac{1}{\sqrt{2} + \sqrt{3}}$. Solution:

   • Multiply by conjugate: $\frac{1}{\sqrt{2} + \sqrt{3}} \times \frac{\sqrt{2} - \sqrt{3}}{\sqrt{2} - \sqrt{3}}$
   • $= \frac{\sqrt{2} - \sqrt{3}}{(\sqrt{2})^2 - (\sqrt{3})^2} = \frac{\sqrt{2} - \sqrt{3}}{2 - 3} = \frac{\sqrt{2} - \sqrt{3}}{-1} = \frac{\sqrt{3} - \sqrt{2}}{1}$