Lecture 6: Simplification, Rational/Irrational Numbers, and Exponents

 This lecture includes simplification of expressions with square roots, classification of numbers, rationalisation of denominators, and laws of exponents.

📘 Lecture 6: Simplification, Rational/Irrational Numbers, and Exponents

🔹 Topic 1: Simplification Involving Square Roots

Let’s begin with examples that show how to simplify algebraic expressions containing square roots.

📍 Example 15: Simplify

1. $(5 + \sqrt{7})(2 + \sqrt{5})$
   Use distributive law (FOIL method):

   $$= 5 \cdot 2 + 5 \cdot \sqrt{5} + \sqrt{7} \cdot 2 + \sqrt{7} \cdot \sqrt{5} = 10 + 5\sqrt{5} + 2\sqrt{7} + \sqrt{35}$$

2. $(5 + \sqrt{5})(5 - \sqrt{5})$
   Identity: $(a + b)(a - b) = a^2 - b^2$

   $$= 25 - 5 = 20$$

3. $(\sqrt{3} + \sqrt{7})^2$
   Identity: $(a + b)^2 = a^2 + 2ab + b^2$

   $$= 3 + 2\sqrt{21} + 7 = 10 + 2\sqrt{21}$$

4. $(\sqrt{11} - \sqrt{7})(\sqrt{11} + \sqrt{7})$
   Identity: $(a - b)(a + b) = a^2 - b^2$

   $$= 11 - 7 = 4$$

🔹 Topic 2: Classifying Numbers as Rational or Irrational

A number is irrational if it cannot be expressed as a fraction $\frac{p}{q}$, where p and q are integers and $q \ne 0$.

📍 Classify the following:

1. $2 - \sqrt{5}$ → Irrational
   (Rational - Irrational = Irrational)

2. $(3 + \sqrt{23}) - \sqrt{23} = 3$ → Rational

3. $\frac{2\sqrt{7}}{7\sqrt{7}} = \frac{2}{7}$ → Rational

4. $\frac{1}{\sqrt{2}}$ → Irrational

5. $2\pi$ → Irrational

🔹 Topic 3: Simplifying Square Root Expressions

Use standard identities and combine like terms.

Examples:

1. $(3 + \sqrt{3})(2 + \sqrt{2})$

2. $(3 + \sqrt{3})(3 - \sqrt{3})$

3. $(\sqrt{5} + \sqrt{2})^2$

4. $(\sqrt{5} - \sqrt{2})(\sqrt{5} + \sqrt{2})$

👉 Use formulas like:

• $(a + b)^2 = a^2 + 2ab + b^2$

• $(a + b)(a - b) = a^2 - b^2$

🔹 Topic 4: Think and Discuss

📌 Why is π irrational, even though it's defined as a ratio (C/d)?

• π is defined as the ratio of circumference to diameter, but it's not a ratio of two integers.

• So while it appears rational in form, it is not expressible exactly as a fraction.

• Its decimal form is non-terminating and non-repeating, making it irrational.

🔹 Topic 5: Number Line Representation

📍 Represent $\sqrt{3}$ on the number line using geometry (similar to the method in Lecture 5).

1. Mark 0 and 1 on the number line.

2. Construct segment 1.5 units from 0, form a right triangle.

3. Use Pythagoras’ Theorem to locate $\sqrt{3}$.

🔹 Topic 6: Rationalising the Denominator

To rationalise means to eliminate the square root from the denominator.

Examples:

1. $\frac{1}{\sqrt{7}} = \frac{\sqrt{7}}{7}$

2. $\frac{1}{\sqrt{7} - \sqrt{6}} = \frac{\sqrt{7} + \sqrt{6}}{7 - 6} = \sqrt{7} + \sqrt{6}$

3. $\frac{1}{\sqrt{5} + \sqrt{2}} = \frac{\sqrt{5} - \sqrt{2}}{3}$

4. $\frac{1}{\sqrt{7} - 2} = \text{Multiply numerator and denominator by } \sqrt{7} + 2$

🔹 Topic 7: Laws of Exponents

Let $a \ne 0$, and m, n are integers.

1. $a^m \cdot a^n = a^{m+n}$

2. $(a^m)^n = a^{mn}$

3. $\frac{a^m}{a^n} = a^{m-n}$

4. $a^m \cdot b^m = (ab)^m$

🔹 Exercise Practice (from Ex 1.5)

1. Find:

• $64^{\frac{2}{3}} = (4^3)^{2/3} = 4^2 = 16$

• $32^{\frac{3}{5}} = (2^5)^{3/5} = 2^3 = 8$

2. Simplify powers using laws of exponents:

• $2^3 \cdot 2^5 = 2^8$

• $\left( \frac{1}{3^7} \right)^{-1} = 3^7$

• $\frac{11^{\frac{1}{2}} \cdot 11^{\frac{1}{4}}}{11^{\frac{3}{4}}} = 11^{(1/2 + 1/4 - 3/4)} = 11^0 = 1$

📝 Homework

1. Rationalise: $\frac{1}{\sqrt{3} + \sqrt{2}}$

2. Classify: $\sqrt{49} + \pi$

3. Simplify: $(\sqrt{3} + 1)^2$

4. Prove: $\sqrt{2}$ is irrational.