EduNes Logo

Less Stress ↓

More Success ↑

EduNes means


Educational Network for Excellence and Success

EduNes Students

Saturday, 31 May 2025

Lecture 7: Worksheet 2: – Number System

 


📘 Lecture 7: Worksheet 2: – Number System

Class: 9 CBSE
Chapter: Number System
Topic: Mixed Problems
Worksheet Title: Consolidation and Skill Practice
Total Marks: 40
Time: 1 Hour


🔹 Section A: Fill in the Blanks (1 mark each)

  1. 50=___\sqrt{50} = \_\_\_

  2. (a3)2=___(a^3)^2 = \_\_\_

  3. 2×8=___\sqrt{2} \times \sqrt{8} = \_\_\_

  4. A number that cannot be written in the form pq\frac{p}{q}, where pp and qq are integers and q0q \neq 0, is called ___.

  5. (14)1/2=___\left(\frac{1}{4}\right)^{-1/2} = \_\_\_


🔹 Section B: Match the Columns (2 marks each)

Match the expression in Column A with its simplified value in Column B.

Column A Column B
a) (53)(5+3)(5 - \sqrt{3})(5 + \sqrt{3}) A. 88
b) 12÷3\sqrt{12} \div \sqrt{3} B. 2222
c) 172\frac{1}{\sqrt{7} - 2} C. 7+23\frac{\sqrt{7} + 2}{3}
d) 23222^3 \cdot 2^{-2} D. 5235^2 - 3


🔹 Section C: Solve the Following (3 marks each)

  1. Simplify:
    (5+2)2(\sqrt{5} + 2)^2

  2. Rationalise the denominator and simplify:
    23+2\frac{2}{\sqrt{3} + \sqrt{2}}

  3. If a=21/3a = 2^{1/3} and b=22/3b = 2^{2/3}, then evaluate aba \cdot b

  4. Classify the following as Rational or Irrational:
    a) 121\sqrt{121}
    b) 327\frac{3\sqrt{2}}{7}
    c) π3\pi - 3


🔹 Section D: Application-Based (4 marks each)

  1. A rope is 75\sqrt{75} metres long. Another rope is 27\sqrt{27} metres long.
    a) Find the total length in simplest form.
    b) Classify the result as rational or irrational.

  2. Simplify using exponent laws:
    a) 51/253/25^{1/2} \cdot 5^{3/2}
    b) (49)3/2\left( \frac{4}{9} \right)^{-3/2}


🔹 Section E: Conceptual & Reasoning (5 marks each)

  1. If a number is written as 2+52 + \sqrt{5}, show that its conjugate is 252 - \sqrt{5}.
    Now multiply the two and determine the result.
    What conclusion can you draw about the product of irrational conjugates?

  2. A student claims:

227\frac{22}{7} is the value of π\pi, so it must be a rational number.”
Do you agree with this claim? Justify your answer and explain the difference between approximation and actual value of irrational numbers.


📌 Challenge Task (5 bonus marks)

Construct a proof that 3\sqrt{3} is irrational using contradiction method (proof by assumption).


Student Instructions:

  • Read each question carefully.

  • Use pencil and scale for diagrams.

  • Try all sections for full understanding.

  • Marks are mentioned next to each question.


📘 Lecture 7 Worksheet 2: – Number System: Solutions


🔹 Section A: Fill in the Blanks

  1. 50=252=52\sqrt{50} = \sqrt{25 \cdot 2} = 5\sqrt{2}

  2. (a3)2=a32=a6(a^3)^2 = a^{3 \cdot 2} = a^6

  3. 2×8=16=4\sqrt{2} \times \sqrt{8} = \sqrt{16} = 4

  4. Irrational number

  5. (14)1/2=(4)1/2=4=2\left(\frac{1}{4}\right)^{-1/2} = (4)^{1/2} = \sqrt{4} = 2


🔹 Section B: Match the Columns

Column A Column B
a) (53)(5+3)(5 - \sqrt{3})(5 + \sqrt{3}) B. 2222
b) 12÷3\sqrt{12} \div \sqrt{3} A. 22
c) 172\frac{1}{\sqrt{7} - 2} C. 7+23\frac{\sqrt{7} + 2}{3}
d) 23222^3 \cdot 2^{-2} D. 232=21=22^{3 - 2} = 2^1 = 2

So matching answers:
a–B, b–A, c–C, d–D


🔹 Section C: Solve the Following

  1. (5+2)2=(5)2+252+22=5+45+4=9+45(\sqrt{5} + 2)^2 = (\sqrt{5})^2 + 2 \cdot \sqrt{5} \cdot 2 + 2^2 = 5 + 4\sqrt{5} + 4 = 9 + 4\sqrt{5}

  2. 23+23232=2(32)(3+2)(32)\frac{2}{\sqrt{3} + \sqrt{2}} \cdot \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} - \sqrt{2}} = \frac{2(\sqrt{3} - \sqrt{2})}{(\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2})}
    Denominator: 32=13 - 2 = 1
    Answer: 2(32)=23222(\sqrt{3} - \sqrt{2}) = 2\sqrt{3} - 2\sqrt{2}

  3. a=21/3,b=22/3ab=21/3+2/3=21=2a = 2^{1/3}, b = 2^{2/3} \Rightarrow a \cdot b = 2^{1/3 + 2/3} = 2^1 = 2

a) 121=11\sqrt{121} = 11 – Rational
b) 327\frac{3\sqrt{2}}{7} – Irrational
c) π3\pi - 3 – Irrational (since π is irrational)


🔹 Section D: Application-Based

a) 75+27=253+93=53+33=83\sqrt{75} + \sqrt{27} = \sqrt{25 \cdot 3} + \sqrt{9 \cdot 3} = 5\sqrt{3} + 3\sqrt{3} = 8\sqrt{3}
b) Since 3\sqrt{3} is irrational, 838\sqrt{3} is also irrational.

a) 51/253/2=51/2+3/2=52=255^{1/2} \cdot 5^{3/2} = 5^{1/2 + 3/2} = 5^2 = 25
b) (49)3/2=(94)3/2=(9/4)3=(3/2)3=27/8\left( \frac{4}{9} \right)^{-3/2} = \left( \frac{9}{4} \right)^{3/2} = \left( \sqrt{9}/\sqrt{4} \right)^3 = (3/2)^3 = 27/8


🔹 Section E: Conceptual & Reasoning

Given number = 2+52 + \sqrt{5}, its conjugate is 252 - \sqrt{5}
Product:
(2+5)(25)=22(5)2=45=1(2 + \sqrt{5})(2 - \sqrt{5}) = 2^2 - (\sqrt{5})^2 = 4 - 5 = -1
Conclusion: The product of conjugates (a+b)(ab)=a2b(a + \sqrt{b})(a - \sqrt{b}) = a^2 - b, always gives a rational number.

No, we do not agree.

  • π227\pi \approx \frac{22}{7}, but this is only an approximation.

  • Actual value of π\pi is non-terminating, non-repeating = irrational

  • Rational approximations are used for calculations, but the true nature of π\pi remains irrational.


📌 Challenge Task: Proof 3\sqrt{3} is irrational

Assume 3\sqrt{3} is rational.
Then 3=pq\sqrt{3} = \frac{p}{q}, where p, q are integers with no common factor and q0q \neq 0.
Squaring both sides:
3=p2q2p2=3q23 = \frac{p^2}{q^2} \Rightarrow p^2 = 3q^2
So p² is divisible by 3 → p is divisible by 3 → p = 3k
Then p2=9k2=3q2q2=3k2qp^2 = 9k^2 = 3q^2 \Rightarrow q^2 = 3k^2 \Rightarrow q is also divisible by 3.

So both p and q are divisible by 3, contradicting that they have no common factor.
Hence, 3\sqrt{3} is irrational.


Would you like this answer sheet converted to a printable PDF format with answer boxes, marking scheme, and school branding (like Edunes Online Education logo)?

Solved Worksheet 1: Number System



Solved Worksheet: Number System – Mixed Problems

Class: 9 CBSE
Total Marks: 40 | Time: 60 Minutes


🔹 Section A: Multiple Choice Questions (1 mark each)

  1. (75)(7+5)=75=2(\sqrt{7} - \sqrt{5})(\sqrt{7} + \sqrt{5}) = 7 - 5 = \boxed{2}Answer: A

  2. Rational number: 34\boxed{\frac{3}{4}}Answer: C

  3. (23)2=4×3=12(2\sqrt{3})^2 = 4 \times 3 = \boxed{12}Answer: A

  4. 163/4=(24)3/4=23=816^{3/4} = (2^4)^{3/4} = 2^3 = \boxed{8}Answer: B

  5. (152)1=52=25\left( \frac{1}{5^2} \right)^{-1} = 5^2 = \boxed{25}Answer: B


🔹 Section B: Very Short Answer (2 marks each)

  1. Rationalise:

13+1×3131=3131=312\frac{1}{\sqrt{3} + 1} \times \frac{\sqrt{3} - 1}{\sqrt{3} - 1} = \frac{\sqrt{3} - 1}{3 - 1} = \boxed{\frac{\sqrt{3} - 1}{2}}
  1. 3113 - \sqrt{11}: Irrational, because 11\sqrt{11} is irrational and difference with rational number remains irrational. → \boxed{\text{Irrational}}

  2. (5+2)2=25+2×5×2+2=27+102(5 + \sqrt{2})^2 = 25 + 2 \times 5 \times \sqrt{2} + 2 = \boxed{27 + 10\sqrt{2}}

  3. Geometrical Representation of 2\sqrt{2}:
    Draw a right-angled triangle with both legs of 1 unit. The hypotenuse is 2\sqrt{2}.
    Using compass, draw an arc from origin with radius 2\sqrt{2}. The intersection point on number line is 2\boxed{\sqrt{2}}.

a) 271/3=327^{1/3} = \boxed{3}
b) 813/4=(34)3/4=33=2781^{3/4} = (3^4)^{3/4} = 3^3 = \boxed{27}


🔹 Section C: Short Answer (3 marks each)

5+252×5+25+2=(5+2)252=5+2+2103=7+2103\frac{\sqrt{5} + \sqrt{2}}{\sqrt{5} - \sqrt{2}} \times \frac{\sqrt{5} + \sqrt{2}}{\sqrt{5} + \sqrt{2}} = \frac{(\sqrt{5} + \sqrt{2})^2}{5 - 2} = \frac{5 + 2 + 2\sqrt{10}}{3} = \boxed{\frac{7 + 2\sqrt{10}}{3}}

a) (3233)÷34=32+34=31=3(3^2 \cdot 3^3) \div 3^4 = 3^{2+3-4} = 3^1 = \boxed{3}

b) (23)224=2624=22=4(2^3)^2 \cdot 2^{-4} = 2^{6} \cdot 2^{-4} = 2^{2} = \boxed{4}

a) 49=7\sqrt{49} = 7 → \boxed{\text{Rational}}
b) 27\frac{2}{\sqrt{7}} → Denominator irrational → \boxed{\text{Irrational}}
c) 3+7\sqrt{3} + \sqrt{7} → sum of irrationals not simplifiable → \boxed{\text{Irrational}}


🔹 Section D: Long Answer (4 marks each)

(2+3)2(23)2=[4+43+3][443+3]=7+43(743)=83(2 + \sqrt{3})^2 - (2 - \sqrt{3})^2 = [4 + 4\sqrt{3} + 3] - [4 - 4\sqrt{3} + 3] = 7 + 4\sqrt{3} - (7 - 4\sqrt{3}) = \boxed{8\sqrt{3}}
352+25+2=3(5+2)+2(52)(5)2(2)2=35+32+252252=55+23=55+23\frac{3}{\sqrt{5} - \sqrt{2}} + \frac{2}{\sqrt{5} + \sqrt{2}} \\ = \frac{3(\sqrt{5} + \sqrt{2}) + 2(\sqrt{5} - \sqrt{2})}{(\sqrt{5})^2 - (\sqrt{2})^2} \\ = \frac{3\sqrt{5} + 3\sqrt{2} + 2\sqrt{5} - 2\sqrt{2}}{5 - 2} \\ = \frac{5\sqrt{5} + \sqrt{2}}{3} = \boxed{\frac{5\sqrt{5} + \sqrt{2}}{3}}
  1. Prove 2\sqrt{2} is irrational – contradiction method:

Assume 2=pq\sqrt{2} = \frac{p}{q}, where p,qp, q are integers, gcd(p,q)=1\gcd(p, q) = 1

Squaring:
2=p2q2p2=2q22 = \frac{p^2}{q^2} \Rightarrow p^2 = 2q^2

So, p2p^2 even ⇒ pp even ⇒ p=2kp = 2k
Then: p2=4k2=2q2q2=2k2p^2 = 4k^2 = 2q^2 ⇒ q^2 = 2k^2qq also even

pp and qq both even ⇒ Contradiction to assumption that they are coprime.

2 is irrational\boxed{\sqrt{2} \text{ is irrational}}

a)

(827)2/3=82/3272/3=(23)2/3(33)2/3=2232=4949\left( \frac{8}{27} \right)^{2/3} = \frac{8^{2/3}}{27^{2/3}} = \frac{(2^3)^{2/3}}{(3^3)^{2/3}} = \frac{2^2}{3^2} = \frac{4}{9} \Rightarrow \boxed{\frac{4}{9}}

b)

82/3÷24/3=(23)2/3÷24/3=22÷24/3=224/3=22/3=22/38^{2/3} \div 2^{4/3} = (2^3)^{2/3} \div 2^{4/3} = 2^2 \div 2^{4/3} = 2^{2 - 4/3} = 2^{2/3} = \boxed{2^{2/3}}


🔹 Section E: Challenge Question (5 marks)

Rohit's statement:
“π is defined as the ratio of circumference to diameter ⇒ it must be rational” is incorrect.

Explanation:

  • Though π=Cd\pi = \frac{C}{d}, both circumference and diameter are real numbers.

  • The ratio is not always rational.

  • π has a non-terminating, non-repeating decimal expansion, and cannot be expressed as a fraction ⇒ Irrational.

Approximation Check:

Radius r=7 cmr = 7 \text{ cm}
Circumference C=2πr2×227×7=44 cmC = 2\pi r \approx 2 \times \frac{22}{7} \times 7 = 44 \text{ cm}

True value using π3.1416\pi \approx 3.1416:
C=2×3.1416×7=43.982444 cmC = 2 \times 3.1416 \times 7 = 43.9824 \approx 44 \text{ cm}

Hence, 227\frac{22}{7} is a very good rational approximation of π, but π itself remains irrational.


Would you like a print-ready PDF version of this worksheet + solution set with your name/school/brand/logo at the top?

Worksheet: Number System – Mixed Problems



📄 Worksheet 1: Number System – Mixed Problems

Class: 9 CBSE
Chapter: Number System
Topic: Simplification, Rational/Irrational, Rationalisation, Exponents
Total Marks: 40
Time: 60 Minutes


🔹 Section A: Multiple Choice Questions (1 mark each)

Choose the correct option and write the answer.

  1. (75)(7+5)(\sqrt{7} - \sqrt{5})(\sqrt{7} + \sqrt{5}) equals:

    • A. 2

    • B. 2\sqrt{2}

    • C. 12

    • D. 35\sqrt{35}

  2. Which of the following is a rational number?

    • A. 2\sqrt{2}

    • B. 13\frac{1}{\sqrt{3}}

    • C. 34\frac{3}{4}

    • D. π\pi

  3. (23)2=(2\sqrt{3})^2 =

    • A. 12

    • B. 6

    • C. 434\sqrt{3}

    • D. 9

  4. 163/416^{3/4} is equal to:

    • A. 4

    • B. 8

    • C. 16

    • D. 2

  5. (152)1=\left( \frac{1}{5^2} \right)^{-1} =

    • A. 5

    • B. 25

    • C. 125\frac{1}{25}

    • D. 25-25


🔹 Section B: Very Short Answer (2 marks each)

Answer in one step wherever possible.

  1. Rationalise: 13+1\frac{1}{\sqrt{3} + 1}

  2. Classify as rational or irrational: 3113 - \sqrt{11}

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

  4. Represent 2\sqrt{2} geometrically on a number line. (Sketch or describe method)

  5. Find:
    a) 271/327^{1/3}
    b) 813/481^{3/4}


🔹 Section C: Short Answer Questions (3 marks each)

Show working where required.

  1. Simplify and express in simplest form:
    5+252\frac{\sqrt{5} + \sqrt{2}}{\sqrt{5} - \sqrt{2}}

  2. Evaluate:
    a) (3233)÷34(3^2 \cdot 3^3) \div 3^4
    b) (23)224(2^3)^2 \cdot 2^{-4}

  3. Without actual calculation, state whether the following are rational or irrational. Justify:

    • a) 49\sqrt{49}

    • b) 27\frac{2}{\sqrt{7}}

    • c) 3+7\sqrt{3} + \sqrt{7}


🔹 Section D: Long Answer Questions (4 marks each)

Explain each step clearly.

  1. Simplify the expression:

(2+3)2(23)2(2 + \sqrt{3})^2 - (2 - \sqrt{3})^2
  1. Rationalise and simplify:

352+25+2\frac{3}{\sqrt{5} - \sqrt{2}} + \frac{2}{\sqrt{5} + \sqrt{2}}
  1. Prove that 2\sqrt{2} is irrational.
    (Use contradiction method)

  2. Evaluate:

(827)2/3\left( \frac{8}{27} \right)^{2/3}

Then use exponent laws to show:

82/3÷24/38^{2/3} \div 2^{4/3}


🔹 Section E: Challenge Question (5 marks)

This tests higher-order thinking.

  1. Rohit says that since π is defined as the ratio of circumference to diameter (C/d), it must be a rational number. Do you agree? Justify with reasoning. Then calculate an approximate value of π using a circle of radius 7 cm (Use π227\pi \approx \frac{22}{7}) and check the accuracy of this approximation.


Instructions for Students:

  • Solve each section step by step.

  • For geometry-related questions, draw figures neatly.

  • Highlight final answers.

  • Use separate sheets if necessary.

Friday, 30 May 2025

EDUNES LANDING PAGE

Edunes Landing

Worksheet: Number System – Mixed Problems

Class 9 Worksheet - Number System

📄 Worksheet: Number System – Mixed Problems

Class: 9 CBSE

Chapter: Number System

Topic: Simplification, Rational/Irrational, Rationalisation, Exponents

Total Marks: 40

Time: 60 Minutes

🔹 Section A: Multiple Choice Questions (1 mark each)

Choose the correct option and write the answer.

1. \( (\sqrt{7} - \sqrt{5})(\sqrt{7} + \sqrt{5}) \) equals:
  • 2
  • \( \sqrt{2} \)
  • 12
  • \( \sqrt{35} \)
2. Which of the following is a rational number?
  • \( \sqrt{2} \)
  • \( \frac{1}{\sqrt{3}} \)
  • \( \frac{3}{4} \)
  • \( \pi \)
3. \( (2\sqrt{3})^2 = \)
  • 12
  • 6
  • \( 4\sqrt{3} \)
  • 9
4. \( 16^{3/4} \) is equal to:
  • 4
  • 8
  • 16
  • 2
5. \( \left( \frac{1}{5^2} \right)^{-1} = \)
  • 5
  • 25
  • \( \frac{1}{25} \)
  • -25

🔹 Section B: Very Short Answer (2 marks each)

Answer in one step wherever possible.

  1. Rationalise: \( \frac{1}{\sqrt{3} + 1} \)
  2. Classify as rational or irrational: \( 3 - \sqrt{11} \)
  3. Simplify: \( (5 + \sqrt{2})^2 \)
  4. Represent \( \sqrt{2} \) geometrically on a number line. (Sketch or describe method)
  5. Find:
    • a) \( 27^{1/3} \)
    • b) \( 81^{3/4} \)

🔹 Section C: Short Answer Questions (3 marks each)

  1. Simplify and express in simplest form: \( \frac{\sqrt{5} + \sqrt{2}}{\sqrt{5} - \sqrt{2}} \)
  2. Evaluate:
    • a) \( (3^2 \cdot 3^3) \div 3^4 \)
    • b) \( (2^3)^2 \cdot 2^{-4} \)
  3. Without actual calculation, state whether the following are rational or irrational. Justify:
    • a) \( \sqrt{49} \)
    • b) \( \frac{2}{\sqrt{7}} \)
    • c) \( \sqrt{3} + \sqrt{7} \)

🔹 Section D: Long Answer Questions (4 marks each)

  1. Simplify the expression:
    \( (2 + \sqrt{3})^2 - (2 - \sqrt{3})^2 \)
  2. Rationalise and simplify:
    \( \frac{3}{\sqrt{5} - \sqrt{2}} + \frac{2}{\sqrt{5} + \sqrt{2}} \)
  3. Prove that \( \sqrt{2} \) is irrational. (Use contradiction method)
  4. Evaluate:
    • \( \left( \frac{8}{27} \right)^{2/3} \)
    • Then use exponent laws to show: \( 8^{2/3} \div 2^{4/3} \)

🔹 Section E: Challenge Question (5 marks)

18. Rohit says that since \( \pi \) is defined as the ratio of circumference to diameter (C/d), it must be a rational number. Do you agree? Justify with reasoning. Then calculate an approximate value of \( \pi \) using a circle of radius 7 cm (Use \( \pi \approx \frac{22}{7} \)) and check the accuracy of this approximation.

✅ Instructions for Students:

  • Solve each section step by step.
  • For geometry-related questions, draw figures neatly.
  • Highlight final answers.
  • Use separate sheets if necessary.

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

    =52+55+72+75=10+55+27+35= 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+5)(55)(5 + \sqrt{5})(5 - \sqrt{5})
    Identity: (a+b)(ab)=a2b2(a + b)(a - b) = a^2 - b^2

    =255=20= 25 - 5 = 20
  3. (3+7)2(\sqrt{3} + \sqrt{7})^2
    Identity: (a+b)2=a2+2ab+b2(a + b)^2 = a^2 + 2ab + b^2

    =3+221+7=10+221= 3 + 2\sqrt{21} + 7 = 10 + 2\sqrt{21}
  4. (117)(11+7)(\sqrt{11} - \sqrt{7})(\sqrt{11} + \sqrt{7})
    Identity: (ab)(a+b)=a2b2(a - b)(a + b) = a^2 - b^2

    =117=4= 11 - 7 = 4


🔹 Topic 2: Classifying Numbers as Rational or Irrational

A number is irrational if it cannot be expressed as a fraction pq\frac{p}{q}, where p and q are integers and q0q \ne 0.

📍 Classify the following:

  1. 252 - \sqrt{5}Irrational
    (Rational - Irrational = Irrational)

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

  3. 2777=27\frac{2\sqrt{7}}{7\sqrt{7}} = \frac{2}{7}Rational

  4. 12\frac{1}{\sqrt{2}}Irrational

  5. 2π2\piIrrational


🔹 Topic 3: Simplifying Square Root Expressions

Use standard identities and combine like terms.

Examples:

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

  2. (3+3)(33)(3 + \sqrt{3})(3 - \sqrt{3})

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

  4. (52)(5+2)(\sqrt{5} - \sqrt{2})(\sqrt{5} + \sqrt{2})

👉 Use formulas like:

  • (a+b)2=a2+2ab+b2(a + b)^2 = a^2 + 2ab + b^2

  • (a+b)(ab)=a2b2(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 3\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 3\sqrt{3}.


🔹 Topic 6: Rationalising the Denominator

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

Examples:

  1. 17=77\frac{1}{\sqrt{7}} = \frac{\sqrt{7}}{7}

  2. 176=7+676=7+6\frac{1}{\sqrt{7} - \sqrt{6}} = \frac{\sqrt{7} + \sqrt{6}}{7 - 6} = \sqrt{7} + \sqrt{6}

  3. 15+2=523\frac{1}{\sqrt{5} + \sqrt{2}} = \frac{\sqrt{5} - \sqrt{2}}{3}

  4. 172=Multiply numerator and denominator by 7+2\frac{1}{\sqrt{7} - 2} = \text{Multiply numerator and denominator by } \sqrt{7} + 2


🔹 Topic 7: Laws of Exponents

Let a0a \ne 0, and m, n are integers.

  1. aman=am+na^m \cdot a^n = a^{m+n}

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

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

  4. ambm=(ab)ma^m \cdot b^m = (ab)^m


🔹 Exercise Practice (from Ex 1.5)

1. Find:

  • 6423=(43)2/3=42=1664^{\frac{2}{3}} = (4^3)^{2/3} = 4^2 = 16

  • 3235=(25)3/5=23=832^{\frac{3}{5}} = (2^5)^{3/5} = 2^3 = 8

2. Simplify powers using laws of exponents:

  • 2325=282^3 \cdot 2^5 = 2^8

  • (137)1=37\left( \frac{1}{3^7} \right)^{-1} = 3^7

  • 111211141134=11(1/2+1/43/4)=110=1\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: 13+2\frac{1}{\sqrt{3} + \sqrt{2}}

  2. Classify: 49+π\sqrt{49} + \pi

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

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



Lecture 5: Geometrical Representation of Square Roots



📘 Lecture 5: Geometrical Representation of Square Roots & Properties of Square Roots


🔹 Topic 1: What does √a mean?

For any real number a>0a > 0,
a=b means b2=a and b>0\sqrt{a} = b \text{ means } b^2 = a \text{ and } b > 0

This definition works for both:

  • Natural numbers (e.g., √4 = 2, because 2² = 4)

  • Real numbers (e.g., √3.5)


🔹 Topic 2: How to Represent √x Geometrically?

We now explore how to construct x\sqrt{x} geometrically when xx is a positive real number.


📐 Example: Constructing √3.5 geometrically


Refer to Fig. 1.11

Steps:

  1. Draw a line segment AB = 3.5 units on a number line.

  2. From point B, mark 1 unit to the right. Let that point be C.

  3. Find midpoint of AC, mark it as O.

  4. Draw a semicircle with center O and radius OC.

  5. Draw a perpendicular line from point B to the semicircle. Let it intersect at point D.

  6. Then, BD = √3.5

✅ This gives a geometric way to represent square root of 3.5.


📐 General Case: Represent √x geometrically

Refer to Fig. 1.12

Steps for any real number x>0x > 0:

  1. Mark AB = x units.

  2. From B, mark 1 unit to the right, name the point as C.

  3. Find the midpoint O of AC.

  4. Draw a semicircle with diameter AC.

  5. Draw a perpendicular from B to the semicircle and mark intersection point as D.

  6. Then, BD = √x


🧠 Proof using Pythagoras Theorem

In △OBD, ∠OBD = 90°, so it’s a right triangle.

Let’s calculate:

  • Radius of semicircle =
    OC=x+12OC = \frac{x + 1}{2}

  • OB =
    xx+12=x12x - \frac{x + 1}{2} = \frac{x - 1}{2}

Using Pythagoras:

BD2=OD2OB2=(x+12)2(x12)2=(x+1)2(x1)24=4x4=xBD^2 = OD^2 - OB^2 = \left( \frac{x+1}{2} \right)^2 - \left( \frac{x-1}{2} \right)^2 = \frac{(x+1)^2 - (x-1)^2}{4} = \frac{4x}{4} = x

Thus,

BD=xBD = \sqrt{x}


📘 Topic 3: Properties of Square Roots

Let a,b>0a, b > 0 be real numbers.

  1. ab=ab\sqrt{ab} = \sqrt{a} \cdot \sqrt{b}

  2. ab=ab\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}

  3. (a+b)(ab)=ab(\sqrt{a} + \sqrt{b})(\sqrt{a} - \sqrt{b}) = a - b

  4. (a+b)(ab)=a2b(a + \sqrt{b})(a - \sqrt{b}) = a^2 - b

  5. (a+b)(c+d)=ac+ad+bc+bd(\sqrt{a} + \sqrt{b})(\sqrt{c} + \sqrt{d}) = \sqrt{ac} + \sqrt{ad} + \sqrt{bc} + \sqrt{bd}

  6. (a+b)2=a+2ab+b(\sqrt{a} + \sqrt{b})^2 = a + 2\sqrt{ab} + b


📚 Exercise for Practice

  1. Geometrically construct 2.5\sqrt{2.5} using the method described.

  2. Prove that (5+3)2=5+215+3(\sqrt{5} + \sqrt{3})^2 = 5 + 2\sqrt{15} + 3

  3. Use Pythagoras Theorem to verify that your geometric construction is correct.

  4. Show that 82=16\sqrt{8} \cdot \sqrt{2} = \sqrt{16}


📌 Key Takeaway

We can visualize square roots of real numbers on a number line using semicircles and perpendicular lines, and use algebraic identities to simplify square root expressions.