Samacheer Kalvi Class 9 Maths Solutions Chapter 2 Real Numbers Exercise 2.3

Get the most accurate TN Board Solutions for Class 9 Maths Chapter 02 Real Numbers here. Updated for the 2026-27 academic session, these solutions are based on the latest TN Board textbooks for Class 9 Maths. Our expert-created answers for Class 9 Maths are available for free download in PDF format.

Detailed Chapter 02 Real Numbers TN Board Solutions for Class 9 Maths

For Class 9 students, solving TN Board textbook questions is the most effective way to build a strong conceptual foundation. Our Class 9 Maths solutions follow a detailed, step-by-step approach to ensure you understand the logic behind every answer. Practicing these Chapter 02 Real Numbers solutions will improve your exam performance.

Class 9 Maths Chapter 02 Real Numbers TN Board Solutions PDF

 

Question 1. Represent the following irrational numbers on the number line.

(i) \( \sqrt{3} \)
Answer:

A B C O E 3 cm 1 cm D

**Steps of construction for \( \sqrt{3} \):**
1. Draw a straight line and mark point A. Then, mark point B so that the distance AB is 3 cm.
2. Mark another point C on this line. Ensure the distance BC is 1 cm.
3. Find the middle point of AC by drawing a perpendicular bisector of AC. Let this midpoint be 'O'.
4. With O as the center and OC (or OA) as the radius, draw a semicircle. This semicircle forms the upper part of the construction.
5. Draw a line from point B that is perpendicular to the main line (AB). This perpendicular line should touch the semicircle at point D.
6. Now, the length of BD will be \( \sqrt{3} \). You can transfer this length to the number line from B, and the point it reaches will be E, so BE is equal to BD, which is \( \sqrt{3} \).
In simple words: To show \( \sqrt{3} \) on a number line, you first create a special triangle where one side is \( \sqrt{3} \) long. You then use a compass to draw an arc from the point B, with the length of the triangle's side BD, to find where \( \sqrt{3} \) is on the number line.

🎯 Exam Tip: Always clearly label all points (A, B, C, D, O, E) and measurements (like 3 cm, 1 cm) on your construction diagram to get full marks. Remember that the length of BD will always represent the square root of the length AB.

 

(ii) \( \sqrt{4.7} \)
Answer:

A B C O E 4.7 cm 1 cm D

**Steps of construction for \( \sqrt{4.7} \):**
1. Draw a straight line and mark point A. Then, mark point B so that the distance AB is 4.7 cm.
2. Mark another point C on this line. Ensure the distance BC is 1 cm.
3. Find the middle point of AC by drawing a perpendicular bisector of AC. Let this midpoint be 'O'.
4. With O as the center and OC (or OA) as the radius, draw a semicircle. This semicircle is key to finding the square root geometrically.
5. Draw a line from point B that is perpendicular to the main line (AB). This perpendicular line should touch the semicircle at point D.
6. Now, the length of BD will be \( \sqrt{4.7} \). You can transfer this length to the number line from B, and the point it reaches will be E, so BE is equal to BD, which is \( \sqrt{4.7} \).
In simple words: The process to represent \( \sqrt{4.7} \) is similar to \( \sqrt{3} \), but you start with a segment AB that is 4.7 cm long instead of 3 cm. The steps for drawing the semicircle and the perpendicular line remain the same to find the square root.

🎯 Exam Tip: When constructing square roots, ensure your line segment AB is accurately measured for the given value. Small errors in initial measurement can lead to inaccuracies in the final square root representation.

 

(iii) \( \sqrt{6.5} \)
Answer:

A B C O E 6.5 cm 1 cm D

**Steps of construction for \( \sqrt{6.5} \):**
1. Draw a straight line and mark point A. Then, mark point B so that the distance AB is 6.5 cm.
2. Mark another point C on this line. Ensure the distance BC is 1 cm.
3. Find the middle point of AC by drawing a perpendicular bisector of AC. Let this midpoint be 'O'.
4. With O as the center and OC (or OA) as the radius, draw a semicircle. This semicircle helps in creating the geometric relationship needed for the square root.
5. Draw a line from point B that is perpendicular to the main line (AB). This perpendicular line should touch the semicircle at point D.
6. Now, the length of BD will be \( \sqrt{6.5} \). You can transfer this length to the number line from B, and the point it reaches will be E, so BE is equal to BD, which is \( \sqrt{6.5} \).
In simple words: For \( \sqrt{6.5} \), you follow the same drawing method as for \( \sqrt{3} \) and \( \sqrt{4.7} \), but you make the first segment AB exactly 6.5 cm long. This setup then naturally helps you find the length of \( \sqrt{6.5} \) on the number line.

🎯 Exam Tip: Always draw construction lines faintly and the final desired representation (e.g., the arc for \( \sqrt{x} \)) more boldly. Use a sharp pencil and proper geometric tools for accuracy.

 

Question 2. Find any two irrational numbers between

(i) 0.3010011000111.... and 0.3020020002....
Answer: Two irrational numbers between the given numbers are 0.301202200222....... and 0.301303300333........ There can be many such numbers, as irrational numbers fill the number line without repeating patterns.
In simple words: Look for numbers that start like the given ones but have digits that keep changing without repeating in a pattern. For example, 0.301 followed by random or increasing non-repeating digits.

🎯 Exam Tip: To find irrational numbers between two given numbers, ensure they are non-terminating and non-repeating. You can achieve this by adding non-repeating digit sequences like '010011000111...' or similar varying patterns.

 

(ii) \( \frac{6}{7} \) and \( \frac{12}{13} \)
Answer: First, convert the fractions to their decimal forms:
\( \frac{6}{7} = 0.\overline{857142} \)
\( \frac{12}{13} = 0.\overline{923076} \)
The two irrational numbers between these are 0.8616611666111........ and 0.8717711777111......... These numbers are non-repeating and non-terminating.
In simple words: Change the fractions into decimals first. Then, pick decimals that fall between them but have endless, mixed-up numbers after the decimal point.

🎯 Exam Tip: When dealing with fractions, always convert them to decimal form first to easily identify the range for finding irrational numbers.

 

(iii) \( \sqrt{2} \) and \( \sqrt{3} \)
Answer: First, find the approximate decimal values of \( \sqrt{2} \) and \( \sqrt{3} \):
\( \sqrt{2} = 1.414 \)
\( \sqrt{3} = 1.732 \)
The two irrational numbers between \( \sqrt{2} \) and \( \sqrt{3} \) are 1.515511555....... and 1.616611666........... Many other such numbers exist in this range.
In simple words: Find out what \( \sqrt{2} \) and \( \sqrt{3} \) are as decimals. Then choose any two decimals between these values that do not repeat in a pattern and go on forever.

🎯 Exam Tip: Remember that between any two distinct rational numbers, there are infinitely many irrational numbers, and vice-versa. Always show at least three decimal places for square roots in such problems.

 

Question 3. Find any two rational numbers between 2.2360679......... and 2.236505500..........
Answer: The two rational numbers are 2.2362 and 2.2363. These numbers terminate, meaning their decimal expansion ends. There are many possible rational answers between the given irrational numbers.
In simple words: Find simple decimal numbers that fit between the two given numbers. Rational numbers can be written as fractions or as decimals that either end or repeat.

🎯 Exam Tip: To find rational numbers between two decimals, simply pick any terminating decimal that lies within the given range. For example, if you need a rational number between 2.2360 and 2.2365, numbers like 2.2361 or 2.2364 are valid.

TN Board Solutions Class 9 Maths Chapter 02 Real Numbers

Students can now access the TN Board Solutions for Chapter 02 Real Numbers prepared by teachers on our website. These solutions cover all questions in exercise in your Class 9 Maths textbook. Each answer is updated based on the current academic session as per the latest TN Board syllabus.

Detailed Explanations for Chapter 02 Real Numbers

Our expert teachers have provided step-by-step explanations for all the difficult questions in the Class 9 Maths chapter. Along with the final answers, we have also explained the concept behind it to help you build stronger understanding of each topic. This will be really helpful for Class 9 students who want to understand both theoretical and practical questions. By studying these TN Board Questions and Answers your basic concepts will improve a lot.

Benefits of using Maths Class 9 Solved Papers

Using our Maths solutions regularly students will be able to improve their logical thinking and problem-solving speed. These Class 9 solutions are a guide for self-study and homework assistance. Along with the chapter-wise solutions, you should also refer to our Revision Notes and Sample Papers for Chapter 02 Real Numbers to get a complete preparation experience.

FAQs

Where can I find the latest Samacheer Kalvi Class 9 Maths Solutions Chapter 2 Real Numbers Exercise 2.3 for the 2026-27 session?

The complete and updated Samacheer Kalvi Class 9 Maths Solutions Chapter 2 Real Numbers Exercise 2.3 is available for free on StudiesToday.com. These solutions for Class 9 Maths are as per latest TN Board curriculum.

Are the Maths TN Board solutions for Class 9 updated for the new 50% competency-based exam pattern?

Yes, our experts have revised the Samacheer Kalvi Class 9 Maths Solutions Chapter 2 Real Numbers Exercise 2.3 as per 2026 exam pattern. All textbook exercises have been solved and have added explanation about how the Maths concepts are applied in case-study and assertion-reasoning questions.

How do these Class 9 TN Board solutions help in scoring 90% plus marks?

Toppers recommend using TN Board language because TN Board marking schemes are strictly based on textbook definitions. Our Samacheer Kalvi Class 9 Maths Solutions Chapter 2 Real Numbers Exercise 2.3 will help students to get full marks in the theory paper.

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Yes, we provide bilingual support for Class 9 Maths. You can access Samacheer Kalvi Class 9 Maths Solutions Chapter 2 Real Numbers Exercise 2.3 in both English and Hindi medium.

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