NCERT Solutions for Class 9 Maths Chapter 1 Number Systems
Neha Tanna29 Jul, 2024
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NCERT Solutions for Class 9 Maths Chapter 1: NCERT Solutions for Class 9 Maths Chapter 1 Number systems are developed by our highly qualified academics. NCERT Math Solutions aids students in effectively and deftly resolving difficulties in preparation for board exams.
They also put a lot of effort into making maths problems simple enough for pupils to comprehend. The goal of the NCERT Solutions for Class 9 Maths Chapter 1 is to provide students with thorough and methodical explanations for each answer to the questions included in the exercises in this chapter.NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Overview
NCERT Solutions for Class 9 Maths Chapter 1, "Number Systems," explores the various types of numbers and their properties. The chapter begins with a review of the real number system, including rational and irrational numbers. Rational numbers are those that can be expressed as fraction, whereas irrational numbers cannot be written as simple fractions and have non-terminating, non-repeating decimal expansions. The NCERT Solutions for Class 9 Maths Chapter 1 then delves into the decimal representation of real numbers, highlighting how to identify if a number is rational or irrational. It also explains the concept of real numbers on the number line, ensuring students understand how to locate and represent these numbers accurately. Next, the chapter introduces operations on real numbers, including addition, subtraction, multiplication, and division. It emphasizes the importance of these operations in simplifying complex expressions and solving equations. The idea of the rationalization of denominators is also discussed, helping students manage irrational numbers more effectively.NCERT Solutions for Class 9 Maths Chapter 1 PDF
You can find the PDF link for NCERT Solutions for Class 9 Maths Chapter 1 below. This PDF contains detailed solutions and explanations for understanding every concept given in the chapter.NCERT Solutions for Class 9 Maths Chapter 1 PDF
NCERT Solutions for Class 9 Maths Chapter 1 Exercise 1.1
Below we have provided NCERT Solutions for Class 9 Maths Chapter 1 Number Systems -1. Is zero a rational number? Can you write it in the form p/q where p and q are integers and q ≠ 0?
Solution: We know that a number is said to be rational if it can be written in the form p/q, where p and q are integers and q ≠ 0. Taking the case of ‘0’, Zero can be written in the form 0/1, 0/2, 0/3 … as well as, 0/1, 0/2, 0/3 .. Since it satisfies the necessary condition, we can conclude that 0 can be written in the p/q form, where q can either be a positive or negative number. Hence, 0 is a rational number.2. Find six rational numbers between 3 and 4.
Solution: There are infinite rational numbers between 3 and 4. As we have to find 6 rational numbers between 3 and 4, we will multiply both the numbers, 3 and 4, with 6+1 = 7 (or any number greater than 6) i.e., 3 × (7/7) = 21/7 and, 4 × (7/7) = 28/7. The numbers in between 21/7 and 28/7 will be rational and will fall between 3 and 4. Hence, 22/7, 23/7, 24/7, 25/7, 26/7, 27/7 are the 6 rational numbers between 3 and 4.3. Find five rational numbers between 3/5 and 4/5.
Solution: There are infinite rational numbers between 3/5 and 4/5. To find out 5 rational numbers between 3/5 and 4/5, we will multiply both the numbers 3/5 and 4/5 with 5+1=6 (or any number greater than 5) i.e., (3/5) × (6/6) = 18/30 and, (4/5) × (6/6) = 24/30 The numbers in between18/30 and 24/30 will be rational and will fall between 3/5 and 4/5. Hence,19/30, 20/30, 21/30, 22/30, 23/30 are the 5 rational numbers between 3/5 and 4/54. State whether the following statements are true or false. Give reasons for your answers.
(i) Every natural number is a whole number.
Solution:True
Natural numbers- Numbers starting from 1 to infinity (without fractions or decimals) i.e., Natural numbers = 1,2,3,4… Whole numbers – Numbers starting from 0 to infinity (without fractions or decimals) i.e., Whole numbers = 0,1,2,3… Or, we can say that whole numbers have all the elements of natural numbers and zero. Every natural number is a whole number; however, every whole number is not a natural number.(ii) Every integer is a whole number.
Solution:False
Integers- Integers are set of numbers that contain positive, negative and 0; excluding fractional and decimal numbers. i.e., integers= {…-4,-3,-2,-1,0,1,2,3,4…} Whole numbers- Numbers starting from 0 to infinity (without fractions or decimals) i.e., Whole numbers= 0,1,2,3…. Hence, we can say that integers include whole numbers as well as negative numbers. Every whole number is an integer; however, every integer is not a whole number.(iii) Every rational number is a whole number.
Solution:False
Rational numbers- All numbers in the form p/q, where p and q are integers and q≠0. i.e., Rational numbers = 0, 19/30 , 2, 9/-3, -12/7… Whole numbers- Numbers starting from 0 to infinity (without fractions or decimals) i.e., Whole numbers= 0,1,2,3…. Hence, we can say that integers include whole numbers as well as negative numbers. All whole numbers are rational, however, all rational numbers are not whole numbers.NCERT Solutions for Class 9 Maths Chapter 1 Exercise 1.2
1. State whether the following statements are true or false. Justify your answers.
(i) Every irrational number is a real number.
Solution:True
Irrational Numbers – A number is said to be irrational, if it cannot be written in the p/q, where p and q are integers and q ≠ 0. i.e., Irrational numbers = π, e, √3, 5+√2, 6.23146…. , 0.101001001000…. Real numbers – The collection of both rational and irrational numbers are known as real numbers. i.e., Real numbers = √2, √5, , 0.102… Every irrational number is a real number, however, every real number is not an irrational number.(ii) Every point on the number line is of the form √m where m is a natural number.
Solution:False
The statement is false since as per the rule, a negative number cannot be expressed as square roots. E.g., √9 =3 is a natural number. But √2 = 1.414 is not a natural number. Similarly, we know that there are negative numbers on the number line, but when we take the root of a negative number it becomes a complex number and not a natural number. E.g., √-7 = 7i, where i = √-1 The statement that every point on the number line is of the form √m, where m is a natural number is false.(iii) Every real number is an irrational number.
Solution:False
The statement is false. Real numbers include both irrational and rational numbers. Therefore, every real number cannot be an irrational number. Real numbers – The collection of both rational and irrational numbers are known as real numbers. i.e., Real numbers = √2, √5, , 0.102… Irrational Numbers – A number is said to be irrational, if it cannot be written in the p/q, where p and q are integers and q ≠ 0. i.e., Irrational numbers = π, e, √3, 5+√2, 6.23146…. , 0.101001001000…. Every irrational number is a real number, however, every real number is not irrational.2. Are the square roots of all positive integers irrational? If not, give an example of the square root of a number that is a rational number.
Solution: No, the square roots of all positive integers are not irrational. For example, √4 = 2 is rational. √9 = 3 is rational. Hence, the square roots of positive integers 4 and 9 are not irrational. ( 2 and 3, respectively).3. Show how √5 can be represented on the number line.
Solution: Step 1: Let line AB be of 2 unit on a number line. Step 2: At B, draw a perpendicular line BC of length 1 unit. Step 3: Join CA Step 4: Now, ABC is a right angled triangle. Applying Pythagoras theorem, AB 2 +BC 2 = CA 2 2 2 +1 2 = CA 2 = 5 ⇒ CA = √5 . Thus, CA is a line of length √5 unit. Step 4: Taking CA as a radius and A as a center draw an arc touching the number line. The point at which number line get intersected by arc is at √5 distance from 0 because it is a radius of the circle whose center was A. Thus, √5 is represented on the number line as shown in the figure.
4. Classroom activity (Constructing the ‘square root spiral’) : Take a large sheet of paper and construct the ‘square root spiral’ in the following fashion. Start with a point O and draw a line segment OP1 of unit length. Draw a line segment P1P2 perpendicular to OP 1 of unit length (see Fig. 1.9). Now draw a line segment P 2 P 3 perpendicular to OP 2 . Then draw a line segment P 3 P 4 perpendicular to OP 3 . Continuing in Fig. 1.9 :
Constructing this manner, you can get the line segment P n-1 Pn by square root spiral drawing a line segment of unit length perpendicular to OP n-1 . In this manner, you will have created the points P 2 , P 3 ,….,Pn,… ., and joined them to create a beautiful spiral depicting √2, √3, √4, …
Solution:
Step 1: Mark a point O on the paper. Here, O will be the center of the square root spiral.
Step 2: From O, draw a straight line, OA, of 1cm horizontally.
Step 3: From A, draw a perpendicular line, AB, of 1 cm.
Step 4: Join OB. Here, OB will be of √2
Step 5: Now, from B, draw a perpendicular line of 1 cm and mark the end point C.
Step 6: Join OC. Here, OC will be of √3
Step 7: Repeat the steps to draw √4, √5, √6….
NCERT Solutions for Class 9 Maths Chapter 1 Exercise 1.3
1. Write the following in decimal form and say what kind of decimal expansion each has :
(i) 36/100 Solution:
= 0.36 (Terminating)
(ii)1/11
Solution:
Solution:
= 4.125 (Terminating)
(iv) 3/13
Solution:
(v) 2/11
Solution:
= 0.8225 (Terminating)
2. You know that 1/7 = 0.142857. Can you predict what the decimal expansions of 2/7, 3/7, 4/7, 5/7, 6/7 are, without actually doing the long division? If so, how?
[Hint: Study the remainders while finding the value of 1/7 carefully.]
Solution:
3. Express the following in the form p/q, where p and q are integers and q 0.
(i)
Solution:
Assume that
x
= 0.666…
Then,10
x
= 6.666…
10
x
= 6 +
x
9
x
= 6
x
= 2/3
(ii)
Solution:
Assume that
x
= 0.001001…
Then, 1000
x
= 1.001001…
1000
x
= 1 +
x
999
x
= 1
x
= 1/999
4. Express 0.99999…. in the form p/q . Are you surprised by your answer? With your teacher and classmates discuss why the answer makes sense.
Solution: Assume that x = 0.9999…..Eq (a) Multiplying both sides by 10, 10 x = 9.9999…. Eq. (b) Eq.(b) – Eq.(a), we get 10 x = 9.9999 – x = -0.9999… _____________ 9 x = 9 x = 1 The difference between 1 and 0.999999 is 0.000001 which is negligible. Hence, we can conclude that, 0.999 is too much near 1, therefore, 1 as the answer can be justified.5. What can the maximum number of digits be in the repeating block of digits in the decimal expansion of 1/17 ? Perform the division to check your answer.
Solution: 1/17 Dividing 1 by 17:
There are 16 digits in the repeating block of the decimal expansion of 1/17.
6. Look at several examples of rational numbers in the form p/q (q ≠ 0), where p and q are integers with no common factors other than 1 and having terminating decimal representations (expansions). Can you guess what property q must satisfy?
Solution: We observe that when q is 2, 4, 5, 8, 10… Then the decimal expansion is terminating. For example: 1/2 = 0. 5, denominator q = 2 1 7/8 = 0. 875, denominator q =2 3 4/5 = 0. 8, denominator q = 5 1 We can observe that the terminating decimal may be obtained in the situation where prime factorization of the denominator of the given fractions has the power of only 2 or only 5 or both.7. Write three numbers whose decimal expansions are non-terminating non-recurring.
Solution: We know that all irrational numbers are non-terminating non-recurring. three numbers with decimal expansions that are non-terminating non-recurring are:- √3 = 1.732050807568
- √26 =5.099019513592
- √101 = 10.04987562112
8. Find three different irrational numbers between the rational numbers 5/7 and 9/11.
Solution:
Three different irrational numbers are:
- 0.73073007300073000073…
- 0.75075007300075000075…
- 0.76076007600076000076…
9. Classify the following numbers as rational or irrational according to their type:
(i)√23 Solution: √23 = 4.79583152331… Since the number is non-terminating and non-recurring therefore, it is an irrational number. (ii)√225 Solution: √225 = 15 = 15/1 Since the number can be represented in p/q form, it is a rational number.(iii) 0.3796
Solution: Since the number,0.3796, is terminating, it is a rational number.(iv) 7.478478
Solution: The number,7.478478, is non-terminating but recurring, it is a rational number.(v) 1.101001000100001…
Solution: Since the number,1.101001000100001…, is non-terminating non-repeating (non-recurring), it is an irrational number.NCERT Solutions for Class 9 Maths Chapter 1 Exercise 1.4
1. Visualise 3.765 on the number line, using successive magnification.
Solution:
NCERT Solutions for Class 9 Maths Chapter 1 Exercise 1.5
1. Classify the following numbers as rational or irrational:
(i) 2 –√5
Solution: We know that, √5 = 2.2360679… Here, 2.2360679…is non-terminating and non-recurring. Now, substituting the value of √5 in 2 –√5, we get, 2-√5 = 2-2.2360679… = -0.2360679 Since the number, – 0.2360679…, is non-terminating non-recurring, 2 –√5 is an irrational number.(ii) (3 +√23)- √23
Solution: (3 + √ 23) –√23 = 3+ √ 23–√23 = 3 = 3/1 Since the number 3/1 is in p/q form, ( 3 +√23)- √23 is rational.(iii) 2√7/7√7
Solution: 2√7/7√7 = ( 2/7)× (√7/√7) We know that (√7/√7) = 1 Hence, ( 2/7)× (√7/√7) = (2/7)×1 = 2/7 Since the number, 2/7 is in p/q form, 2√7/7√7 is rational.(iv) 1/√2
Solution: Multiplying and dividing numerator and denominator by √2 we get, (1/√2) ×(√2/√2)= √2/2 ( since √2×√2 = 2) We know that, √2 = 1.4142… Then, √2/2 = 1.4142/2 = 0.7071.. Since the number , 0.7071..is non-terminating non-recurring, 1/√2 is an irrational number.(v) 2
Solution: We know that, the value of = 3.1415 Hence, 2 = 2×3.1415.. = 6.2830… Since the number, 6.2830…, is non-terminating non-recurring, 2 is an irrational number.2. Simplify each of the following expressions:
(i) (3+√3)(2+√2)
Solution: (3+√3)(2+√2 ) Opening the brackets, we get, (3×2)+(3×√2)+(√3×2)+(√3×√2) = 6+3√2+2√3+√6(ii) (3+√3)(3-√3 )
Solution: (3+√3)(3-√3 ) = 3 2 -(√3) 2 = 9-3 = 6(iii) (√5+√2) 2
Solution: (√5+√2) 2 = √5 2 +(2×√5×√2)+ √2 2 = 5+2×√10+2 = 7+2√10(iv) (√5-√2)(√5+√2)
Solution: (√5-√2)(√5+√2) = (√5 2 -√2 2 ) = 5-2 = 33. Recall, π is defined as the ratio of the circumference (say c) of a circle to its diameter, (say d). That is, π =c/d. This seems to contradict the fact that π is irrational. How will you resolve this contradiction?
Solution: There is no contradiction. When we measure a value with a scale, we only obtain an approximate value. We never obtain an exact value. Therefore, we may not realize whether c or d is irrational. The value of π is almost equal to 22/7 or 3.142857…4. Represent (√9.3) on the number line.
Solution: Step 1: Draw a 9.3 units long line segment, AB. Extend AB to C such that BC=1 unit. Step 2: Now, AC = 10.3 units. Let the centre of AC be O. Step 3: Draw a semi-circle of radius OC with centre O. Step 4: Draw a BD perpendicular to AC at point B intersecting the semicircle at D. Join OD. Step 5: OBD, obtained, is a right angled triangle. Here, OD 10.3/2 (radius of semi-circle), OC = 10.3/2 , BC = 1 OB = OC – BC ⟹ (10.3/2)-1 = 8.3/2 Using Pythagoras theorem, We get, OD 2 =BD 2 +OB 2 ⟹ (10.3/2) 2 = BD 2 +(8.3/2) 2 ⟹ BD 2 = (10.3/2) 2 -(8.3/2) 2 ⟹ (BD) 2 = (10.3/2)-(8.3/2)(10.3/2)+(8.3/2) ⟹ BD 2 = 9.3 ⟹ BD = √9.3 Thus, the length of BD is √9.3. Step 6: Taking BD as radius and B as centre draw an arc which touches the line segment. The point where it touches the line segment is at a distance of √9.3 from O as shown in the figure.
5. Rationalize the denominators of the following:
(i) 1/√7
Solution: Multiply and divide 1/√7 by √7 (1×√7)/(√7×√7) = √7/7(ii) 1/(√7-√6)
Solution: Multiply and divide 1/(√7-√6) by (√7+√6) [1/(√7-√6)]×(√7+√6)/(√7+√6) = (√7+√6)/(√7-√6)(√7+√6) = (√7+√6)/√7 2 -√6 2 [denominator is obtained by the property, (a+b)(a-b) = a 2 -b 2 ] = (√7+√6)/(7-6) = (√7+√6)/1 = √7+√6(iii) 1/(√5+√2)
Solution: Multiply and divide 1/(√5+√2) by (√5-√2) [1/(√5+√2)]×(√5-√2)/(√5-√2) = (√5-√2)/(√5+√2)(√5-√2) = (√5-√2)/(√5 2 -√2 2 ) [denominator is obtained by the property, (a+b)(a-b) = a 2 -b 2 ] = (√5-√2)/(5-2) = (√5-√2)/3(iv) 1/(√7-2)
Solution: Multiply and divide 1/(√7-2) by (√7+2) 1/(√7-2)×(√7+2)/(√7+2) = (√7+2)/(√7-2)(√7+2) = (√7+2)/(√7 2 -2 2 ) [denominator is obtained by the property, (a+b)(a-b) = a 2 -b 2 ] = (√7+2)/(7-4) = (√7+2)/3NCERT Solutions for Class 9 Maths Chapter 1 Exercise 1.6
1. Find:
(i)64 1/2
Solution: 64 1/2 = (8×8) 1/2 = (8 2 ) ½ = 8 1 [⸪2×1/2 = 2/2 =1] = 8(ii)32 1/5
Solution: 32 1/5 = (2 5 ) 1/5 = (2 5 ) ⅕ = 2 1 [⸪5×1/5 = 1] = 2(iii)125 1/3
Solution: (125) 1/3 = (5×5×5) 1/3 = (5 3 ) ⅓ = 5 1 (3×1/3 = 3/3 = 1) = 52. Find:
(i) 9 3/2
Solution: 9 3/2 = (3×3) 3/2 = (3 2 ) 3/2 = 3 3 [⸪2×3/2 = 3] =27(ii) 32 2/5
Solution: 32 2/5 = (2×2×2×2×2) 2/5 = (2 5 ) 2⁄5 = 2 2 [⸪5×2/5= 2] = 4(iii)16 3/4
Solution: 16 3/4 = (2×2×2×2) 3/4 = (2 4 ) 3⁄4 = 2 3 [⸪4×3/4 = 3] = 8(iv) 125 -1/3
125 -1/3 = (5×5×5) -1/3 = (5 3 ) -1⁄3 = 5 -1 [⸪3×-1/3 = -1] = 1/53. Simplify :
(i) 2 2/3 ×2 1/5
Solution: 2 2/3 ×2 1/5 = 2 (2/3)+(1/5) [⸪Since, a m ×a n =a m+n ____ Laws of exponents] = 2 13/15 [⸪2/3 + 1/5 = (2×5+3×1)/(3×5) = 13/15](ii) (1/3 3 ) 7
Solution: (1/3 3 ) 7 = (3 -3 ) 7 [⸪Since,(a m ) n = a m x n ____ Laws of exponents] = 3 -21(iii) 11 1/2 /11 1/4
Solution: 11 1/2 /11 1/4 = 11 (1/2)-(1/4) = 11 1/4 [⸪(1/2) – (1/4) = (1×4-2×1)/(2×4) = 4-2)/8 = 2/8 = ¼ ](iv) 7 1/2 ×8 1/2
Solution: 7 1/2 ×8 1/2 = (7×8) 1/2 [⸪Since, (a m ×b m = (a×b) m ____ Laws of exponents] = 56 1/2Benefits of NCERT Solutions for Class 9 Maths Chapter 1
NCERT Solutions for Class 9 Maths Chapter 1 "Number Systems" offer several benefits to students:Conceptual Clarity : These solutions help in understanding the fundamental concepts of number systems, including rational and irrational numbers, their properties, and how to represent them on the number line. This clarity is crucial for building a strong mathematical foundation.
Step-by-Step Solutions : The solutions provide detailed, step-by-step explanations for each problem, making it easier for students to follow along and understand the logical progression of mathematical reasoning and problem-solving techniques.
Exam Preparation : By practicing these solutions, students become familiar with the types of questions that can appear in exams, the format of the answers, and the method of presenting solutions, which can help in scoring higher marks.
Homework Assistance : These solutions can be a valuable resource for completing homework assignments accurately and efficiently, ensuring that students reinforce their classroom learning and practice consistently.
Problem-Solving Skills : Working through various problems in the solutions helps in enhancing analytical and problem-solving skills. Students learn different approaches to tackle a problem, which can be applied to other mathematical topics as well.
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