NCERT solutions Class 9 Maths Chapter 1 Number Systems

NCERT solutions for Class 9 Maths Chapter 1 – Number Systems (updated)

Exercise 1.1 Page: 5

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 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 4/5

4. 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 includes whole numbers as well as negative numbers.

Every whole numbers are rational, however, every rational numbers are not whole numbers.

Exercise 1.2 Page: 8

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 numbers are not irrational numbers.

(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, the 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,

AB2+BC2 = CA2

22+12 = CA2 = 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.

Ncert solution class 9 chapter 1-1

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 OP1 of unit length (see Fig. 1.9). Now draw a line segment P2P3 perpendicular to OP2. Then draw a line segment P3P4 perpendicular to OP3. Continuing in Fig. 1.9 :

Ncert solution class 9 chapter 1-2

Constructing this manner, you can get the line segment Pn-1Pn by square root spiral drawing a line segment of unit length perpendicular to OPn-1. In this manner, you will have created the points P2, P3,….,Pn,… ., and joined them to create a beautiful spiral depicting √2, √3, √4, …

Solution:

Ncert solution class 9 chapter 1-3

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….

Exercise 1.3 Page: 14

1. Write the following in decimal form and say what kind of decimal expansion each has :

(i) 36/100

Solution:

NCERT Solution For Class 9 Maths Ex-1.3-1

= 0.36 (Terminating)

(ii)1/11

Solution:

NCERT Solution For Class 9 Maths Ex-1.3-2
Ncert solution class 9 chapter 1-4
Ncert solution class 9 chapter 1-5

Solution:

Ncert solution class 9 chapter 1-6
NCERT Solution For Class 9 Maths Ex-1.3-3

= 4.125 (Terminating)

(iv) 3/13

Solution:

NCERT Solution For Class 9 Maths Ex-1.3-4
Ncert solution class 9 chapter 1-7

(v) 2/11

Solution:

NCERT Solution For Class 9 Maths Ex-1.3-5

Ncert solution class 9 chapter 1-8

(vi) 329/400

Solution:

NCERT Solution For Class 9 Maths Ex-1.3-6

= 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:

Ncert solution class 9 chapter 1-9

3. Express the following in the form p/q, where p and q are integers and q 0.

(i) 

Solution:

Ncert solution class 9 chapter 1-11

Assume that  x = 0.666…

Then,10x = 6.666…

10x = 6 + x

9x = 6

x = 2/3

(ii) 0.4\overline{7}0.47

Solution:0.4\overline{7} = 0.4777..0.47=0.4777..

= (4/10)+(0.777/10)

Assume that x = 0.777…

Then, 10x = 7.777…

10x = 7 + x

x = 7/9

(4/10)+(0.777../10) = (4/10)+(7/90) ( x = 7/9 and x = 0.777…0.777…/10 = 7/(9×10) = 7/90 )

= (36/90)+(7/90) = 43/90

Ncert solution class 9 chapter 1-14

Solution:

Ncert solution class 9 chapter 1-15

Assume that  x = 0.001001…

Then, 1000x = 1.001001…

1000x = 1 + x

999x = 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,

10x = 9.9999…. Eq. (b)

Eq.(b) – Eq.(a), we get

(10x = 9.9999)-(x = 0.9999…)

9x = 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:

NCERT Solution For Class 9 Maths Ex-1.3-7
Ncert solution class 9 chapter 1-16

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 = 21

7/8 = 0. 875, denominator q =23

4/5 = 0. 8, denominator q = 51

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:

  1. √3 = 1.732050807568
  2. √26 =5.099019513592
  3. √101 = 10.04987562112

8. Find three different irrational numbers between the rational numbers 5/7 and 9/11.

Solution:

Ncert solution class 9 chapter 1-17

Three different irrational numbers are:

  1. 0.73073007300073000073…
  2. 0.75075007300075000075…
  3. 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 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.

Exercise 1.4 Page: 18

1. Visualise 3.765 on the number line, using successive magnification.

Solution:

Ncert solutions class 9 chapter 1-18
Ncert solution class 9 chapter 1-19
Ncert solutions class 9 chapter 1-20

Exercise 1.5 Page: 24

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 ) = 32-(√3)2 = 9-3

= 6

(iii) (√5+√2)2

Solution:

(√5+√2)= √52+(2×√5×√2)+ √22

= 5+2×√10+2 = 7+2√10

(iv) (√5-√2)(√5+√2)

Solution:

(√5-√2)(√5+√2) = (√52-√22) = 5-2 = 3

3. 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,

OD2=BD2+OB2

⟹ (10.3/2)2 = BD2+(8.3/2)2

⟹ BD2 = (10.3/2)2-(8.3/2)2

⟹ (BD)= (10.3/2)-(8.3/2)(10.3/2)+(8.3/2)

⟹ BD2 = 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.

Ncert solutions class 9 chapter 1-21

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)/√72-√6[denominator is obtained by the property, (a+b)(a-b) = a2-b2]

= (√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)/(√52-√22) [denominator is obtained by the property, (a+b)(a-b) = a2-b2]

= (√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)/(√72-22) [denominator is obtained by the property, (a+b)(a-b) = a2-b2]

= (√7+2)/(7-4)

= (√7+2)/3

Exercise 1.6 Page: 26

1. Find:

(i)641/2

Solution:

641/2 = (8×8)1/2

= (82)½

= 81 [⸪2×1/2 = 2/2 =1]

= 8

(ii)321/5

Solution:

321/5 = (25)1/5

= (25)

= 21 [⸪5×1/5 = 1]

= 2

(iii)1251/3

Solution:

(125)1/3 = (5×5×5)1/3

= (53)

= 51 (3×1/3 = 3/3 = 1)

= 5

2. Find:

(i) 93/2

Solution:

93/2 = (3×3)3/2

= (32)3/2

= 33 [⸪2×3/2 = 3]

=27

(ii) 322/5

Solution:

322/5 = (2×2×2×2×2)2/5

= (25)2⁄5

= 22 [⸪5×2/5= 2]

= 4

(iii)163/4

Solution:

163/4 = (2×2×2×2)3/4

= (24)3⁄4

= 23 [⸪4×3/4 = 3]

= 8

(iv) 125-1/3

125-1/3 = (5×5×5)-1/3

= (53)-1⁄3

= 5-1 [⸪3×-1/3 = -1]

= 1/5

3. Simplify:

(i) 22/3×21/5

Solution:

22/3×21/5 = 2(2/3)+(1/5) [⸪Since, am×an=am+n____ Laws of exponents]

= 213/15 [⸪2/3 + 1/5 = (2×5+3×1)/(3×5) = 13/15]

(ii) (1/33)7

Solution:

(1/33)= (3-3)7 [⸪Since,(am)= am x n____ Laws of exponents]

= 3-21

(iii) 111/2/111/4

Solution:

111/2/111/4 = 11(1/2)-(1/4)

= 111/4 [⸪(1/2) – (1/4) = (1×4-2×1)/(2×4) = 4-2)/8 = 2/8 = ¼ ]

(iv) 71/2×81/2

Solution:

71/2×81/2 = (7×8)1/2 [⸪Since, (am×b= (a×b)m ____ Laws of exponents]

= 561/2

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ट्रेन के बीच में ही AC कोच क्यों लगाए जाते हैं? Rbse books for class 1 to 12 hindi medium 2021-22
ट्रेन के बीच में ही AC कोच क्यों लगाए जाते हैं? Rbse books for class 1 to 12 hindi medium 2021-22