NCERT Solution for Class 9 Maths Chapter-1 Number System

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.

 

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.

 

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 number 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 between 18/30 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 ⅘

 

(i) 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) 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) 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.

 

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

 2*2+1*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.

(i) 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, π, 0.102… 

Every irrational number is a real number, however, every real number is not an irrational number. 

(ii) 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) 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.

 

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). 

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 1 cm 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….

 

√2 = 1.414213562 ………..
√3 = 1.732050808 …….
√5 = 2.23606797 …….

 

(i) We have,

36/100= 0.36

Thus, the decimal expansion of 36/100 is terminating.

(ii) Dividing 1 by 11, we have

Thus, the decimal expansion of 1/11 is non-terminating repeating.

Dividing 33 by 8, we get

 = 4.125. Thus, the answer is terminating.

(iv)3/13

Dividing 3 by 13, we get

Here, the repeating block of digits is 230769
∴ 3/13 = 0.23076923…
Thus, the decimal expansion of 3/13 is non-terminating repeating.

(v) 2/11

Dividing 2 by 11, we get

Here, the repeating block of digits is 18.
∴ 2/11 = 0.1818… = 0.18

Thus, the decimal expansion of 2/11 is non-terminating repeating.

(vi)329/400 

Dividing 329 by 400, we get

 ∴ 329/400 = 0.8225. Thus, the decimal expansion of 329/400is terminating.

 

Solution:  Assume that x = 0.666… 

Then ,10x = 6.666… 

10x = 6 + x

 9x = 6 

x = ⅔

= (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 

Assume that x = 0.001001… 

Then, 1000x = 1.001001… 

1000x = 1 + x 

999x = 1

 x = 1/999

 

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 very near 1, therefore, 1 as the answer can be justified.

 

1/17 

Dividing 1 by 17:

Dividing 1 by 17, we have

The remainder I is the same digit from which we started the division.

∴ 1/17= 0.0588235294117647

Thus, there are 16 digits in the repeating block in the decimal expansion of 1/17.

Hence, our answer is verified.

 

Let us look decimal expansion of the following terminating rational numbers:

We observe that the prime factorisation of q (i.e. denominator) has only powers of 2 or powers of 5 or powers of both.

 

We have,

Three irrational numbers between 0.714285 and 0.81 are
(i) 0.750750075000 …..
(ii) 0.767076700767000 ……
(iii) 0.78080078008000 ……

 

(i) ∵ 23 is not a perfect square.
∴ √23  is an irrational number.

(ii) ∵ 225 = 15 x 15 = 152
∴ 225 is a perfect square.
Thus, √225 is a rational number.

(iii) ∵ 0.3796 is a terminating decimal.
∴ It is a rational number.

(iv) 7.478478 

The number,7.478478, is non-terminating but recurring

∴ It is a rational number.(v) Since, 1.101001000100001… is a non terminating, non-repeating decimal number.
∴ It is an irrational number.

 

4.2626 lies between 4 and 5.

3.765 lies between 3 and 4.

 

(i) Since, it is a difference of a rational and an irrational number.
∴ 2 – √5 is an irrational number.

(ii) 3 +  √23- √23

(3 +√23) –√23 = 3+√23–√23

= 3

= 3/1 

Since the number 3/1 is in p/q form, (3 +√23)- √23 is rational
which is a rational number.

(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.  , which is a rational number.

(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(v)

 ∵ 2π = 2 x π = Product of a rational and an irrational number is an irrational number.
∴ 2π is an irrational number.

 

(i) (3 + √3)(2 + √2)
= 2(3 + √3) + √2(3 + √3)
= 6 + 2√3 + 3√2 + √6
Thus, (3 + √3)(2 + √2) = 6 + 2√3 + 3√2 + √6

(ii) (3 + √3)(3 – √3) = (3)2 – (√3)2
= 9 – 3 = 6
Thus, (3 + √3)(3 – √3) = 6

(iii) (√5 + √2)2 = (√5)2 + (√2)2 + 2(√5)(√2)
= 5 + 2 + 2√10 = 7 + 2√10
Thus, (√5 + √2 )2 = 7 + 2√10

(iv) (√5 – √2)(√5 + √2) = (√5)2 – (√2)2 = 5 – 2 = 3
Thus, (√5 – √2) (√5 + √2) = 3

 

When we measure the length of a line with a scale or with any other device, we only get an approximate ational value, i.e. c and d both are irrational.
∴ c/d is irrational and hence π is irrational.
Thus, there is no contradiction in saying that it is irrational.

 

Draw a line segment AB = 9.3 units and extend it to C such that BC = 1 unit.
Find mid point of AC and mark it as O.
Draw a semicircle taking O as centre and AO as radius. Draw BD ⊥ AC.
Draw an arc taking B as centre and BD as radius meeting AC produced at E such that  BE = BD = √9.3 units.

(i) 64 = 8 x 8 = 82
∴ (64)1/2 = (82)1/2 = 82 x 1/2 = 8 [(am)n = am x n]
(ii) 32 = 2 x 2x 2 x 2 x 2 = 25
∴ (32)1/5= (25)1/5= 25 x 1/5 = 2 [(am)n= am x n]
(iii) 125 = 5 x 5 x 5 = 53
∴ (125)1/3 = (53)1/3 = 53 x 1/3 = 5 [(am)n = am x n]

 

(i) 9 = 3 x 3 = 32
∴ (9)3/2= (32)3/2 = 32 x 3/2 = 33 = 27
[(am)n = amn]
(ii) 32 = 2 x 2 x 2 x 2 x 2= 25
∴ (32)2/5 = (25)2/5 = 25 x 2/5 = 22 = 4
[(am)n = amn]
(iii) 16 = 2 x 2 x 2 x 2 = 24
∴ (16)3/4 = (24)3/4 = 24 x 3/4 = 23 = 8
[(am)n = amn]
(iv) 125 = 5 x 5 x 5 = 53
∴ (125)-1/3 = (53)-1/3 = 53 x (-1/3) = 5-1
=1/5

 

(i) 22/3. 21/5 = 22/3 + 1/5 = 213/15

[am. an = am + n

(iv) 71/2.81/2 = (7 x 8)1/2 = (56)1/2 [am x bm = (ab)m]