Grade 10 NCERT Math Solution: Chapter 2 - Polynomials

Exercise 2.1

1. The graphs y = p(x) are given in the following figure, for some polynomials p(x). Find the number of zeros of p(x), in each case.

Solution:

(i)

The number of zeros is 0 as the graph does not intersect the x-axis at any point.

(ii)

The number of zeros is 1 as the graph intersects x-axis at only one point.

(iii)

The number of zeros is 3 as the graph intersects x-axis at 3 points.

(iv)

The number of zeros is 2 as the graph intersects x-axis at 2 points.

(v)

The number of zeros is 4 as the graph intersects x-axis at 4 points.

(vi)

The number of zeros is 3 as the graph intersects x-axis at 3 points.


Exercise 2.2:


1. Find the zeros of the following quadratic polynomials and verify the relationship between the zeros and coefficients.

(i) x² ˗ 2x ˗ 8

(ii) 4s² ˗ 4s + 1

(iii) 6x² ˗ 3 ˗ 7x

(iv) 4u² + 8u

(v) t² ˗ 15

(vi) 3x² ˗ x ˗ 4


Solution:

(i)

x² ˗ 2x ˗ 8 = x² ˗ 4x + 2x ˗ 8                          [ ˗ 4 + 2 = ˗ 2 and ˗ 4 × 2 = ˗ 8 ]
                = x(x ˗ 4) + 2(x ˗ 4)
                = (x ˗ 4)(x + 2)

To find the zero, equate the polynomial to 0.

(x ˗ 4)(x + 2) = 0

⇒ x = 4 and x = ˗ 2.

Hence the zeros of x² ˗ 2x ˗ 8 is 4 and ˗ 2.


Verifying the relationship between zeros and coefficients.

(ii)

4s² ˗ 4s + 1 = ( 2s ˗ 1 )²
           
To find the zero, equate the polynomial to 0.

( 2s ˗ 1 )² = 0

⟹ ( 2s ˗ 1 )( 2s ˗ 1 ) = 0

Verifying the relationship between zeros and coefficients.

(iii)

6x² ˗ 3 ˗ 7x = 6x² ˗ 7x ˗ 3
                   = 6x² + 2x ˗ 9x ˗ 3                         [ 2 ˗ 9 = ˗7 and 2 × ( ˗ 9 ) = ˗ 18 ]
                   = 2x(3x + 1) ˗ 3(3x + 1)
                   = (3x + 1)(2x ˗ 3)
                 
       
To find the zero, equate the polynomial to 0.

Verifying the relationship between zeros and coefficients.

(iv)

4u² + 8u = 4u(u + 2 )

To find the zero, equate the polynomial to 0.

4u(u + 2 ) = 0

4u = 0 and  u + 2 = 0

u = 0  and u = ˗2

Verifying the relationship between zeros and coefficients.

(v)

t² ˗ 15 = ( t + √15 ) ( t ˗ √15 )

To find the zero, equate the polynomial to 0.

( t + √15 ) ( t ˗ √15 ) = 0

t + √15 = 0  and  t ˗ √15 = 0

t = ˗√15   and  t =  √15

Verifying the relationship between zeros and coefficients.

(vi)

3x² ˗ x ˗ 4 = 3x² + 3x ˗ 4x ˗ 4
                 = 3x( x + 1 ) ˗4( x + 1 )
                 = ( x + 1 ) ( 3x ˗ 4 )

To find the zero, equate the polynomial to 0.

( x + 1 ) ( 3x ˗ 4 ) = 0

x + 1 = 0  and  3x ˗ 4 = 0

x = ˗1  and  3x = 4

Verifying the relationship between zeros and coefficients.

2. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.

Solution:

(i)

Let the polynomial be ax² + bx + c and its zeroes be ⍺ and β.

∴ a = 4 , b = ˗1 and c = ˗4.

Hence the quadratic polynomial is 4x² ˗ x ˗ 4.


(ii)

Let the polynomial be ax² + bx + c and its zeroes be ⍺ and β.

∴ a = 3, b = ˗3√2  and c = 1.

Hence the quadratic polynomial is 3x² ˗ 3√2x + 1.


(iii)

Let the polynomial be ax² + bx + c and its zeroes be ⍺ and β.

∴ a = 1 , b = 0  and c = √5.

Hence the quadratic polynomial is x²  + √5.


(iv)

Let the polynomial be ax² + bx + c and its zeroes be ⍺ and β.

∴ a = 1, b = ˗1  and c = 1

Hence the quadratic polynomial is x² ˗ x + 1.


(v)

Let the polynomial be ax² + bx + c and its zeroes be ⍺ and β.

∴ a = 4, b = 1  and c = 1.

Hence the quadratic polynomial is 4x² + x + 1.


(vi)

Let the polynomial be ax² + bx + c and its zeroes be ⍺ and β.

∴ a = 1, b = ˗4  and c = 1.

Hence the quadratic polynomial is x² ˗ 4x + 1.


Exercise 2.3:


1. Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following.

(i)

p(x) = x³ ˗ 3x² + 5x ˗ 3, g(x) = x² ˗ 2.

Solution:

∴ Quotient = x ˗ 3  and Remainder = 7x ˗ 9.


(ii)

p(x) = x⁴ ˗ 3x² + 4x + 5, g(x) = x² + 1 ˗ x               

Solution:

∴ Quotient = x² + x ˗ 3  and Remainder = 8


(iii)

p(x) = x⁴ ˗ 5x + 6, g(x) = 2 ˗ x²

Solution:

∴ Quotient = ˗x² ˗2  and Remainder = ˗5x + 10


2. Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial.

(i) t² ˗ 3, 2t⁴ + 3t³ ˗ 2t² ˗ 9t ˗ 12

Solution:

The remainder is 0 while dividing 2t⁴ + 3t³ ˗ 2t² ˗ 9t ˗ 12 by t² ˗ 3.

∴ t² ˗ 3 is a factor of 2t⁴ + 3t³ ˗ 2t² ˗ 9t ˗ 12.


(ii) x² + 3x + 1, 3x⁴ + 5x³ – 7x² + 2x + 2

Solution:

The remainder is 0 while dividing 3x⁴ + 5x³ – 7x² + 2x + 2 by x² + 3x + 1.

Hence x² + 3x + 1 is a factor of 3x⁴ + 5x³ – 7x² + 2x + 2.


(iii) x³ – 3x + 1, x⁵ – 4x³ + x² + 3x + 1

Solution:

The remainder is 0 while dividing x⁵ – 4x³ + x² + 3x + 1 by x³ – 3x + 1.

Hence x³ – 3x + 1 is a factor of x⁵ – 4x³ + x² + 3x + 1.


3. Obtain all other zeroes of 3x⁴ + 6x³ – 2x² – 10x – 5, if two of its zeroes are √(5/3)
and –√(5/3).

Solution:

p(x) = 3x⁴ + 6x³ – 2x² – 10x – 5
Since the two zeroes are √(5/3) and –√(5/3).

To find the other zeros we equate (x + 1)² to 0.

(x + 1)² = 0

(x + 1)(x + 1) = 0

x = ˗1 and ˗1.

Hence the zeroes of the given polynomial are √(5/3), –√(5/3), ˗1 and ˗1.


4. On dividing x³ ˗ 3x² + x + 2 by a polynomial g(x), the quotient and remainder were x ˗ 2 and
˗2x + 4, respectively. Find g(x).

Solution:

Dividend = x³ ˗ 3x² + x + 2

Divisor = g(x) = ?

Quotient = x ˗ 2

Remainder = ˗2x + 4

We know that

Dividend = Quotient × Divisor + Remainder

x³ ˗ 3x² + x + 2 = (x ˗ 2) × g(x) + (˗2x + 4 )

(x ˗ 2) × g(x) = ( x³ ˗ 3x² + x + 2 ) ˗ (˗2x + 4 )

                      =   x³ ˗ 3x² + x + 2 + 2x ˗ 4

                      =   x³ ˗ 3x² + 3x ˗ 2

∴ g(x) = x² ˗ x + 1


5. Give examples of polynomial p(x), g(x), q(x) and r(x), which satisfy the division algorithm and
(i) deg p(x) = deg q(x)
(ii) deg q(x) = deg r(x)
(iii) deg r(x) = 0

Solution:

(i)

Let us assume the division of 6x² + 2x + 2 by 2

Here, p(x) = 6x² + 2x + 2
g(x) = 2
q(x) = 3x² + x + 1
r(x) = 0
Degree of p(x) and q(x) is same i.e. 2.
Checking for division algorithm,
p(x) = g(x) × q(x) + r(x)
Or, 6x² + 2x + 2 = 2 × (3x² + x + 1) + 0
                           = 6x² + 2x + 2
Hence, division algorithm is satisfied.

(ii)

Let us assume the division of x³ + x by x²,

Here, p(x) = x³ + x
g(x) = x²
q(x) = x and r(x) = x
Clearly, the degree of q(x) and r(x) is the same i.e., 1.
Checking for division algorithm,
p(x) = g(x) × q(x) + r(x)
x³ + x = (x² ) × x + x
          = x³ + x
Thus, the division algorithm is satisfied.


(iii)

Let us assume the division of x³ + 1 by x².

Here, p(x) = x³ + 1
g(x) = x²
q(x) = x and r(x) = 1
Clearly, the degree of r(x) is 0.
Checking for division algorithm,
p(x) = g(x) × q(x) + r(x)
x³ + 1 = (x² ) × x + 1
          = x³ + 1
Thus, the division algorithm is satisfied.

Grade 10 NCERT Math Solution: Chapter 1 - Real Numbers

Exercise 1.1:

1. Use Euclid's division algorithm to find the HCF of

(i) 135 and 225

(ii) 196 and 38220

(iii) 867 and 255

Solution:

(i)

Since 225 > 135 we can apply Euclid's division algorithm.         

225 = 135 × 1 + 90

135 > 90, so we apply Euclid's division algorithm again.

135 = 90 × 1 + 45

90 > 45, so we apply Euclid's division algorithm again.

90 = 45 × 2 + 0

At this stage, the remainder is 0.

So the divisor 45 is the HCF.


(ii)

38220 > 196, so we apply Euclid's division algorithm to find HCF.

38220 = 196 × 195 + 0

The remainder is 0.

Hence the divisor 196 is the HCF.


(iii)

867 > 255, so we apply Euclid's division algorithm to find HCF.

867 = 255 × 3 + 102

255 > 102 so we apply Euclid's division algorithm again.

255 = 102 × 2 + 51

102 > 51 so we apply Euclid's division algorithm again.

102 = 51 × 2 + 0

The remainder is 0.

Hence the divisor 51 is the HCF.




2. Show that any positive odd integer is of the form 6q+1 or 6q+3 or  6q+5, where q is some integer.

Solution:

Let a be any positive integer and b = 6.

By Euclid's division algorithm,

a = 6q + 6, where q is more than or equal to 0 .

The remainder, r = 0,1,2,3,4,5 because 0 ≤ r < b and the value of b is 6.

Hence the total possible forms are 6q+0, 6q+1, 6q+2, 6q+3, 6q+4, and 6q+5 .

6q+0
6 is divisible by 2, so it is an even number.

6q is an even number and 6q+1, 6q+3, 6q+5 leave the remainder 1, 3 and 5 respectively. So 6q+1, 6q+3 and 6q+5 are odd numbers.

6q+2 and 6q+4 leave the remainder 2 and 4 respectively. So 6q+2 and 6q+ 4 are even numbers.

Hence any positive odd integer is of the form 6q+1 or 6q+3 or 6q+5.


3. An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?

Solution:

HCF of 616 and 32 will give the maximum number of columns in which they can march.

616 > 32, so we apply Euclid's division algorithm to find the HCF.

616 = 32 × 19 + 8

32 > 8, so we apply Euclid's division algorithm again.

32 = 8 × 4 + 0

The remainder is 0, so the divisor 8 is the HCF of 616 and 32.

∴ The army contingent can march in 8 columns.


4. Use Euclid's division lemma to show that the square of any positive integer is either of form 3m or 3m+1 for some integer m.
[ Hint: Let x be any positive integer then it is of the form 3q, 3q+1 or 3q+2. Now square each of these and show that they can be rewritten in the form 3m or 3m+1.]

Solution:

Let 'a' be any positive integer and b = 3.

Then a = bq+r for some integer q ≥ 0.

r = 0, 1, 2 because 0 ≤ r < 3.

Hence a = 3q or 3q+1 or 3q+2

Now,
Form 1:

a = 3q
Squaring both sides,
a² = (3q)²
    = 9q²
    = 3(3q²)
    = 3k₁   where k₁ = 3q² and q is some integer.

Form 2:

a  = 3q+1
Squaring both sides,
a² = (3q+1)²                                   [ (a+b)² = a² + 2ab + b² ]
    = 9q² + 6q + 1
    = 3(3q² + 2q) + 1
    = 3k₂+1, where k₂ = (3q² + 2q) and q is some integer.

Form 3:

a  = 3q+2
Squaring both sides,
a² = (3q+2)²
    = 9q² + 12q + 4
    = 9q² + 12q + 3 + 1
    = 3(3q² + 4q + 1) + 1
    = 3k₃+1, where k₃ = (3q² + 4q + 1) and q is some integer.

Hence the square of any positive integer is either of the form 3m or 3m+1.


5. Use Euclid's division lemma to show that the cube of any positive integer is of the form 9m, 9m+1 or 9m+8.

Solution:

Let 'a' be any positive integer and b = 3.

a = 3q+r, where q ≥ 0 and

r = 0, 1, 2 because 0 ≤ r < 3.

∴ a = 3q or 3q+1 or 3q+2

Form 1:

a = 3q
Cubing both sides,
a³ = (3q)³
    = 29q³
    = 9(3q³)
    = 9m, where m is an integer such that m = 3q³

Form 2:

a = 3q+1
Cubing both sides
a³ = (3q+1)³
    = (3q)³ + 1³ + 3(3q)²(1) + 3(3q)(1²)              [ (a+b)³ = a³ + b³ + 3a²b + 3ab² ] 
    = 27q³ + 1 + 3(9q²) + 9q
    = 27q³ + 27q² + 9q + 1
    = 9(3q³ + 3q² + q) + 1
    = 9m₁+1, where m₁ is an integer such that m₁ = 3q³ + 3q² + q

Form 3:

a = 3q + 2
Cubing both sides,
a³ = (3q+2)³
    = (3q)³ + 2³ + 3(3q)²(2) + 3(3q)(2²)
    = 27q³ + 8 + 6(9q²) + 36q
    = 27q³ + 54q² + 36q + 8
    = 9(3q³ + 6q² + 4q) + 8
    = 9m₂+8, where m₂ is an integer such that m₂ = 3q³ + 6q² + 4q

∴ Cube of positive integer is of the form 9m or 9m+1 or 9m+8.



Exercise 1.2:

1. Express each number as product of its prime factors.

(i) 140

(ii) 156

(iii) 3825

(iv) 5005

(v) 7429

Solution:

(i)

∴ 140 = 2 × 2 × 5 × 7  or
          = 2² × 5 × 7


(ii)

∴ 156 = 2 × 2 × 3 × 13  or
          = 2² × 3 × 13


(iii)

∴ 3825 = 5 × 5 × 3 × 3 × 17  or
            = 5² × 3² × 17


(iv)

             






∴ 5005 =  5 × 7 × 11 × 13


(v)

∴ 7429 =  17 × 19 × 23


2. Find the LCM and HCF of the following pairs of integer and verify that LCM × HCF = Product of the two numbers.

(i) 26 and 91

(ii) 510 and 92

(iii) 336 and 54

Solution:

(i)

26 = 2 × 13

91 = 7 × 13

∴ HCF = 13

LCM = 2 × 7 × 13
          = 182

LCM × HCF = 182 × 13
                      = 2366

Product of the two numbers = 26 × 91
                                              = 2366


∴ LCM × HCF = Product of the two numbers


(ii)

510 = 2 × 5 × 3 × 17

92   = 2 × 2 × 23







∴ HCF = 2

LCM = 2 × 2 × 3 × 5 × 17 × 23
          = 23460


LCM × HCF = 23640 × 2
                      = 46920

Product of the two numbers = 510 × 92
                                             = 46920

∴ LCM × HCF = Product of the two numbers.


(iii)

336 = 2 × 2 × 2 × 2 × 3 × 7

54   = 2 × 3 × 3 × 3

∴ HCF = 2 × 3
            = 6

LCM = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 7
          = 3024

LCM × HCF = 3024 × 6
                      = 18144

Product of the two numbers = 336 × 54
                                             = 18144

∴ LCM × HCF = Product of the two numbers.


3. Find the LCM and HCF of the following integers by applying the prime factorization method.

(i) 12, 15 and 21

(ii) 17, 23 and 29

(iii) 8, 9 and 25

Solution:

(i)

12 = 2 × 2 × 3

15 = 3 × 5

21 = 3 × 7

LCM = 2 × 2 × 3 × 5 × 7
          = 420

HCF = 3


(ii)

17 = 17 × 1

23 = 23 × 1

29 = 29 × 1

LCM = 17 × 23 × 29
         =  11339

HCF = 1


(iii)

8 = 2 × 2 × 2

9 = 3 × 3 × 3

25 = 5 × 5

If there is no common prime factor, then HCF = 1.

Because

8 = 2 × 2 × 2 × 1

9 = 3 × 3 × 3 × 1

25 = 5 × 5 × 1

So the HCF of 8, 9 and 25 = 1

LCM = 2 × 2 × 2 × 3 × 3 × 3 × 5 × 5
         = 1800


4. Given that HCF (306, 657) = 9, find LCM (306, 657).

Solution:

LCM × HCF = Product of the two numbers.

            = 

            = 34 × 657

            = 22338

∴ LCM (306, 657) = 22338


5. Check whether 6n can end with the digit 0 for any natural number n.

Solution:

Any number that ends with the digit 0 should be divisible 10. i.e, it will be divisible by 2 and 5.

Because prime factorization of 10 = 2 × 5

Prime factorization of 6 = 2 × 3

So 6n is divisible by 2 but not divisible by 5.

∴ 6n can not end with the digit 0.


6. Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.

Solution:

7 × 11 × 13 + 13 = 13( 7 × 11 × 1 + 1 )                    [ Taking 13 common ]

                           = 13( 77 + 1 )
                           = 13(78)
                           = 1014

∴ 7 × 11 × 13 + 13 = 2 × 3 × 13 × 13

The expression 7 × 11 × 13 + 13 has more than 2 factors, hence it is a composite number.


7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 = 5(7 × 6 ×  4 × 3 × 2 × 1 + 1)            [ Taking 5 common ]
                                              = 5(1008 + 1)
                                              = 5(1009)
                                              = 5045
                                              = 5 × 1009
The expression 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 has factors other than 1 and the number itself, hence it is a composite number.


7. There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point?

Solution:

LCM of 18 and 12 will give the answer for after how many minutes they will meet again at the starting point.

18 = 2 × 3 × 3

12 = 2 × 2 × 3

LCM = 2 × 2 × 3 × 3
         = 36

∴ They will meet again at the starting point after 36 minutes.



Exercise 1.3


1. Prove that √5 is irrational.

Solution:

Let √5 be a rational number.

Let us take two coprime a and b where b ≠ 0

Þ b√5 = a
Squaring both sides

5b² = a² ------------(i)

So a² is divisible and hence a is also divisible 5.

Now we can write

a = 5c

Squaring both sides

a² = 25c²

5b² = 25c²

Þ  b² = 5c²

So b² is divisible by 5 and hence b is also divisible by 5.

But it is contradiction to the fact that a and b are coprime.

Hence √5 is an irrational number.


2. Prove that 3 + 2√5 is irrational.

Solution:

Proof of √5 as an irrational number:

Let √5 be a rational number.

Let us take two coprime a and b where b ≠ 0

Þ b√5 = a
Squaring both sides

5b² = a² ------------(i)

So a² is divisible and hence a is also divisible 5.

Now we can write

a = 5c

Squaring both sides

a² = 25c²

5b² = 25c²

Þ  b² = 5c²

So b² is divisible by 5 and hence b is also divisible by 5.

But it is contradiction to the fact that a and b are coprime.

Hence √5 is an irrational number.


Now, Let 3 + 2√5 be a rational number.

     

So √5 is a rational number. This contradicts to the fact that √5 is irrational number.

Hence our assumption that 3 + 2√5 as rational number is false.

So 3 + 2√5 is irrational.


3. Prove that the following are irrationals.

(ii) 7√5

(iii) 6 + √2

Solution:

(i)

Proof of √2 as irrational number:

Let √2 be a rational number.

a = b√2

Squaring both sides

a² = 2b² ----(i)

So a² is divisible by 2 and hence a is divisible by 2.

Let a = 2c

Squaring both sides

a² = 4c²

2b² = 4c²  [ From (i) ]

b² = 2c²

So b² is divisible by 2 and hence b is divisible by 2.

This contradicts to the fact that a and b are coprime. So our assumption that √2 as a rational number is false.

Hence √2 is irrational.

Now,

Þ  b = a√2

This contradicts to the fact that √2 is irrational.

(ii)

Proof of √5 as an irrational number:

Let √5 be a rational number.

Let us take two coprime a and b where b ≠ 0

Þ b√5 = a
Squaring both sides

5b² = a² ------------(i)

So a² is divisible and hence a is also divisible 5.

Now we can write

a = 5c

Squaring both sides

a² = 25c²

5b² = 25c²

Þ  b² = 5c²

So b² is divisible by 5 and hence b is also divisible by 5.

But it is contradiction to the fact that a and b are coprime.

Hence √5 is an irrational number.


Now,

 So √5 is a rational. But this contradicts to the fact that √5 is irrational.

Hence our assumption that 7√5 as a rational number is false.

So 7√5 is irrational.


(iii)

Proof of √2 as irrational number:

Let √2 be a rational number.

a = b√2

Squaring both sides

a² = 2b² ----(i)

So a² is divisible by 2 and hence a is divisible by 2.

Let a = 2c

Squaring both sides

a² = 4c²

2b² = 4c²  [ From (i) ]

b² = 2c²

So b² is divisible by 2 and hence b is divisible by 2.

This contradicts to the fact that a and b are coprime. So our assumption that √2 as a rational number is false.

Hence √2 is irrational.


Now,

Let 6 + √2 be rational number.

So √2 is a rational number. This contradicts the fact that √2 is irrational.

Hence our assumption that 6 + √2 as rational is false.

So 6 + √2 is irrational.


Exercise 1.4 :

1. Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion.

     


















Solution:

(i)

We factorize the denominator.

∴ 3125 = 5 × 5 × 5 × 5 × 5
            = 5⁵

So the given rational number is terminating.



(ii)

We factorize the denominator of the rational number.

8 = 2 × 2 × 2
  = 2³

So the given rational number is terminating.



(iii)

We factorize the denominator.

455 = 5 × 7 × 13

The denominator has the factor 5 but it also has 7 and 13.

So the rational number is non-terminating.


(iv)

We factorize the denominator.

1600 = 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5
         = 2⁶ × 5²

So the rational number is terminating.


(v)

We factorize the denominator.

343 = 7 × 7 × 7
      = 7³

The denominator has 7 as its factor.

So the rational number is non-terminating.


(vi)

So the rational number is terminating.


(vii)

The denominator has the factors 2 and 5 but also 7.

So the rational number is non-terminating.


(viii)

So the given rational number is terminating.


(ix)

10 = 2 × 5

So the rational number is terminating.


(x)

We factorize the denominator.

The denominator has the factor 2 and 5 but also has 3 .

So the rational number is non-terminating.


2. Write down the decimal expansions of those rational numbers in Question 1 above which have terminating decimal expansions.

3. The following real numbers have decimal expansions as given below. In each case, decide whether they are rational or not. If they are rational, and of the form p, q , what do you say about the prime factors of q.

(i) 43.123456789

(ii) 0.120120012000120000...

Solution:

(i)

The number has terminating decimal expansion.

(ii)

The decimal expansion is neither terminating nor repeating, so it an irrational number.

(iii)

The decimal expansion is non-terminating and repeating.