Real Numbers

i.

∵ The division is exact

∴ it is a terminating decimal.

ii.

∵ The division never ends and the digits ‘18’ is repeated endlessly

∴ it is a non-terminating recurring type decimal.

iii.

∵ The division is exact

∴ it is a terminating decimal.

iv.

∵ The division is exact

∴ it is a terminating decimal.

v.

∵ The division never ends and the digit ‘3’ is repeated endlessly

∴ it is a non-terminating recurring type decimal.

i.

ii.

iii.

iv.

v.

i.

Let

⇒ 10x = 6.66666......

Now,

10x - x = 6.66 – 0.6666

⇒9x = 6

ii.

Let

⇒ 100x = 37.3737......

Now,

100x - x = 37.3737 – 0.3737

⇒ 99x = 37

iii.

Let

⇒ 100x = 317.1717......

Now,

100x - x = 317.1717 – 3.1717

⇒ 99x = 314

iv.

Let

⇒ 100x = 1589.8989......

Now,

100x - x = 1589.8989 – 15.8989

⇒ 99x = 1574

v.

Let

⇒ 1000x = 2514.514514......

Now,

1000x - x = 2514.514514 – 2.514514

⇒ 999x = 2512

Let us assume that 4√2 is a rational number

where, b≠0 and a, b are integers

∵ a, b are integers ∴ 4b is also integer

is rational which cannot be possible

∵ which is an irrational number

∵ it is contradicting our assumption

∴ the assumption was wrong

Hence, 4√2 is an irrational number

Let us assume that 3 + √5 is a rational number

where, b≠0 and a, b are integers

∵ a, b are integers ∴ a – 3b is also integer

is rational which cannot be possible

∵ which is an irrational number

∵ it is contradicting our assumption ∴ the assumption was wrong

Hence, 3 + √5 is an irrational number

By Pythagoras theorem,

(√5)² = 2² + 1²

⇒ (√5)² = 4 + 1

First mark 0 and 2 on the number line. Then, draw a perpendicular of 1 unit from 2. And Join the top of perpendicular and 0. This line would be equal to √5. Now measure the line with compass and marc an arc on the number line with the same measurement. This point is √5.

Also,

By Pythagoras theorem,

(√10)² = 3² + 1²

⇒ (√10)² = 9 + 1

First mark 0 and 3 on the number line. Then, draw a perpendicular of 1 unit from 3. And Join the top of perpendicular and 0. This line would be equal to √10. Now measure the line with compass and marc an arc on the number line with the same measurement. This point is √10.

0.3 and -0.5

To find a rational number x between two rational numbers and , we use

Therefore, to find rational number x (let) between

and

Now if we find a rational number between and it will also be between 0.3 and -0.5 since lies between 0.3 and -0.5.

Therefore, to find rational number y (let) between and

Now if we find a rational number between and it will also be between 0.3 and -0.5 since lies between 0.3 and -0.5.

Therefore, to find rational number z (let) between and

Hence the numbers are -0.2, -0.1 and 0.1

-2.3 and -2.33

To find a rational number x between two rational numbers and , we use

Therefore, to find rational number x (let) between and

⇒ x = -2.315

Now if we find a rational number between and it will also be between -2.3 and -2.33 since -2.315 lies between -2.3 and -2.33

Therefore, to find rational number y (let) between and

⇒ y = -2.3075

Now if we find a rational number between and it will also be between -2.3 and -2.33 since -2.315 lies between -2.3 and -2.33

Therefore, to find rational number z (let) between and

⇒ z = -2.3225

Hence the numbers are -2.3225, -2.3075 and -2.315

5.2 and 5.3

To find a rational number x between two rational numbers and , we use

Therefore, to find rational number x (let) between and

⇒ x = 5.25

Now if we find a rational number between and it will also be between 5.2 and 5.3 since 5.25 lies between 5.2 and 5.3

Therefore, to find rational number y (let) between and

⇒ y = 5.225

Now if we find a rational number between and it will also be between 5.2 and 5.3 since 5.25 lies between 5.2 and 5.3

Therefore, to find rational number z (let) between and

⇒ z = 5.275

Hence the numbers are 5.225, 5.25 and 5.275

-4.5 and 4.6

To find a rational number x between two rational numbers and , we use

Therefore, to find rational number x (let) between and

⇒ x = 0.05

Now if we find a rational number between and it will also be between -4.5 and 4.6 since 0.05 lies between -4.5 and 4.6

Therefore, to find rational number y (let) between and

⇒ y = -2.225

Now if we find a rational number between and it will also be between -4.5 and 4.6 since 0.05 lies between -4.5 and 4.6

Therefore, to find rational number z (let) between and

⇒ z = 2.325

Hence the numbers are -2.225, 0.05and 2.325

In , n is called the order of the surd.

Therefore,

_i.

In this, the order of surd is 3.

_ii.

In this, the order of surd is 5.

_iii.

In this, the order of surd is 4.

_iv.

In this, the order of surd is 2.

_v.

In this, the order of surd is 3

Surds are numbers left in root form (√) to express its exact value. It has an infinite number of non-recurring decimals. Therefore, surds are irrational numbers.

Therefore,

_i.

It is a surd ∵ it cannot be expressed as a rational number.

_ii.

It is a surd ∵ it cannot be expressed as a rational number.

_iii.

It is a surd ∵ it cannot be expressed as a rational number.

_{iv.√256 = √16}² = 16

It is not a surd ∵ it is a rational number.

_v.

It is not a surd ∵ it is a rational number.

vi.

It is a surd ∵ it cannot be expressed as a rational number.

Two or more surds are said to be similar or like surds if they have the same surd-factor.

And,

Two or more surds are said to be dissimilar or unlike when they are not similar.

Therefore,

i. √52, 5√13

√52 = √(2×2×13) = 2√13

5√13

∵ both surds have same surd-factor i.e., √13.

∴ they are like surds.

ii. √68, 5√3

√68 = √(2×2×17) = 2√17

5√3

∵ both surds have different surd-factors √17 and √3.

∴ they are unlike surds.

iii. 4√18, 7√2

4√18 = 4√(2×3×3) = 4×3√2 = 12√2

7√2

∵ both surds have same surd-factor i.e., √2.

∴ they are like surds.

iv. 19√12, 6√3

19√12 = 19√(2×2×3) = 19×2√3 = 38√3

6√3

∵ both surds have same surd-factor i.e., √3.

∴ they are like surds.

v. 5√22, 7√33

∵ both surds have different surd-factors √22 and √33.

∴ they are unlike surds.

vi. 5√5, √75

5√5

√75 = √(5×5×3) = 5√3

∵ both surds have different surd-factors √5 and √3.

∴ they are unlike surds.

i.

ii.

iii.

iv.

v.

i. 7√2 , 5√3

(7√2)² = 7 × 7 × √2 × √2

⇒ (7√2)² = 49 × 2

⇒ (7√2)² = 98

And

(5√3)² = 5 × 5 × √3 × √3

⇒ (5√3)² = 25 × 3

⇒ (5√3)² = 75

Clearly,

98 > 75

∴ 7√2 > 5√3

ii. √247, √274

(√247)² = 247

And

(√274)² = 274

Clearly,

247 < 274

∴ √247 < √274

iii. 2√7, √28

(2√7)² = 2 × 2 × √7 × √7

⇒ (2√7)² = 4 × 7

⇒ (2√7)² = 28

And

(√28)² = 28

Clearly,

28 = 28

∴ 2√7 = √28

iv. 5√5, 7√2

(5√5)² = 5 × 5 × √5 × √5

⇒ (5√5)² = 25 × 5

⇒ (5√5)² = 125

And

(7√2)² = 7 × 7 × √2 × √2

⇒ (7√2)² = 49 × 2

⇒ (7√2)² = 98

Clearly,

125 = 98

∴ 5√5= 7√2

v. 4√42, 9√2

(4√42)² = 4 × 4 × √42 × √42

⇒ (4√42)² = 16 × 42

⇒ (4√42)² = 672

And

(9√2)² = 9 × 9 × √2 × √2

⇒ (9√2)² = 81 × 2

⇒ (9√2)² = 162

Clearly,

672 > 162

∴ 4√42 > 9√2

vi. 5√3, 9

(5√3)² = 5 × 5 × √3 × √3

⇒ (5√3)² = 25 × 3

⇒ (5√3)² = 75

And

(9)² = 9 × 9

⇒ (9)² = 81

Clearly,

75 < 81

∴ 5√3 < 9

vii. 7, 2√5

(2√5)² = 2 × 2 × √5 × √5

⇒ (2√5)² = 4 × 5

⇒ (2√5)² = 20

And

(7)² = 7 × 7

⇒ (7)² = 49

Clearly,

49 > 20

∴ 7 > 2√5

i. 5√3 + 8√3

5√3 + 8√3 = (5 + 8)√3

⇒ 5√3 + 8√3 = 13√3

ii. 9√5 – 4√5 + √125

9√5 – 4√5 + √125 = 9√5 – 4√5 + √(5 × 5 × 5)

⇒ 9√5 – 4√5 + √125 = 9√5 – 4√5 + √(5 × 5 × 5)

⇒ 9√5 – 4√5 + √125 = 9√5 – 4√5 + 5√5

⇒ 9√5 – 4√5 + √125 = (9 – 4 + 5)√5

⇒ 9√5 – 4√5 + √125 = 10√5

iii. 7√48 – √27 – √3

7√48 – √27 – √3 = 7√(2 × 2 × 2 × 2 × 3) – √(3 × 3 × 3) – √3

⇒ 7√48 – √27 – √3 = 7×4√3 – 3√3 – √3

⇒ 7√48 – √27 – √3 = 28√3 – 3√3 – √3

⇒ 7√48 – √27 – √3 = (28 – 3 – 1)√3

⇒ 7√48 – √27 – √3 = 24√3

iv.

i. 3√12 × √18

3√12 × √18 = 3√(2 × 2 × 3) × √(2 × 3 × 3)

⇒3√12 × √18 = 3 × 2√3 × 3√2

⇒ 3√12 × √18 = 6√3 × 3√2

⇒ 3√12 × √18 = 18√6

ii. 3√12 × 7√15

3√12 × 7√15 = 3√(2 × 2 × 3) × 7√(3 × 5)

⇒3√12 × 7√15 = 3 × 2√3 × 7√(3 × 5)

⇒3√12 × 7√15 = 3 × 2 × 7 × √(3 × 3 × 5)

⇒3√12 × 7√15 = 3 × 2 × 7 × 3√5

⇒ 3√12 × 7√15 = 126√5

iii. 3√8 × √5

3√8 × √5 = 3√(2 × 2 × 2) × √5

⇒3√8 × √5 = 3 × 2√2 × √5

⇒ 3√8 × √5 = 3 × 2 ×√(2 × 5)

⇒ 3√8 × √5 = 6√10

iv. 5√8 × 2√8

5√8 × 2√8 = 5√(2 × 2 × 2) × 2√(2 × 2 × 2)

⇒5√8 × 2√8 = 5 × 2√2 × 2 × 2√2

⇒ 5√8 × 2√8 = 5 × 2 × 2 × 2 ×√(2 × 2)

⇒ 5√8 × 2√8 = 5 × 2 × 2 × 2 × 2

⇒ 5√8 × 2√8 = 80

i. √98 ÷ √2

⇒√98 ÷ √2 = 7

ii. √125 ÷ √50

iii. √54 ÷ √27

⇒√54 ÷ √27 = √2

iv. √310 ÷ √5

⇒ √310 ÷ √5 = √62

i. We know that √5 × √5 = 5, ∴ to rationalize the denominator of multiply both numerator and denominator by √5.

ii. We know that √14 × √14 = 14, ∴ to rationalize the denominator of multiply both numerator and denominator by √14.

iii. We know that √7 × √7 = 7, ∴ to rationalize the denominator of multiply both numerator and denominator by √7.

iv. We know that √3 × √3 = 3, ∴ to rationalize the denominator of multiply both numerator and denominator by √3.

v. We know that √3 × √3 = 3, ∴ to rationalize the denominator of multiply both numerator and denominator by √3.

i. √3(√7 – √3)

=√3 × √7 – √3 × √3

[∵√a(√b–√c)=√a×√b–√a×√c]

=√21 – 3

ii. (√5 – √7)√2

=√5 × √2 – √7 × √2

[∵√a(√b–√c)=√a×√b–√a×√c]

=√10 – √14

iii. (3√2 – √3)(4√3 – √2)

=3√2(4√3 – √2) – √3(4√3 – √2)

=3√2×4√3 – 3√2×√2 – √3×4√3 + √3×√2

[∵√a(√b–√c)=√a×√b–√a×√c]

=12√6 – 3×2 – 4×3 + √6

=12√6 – 6 – 12 + √6

=13√6 – 18

i. The rationalizing factor of √7 + √2 is √7 – √2. Therefore, multiply both numerator and denominator by √7 – √2.

[∵ (a-b)(a+b) = a² – b²]

ii. The rationalizing factor of 2√5 – 3√2 is 2√5 + 3√2. Therefore, multiply both numerator and denominator by 2√5 + 3√2.

[∵ (a-b)(a+b) = a² – b²]

iii. The rationalizing factor of 7 + 4√3 is 7 – 4√3. Therefore, multiply both numerator and denominator by 7 – 4√3.

[∵ (a-b)(a+b) = a² – b²]

iv. The rationalizing factor of √5 + √3 is √5 - √3. Therefore, multiply both numerator and denominator by √5 - √3.

[∵ (a-b)(a+b) = a² – b²]

[∵ (a-b)² = a² + b² – 2ab]

Absolute value describes the distance of a number on the number line from 0 without considering which direction from zero the number lies. The absolute value of a number is never negative.

Therefore,

i. |15 - 2| = |13| = 13

ii. |4 - 9| = |-5| = 5

iii. |7| × |-4| = 7 × 4 = 28

i. |3x - 5| = 1

⇒ 3x - 5 = 1 or 3x - 5 = -1

⇒ 3x = 1 + 5 or 3x = -1 + 5

⇒ 3x = 6 or 3x = 4

ii. |7 – 2x| = 5

⇒ 7 – 2x = 5 or 7 – 2x = -5

⇒ 2x = 7 - 5 or 2x = 7 + 5

⇒ 2x = 2 or 2x = 12

⇒ x = 1 or x = 6

iii.

⇒ 8 – x = 2 × 5 or 8 – x = 2 × -5

⇒ 8 – x = 10 or 8 – x = -10

⇒ x = 8 – 10 or x = 8 + 10

⇒ x = -2 or x = 18

iv.

⇒ 20 + x = 4 × 5 or 20 + x = 4 × -5

⇒ 20 + x = 20 or 20 + x = -20

⇒ x = 20 – 20 or x = -20 – 20

⇒ x = 0 or x = -40

An irrational number is a number that cannot be expressed as a fraction for any integers p and q and q ≠ 0.

since it can be written as , it is a rational number.

since it can be written as , it is a rational number.

since it can be written as , it is a rational number.

Since √5 cannot be written as it is an irrational number

Therefore √5 is an irrational number.

An irrational number is a number that cannot be expressed as a fraction for any integers p and q and q ≠ 0.

.

Since it can be written as ,

it is a rational number.

is a rational number because it is a non-terminating but repeating decimal.

is a rational number because it is a non-terminating but repeating decimal.

0.101001000.... is an irrational number because it is a non-terminating and non-`repeating decimal.

Therefore, 0.101001000.... is an irrational number.

A non-terminating recurring decimal representation means that the number will have an infinite number of digits to the right of the decimal point and those digits will repeat themselves.

∵ it does not have an infinite number of digits to the right of the decimal point ∴ it is not a non-terminating recurring decimal.

∵ it does not have an infinite number of digits to the right of the decimal point ∴ it is not a non-terminating recurring decimal.

∵ it has an infinite number of digits to the right of the decimal point which are repeating themselves ∴ it is a non-terminating recurring decimal.

∵ it does not have an infinite number of digits to the right of the decimal point ∴ it is not a non-terminating recurring decimal.

Therefore, is a non-terminating recurring decimal.

Every point of a number line is assumed to correspond to a real number, and every real number to a point. Therefore, Every point on the number line represent a real number.

∵ the denominator of all the above options is 9 ∴ we multiply both numerator and denominator by 0.9 as 10 × 0.9 = 9

If n is not a perfect square number, then √n cannot be expressed as ratio of a and b where a and b are integers and b ≠ 0

Therefore, √n is an Irrational number

Which is a rational number

Therefore, is not a surd.

Therefore, the order of the surd is 6.

A math conjugate is formed by changing the sign between two terms in a binomial. For instance, the conjugate of x + y is x - y.

Now,

2√5 + √3 = √3 + 2√5

Its conjugate pair = √3 - 2√5 = -2√5 + √3

∴ The conjugate pair of 2√5 + √3 = -2√5 + √3

|12 – (13 + 7) × 4| = |12 – 20 × 4| (Solving it according to BODMAS)

⇒ |12 – (13 + 7) × 4| = |12 – 80|

⇒ |12 – (13 + 7) × 4| = |-68|

⇒ |12 – (13 + 7) × 4| = 68

i.

ii.

Let

⇒ 1000x = 29568.568568......

Now,

1000x - x = 29568.568568 – 29.568568

⇒999x = 29539.0

iii.

Let x = 9.315315…

⇒ 1000x = 9315.315315......

Now,

1000x - x = 9315.315315 – 9.315315

⇒999x = 9306.0

iv.

Let x = 357.417417…

⇒ 1000x = 357417.417417…

Now,

1000x - x = 357417.417417 – 357.417417

⇒999x = 357060.0

v.

Let

⇒ 1000x = 30219.219219…

Now,

1000x - x = 30219.219219 – 30.219219

⇒999x = 30189.0

i.

ii.

iii.

√5 = 2.236067977…….

iv.

v.

Let us assume that 5 + √7 is a rational number

where, b≠0 and a, b are integers

∵ a, b are integers ∴ a – 5b and b are also integers

is rational which cannot be possible ∵ which is an irrational number

∵ it is contradicting our assumption ∴ the assumption was wrong

Hence, 5 + √7 is an irrational number

i.

ii.

i. √32

∴ Its rationalizing factor = √2

ii. √50

∴ Its rationalizing factor = √2

iii. √27

∴ Its rationalizing factor = √3

∵ √10 cannot be further simplified

∴ Its rationalizing factor = √10

v. 3√72

∴ Its rationalizing factor = √2

vi. 4√11

∵ √11 cannot be further simplified

∴ Its rationalizing factor = √11

i.

= 4√3 + 3√3 – √3

= 7√3 – √3

= 6√3

ii.

iii.

iv.

v.

i.

ii.

iii.

iv.

v.