Number Systems

Given :

As per the definition, all the rational number as well as irrational number together makes the collection of real number.

Hence, every rational number is a real number.

Option (C) is the correct answer.

Given :

There are infinitely many rational numbers between two rational numbers.

As 2 and 3 are two rational numbers.

And and so on..

Hence option (C ) is correct answer

Given :

Decimal representation of a rational number cannot be non-terminating non-repeating. It is always terminating or it will be periodic i.e. repeating.

Hence option (D) is correct answer.

Given :

The product of any two irrational numbers is sometimes rational, sometimes irrational.

As π and 1/π both are irrational number but when they multiply we get 1 which is a rational number.

And (irrational)

Hence option (D) is correct answer.

Given :



√2 = 1.4142

Hence option (B) is the correct answer.

Given : values to check rational numbers.

Hence option (C) is correct answer.

Given : to check the irrational number.

Because an irrational number is non terminating and non repeating.

In option D the digits are non terminating and non repeating.

Hence option (D) is correct answer.

Given : number

To find rational number between two given number a and b is

Hence, rational number between two given number

Hence option (A) is correct answer.

Given : 1.999...

Let x = 1.999…. (1)

Multiply (1) by 10

10x = 19.99… (2)

Subtract (1) from (2)

10x – x = 19.99…. – 1.999….

9x = 18

x = 2

Hence option (C) is correct answer.

Given : number

Hence option (C) is correct answer.

Given : number

Hence option (B) is correct answer.

Given : number

After rationalising:

Hence option (A) is correct answer.

Given : number

After rationalising:

Hence option (D) is correct answer.

Given : number

After rationalising:

Hence option (B) is correct answer.

Given : number

Hence option (B) is correct answer.

Given : number

= 0.4142

Hence option (C) is correct answer.

Given : number

Hence option (C) is correct answer.

Given : number

= 2

Hence option (B) is correct answer.

Given : number

Hence option (A) is correct answer.

Given : number (256)^0.16×(256)^0.09

= (256)^{0.16 + 0.09}

= (256)^0.25

= (2)²

= 4

Hence option (A) is correct answer.

Given : x

= x

Hence option (A) is correct answer.

Given: x is rational number and y is irrational number.

Yes, x + y is necessarily an irrational number.

Example: Let x = 2, which is rational.

Let y = √2, which is irrational.

Then, x + y = 2 + √2, which is irrational.

∴ Sum of a rational and an irrational number is always irrational.

Given: x is rational number and y is irrational number.

Yes, xy is necessarily an irrational number.

Example: Let x = 2, which is rational.

Let y = √2, which is irrational.

Then, x × y = 2 × √2 = 2√2, which is again irrational.

Also, consider the case when x = 0.

Then xy = 0, which is rational.

∴ Product of a rational and an irrational number is always irrational, only if the rational number is not zero.

False

is not a rational number.

Justification: We can write as .

Here, √2 is irrational and 1/3 is rational number.

Product of a rational and an irrational number is always irrational.

∴ is irrational.

Hence, it is not rational and thus the given statement is false.

False

Justification: Because there are a finite numbers of integers between any two integers.

Example: The integers between 1 and 5 are 2, 3 and 4; which are finite in number.

False

Justification: Because there are infinite number of rationals between any two numbers.

∴ The given statement is false.

True

Justification: All the irrational numbers are the numbers which cannot be written in the form p/q; q≠0, p, q both are integers and there are infinitely many irrationals.

∴ The given statement is true.

False

Justification:

Consider the irrational number x =

Then x² = √3, which is again an irrational number.

∴ The given statement is false.

False

Justification:

Thus, on simplifying, it turns out to be a rational number.

∴ The given statement is false.

False

Justification:

Thus, on simplifying, it turns out to be an irrational number again.

∴ The given statement is false.

The given number is rational number.

Justification:

Thus, √196 is rational.

The given number is irrational number.

Justification:

which is irrational.

Thus, 3√18 is irrational.

The given number is irrational number.

Justification:

(Simplifying the given expression)

which is product of a rational number (1/3) and an irrational number (√3), which results in an irrational number.

Thus, is irrational.

The given number is rational number.

Justification:

(Simplifying the given expression)

(Taking the square root)

which is in form p/q; q≠0, p, q both are integers.

Thus, is rational.

The given number is irrational number.

Justification: (Simplifying the given expression)

which is product of a rational number (-1/5) and an irrational number (√10), which results in an irrational number.

Thus, is irrational.

The given number is rational number.

Justification: (Simplifying the given expression)

(Taking the square root)

which is in form p/q; q≠0, p, q both are integers.

Thus, is rational.

The given number is rational number.

Justification:

Since 0.5918 is non – repeating terminating decimal number, therefore it can be written in the form p/q; q≠0, p, q both are integers.

Thus, 0.5918 is rational.

The given number is rational number.

Justification:

Consider (1 + √5) – (4 + √5) = 1 – 4 + √5 -√5

= 1 – 4

= 3, which is rational.

Thus, (1 + √5) – (4 + √5) is rational.

The given number is rational number.

Justification:

Since 10.124124 is non - terminating recurring decimal number, therefore it can be written in the form p/q; q≠0, p, q both are integers.

Thus, 10.124124 is rational.

The given number is irrational number.

Justification: Since 1.010010001 is non - terminating non - recurring decimal number, therefore it cannot be written in the form p/q; q≠0, p, q both are integers.

Thus, 1.010010001 is irrational.

(i) x² = 5

On taking square root on both sides, we get

⇒ x = ± √5

So, x is an irrational number.

(ii) y² = 9

On taking square root on both sides, we get

⇒ y = ± 3

So, y is a rational number.

(iii) z² = .04

On taking square root on both sides, we get

⇒ z = ± 0.2

So, z is a rational number.

On taking square root on both sides, we get

So, u is an irrational number because √17 is irrational.

(i) –1 and –2

Three rational numbers between –1 and –2 are –1.1, –1.2 and –1.3.

(ii) 0.1 and 0.11

Three rational numbers between 0.1 and 0.11 are 0.101, 0.102 and 0.103.

(iii)

We can write and

.

(iv)

LCM of 4 and 5 is 20.

We can write and

So, three rational numbers between

(i) Rational number: 2.1 and Irrational number: 2.040040004...

(ii) Rational number: 0.03 and Irrational number: 0.007000700007…

(iii) Rational number between

LCM of 3 and 2 is 6.

We can write and

So, rational number between and Irrational number: 0.414114111...

(iv)

Rational number: 0 and Irrational number: 0.151151115...

(v) Rational number: 0.151 and Irrational number: 0.151551555...

(vi) √2 = 1.41 and √3 = 1.732

Rational number: 1.5 and Irrational number: 1.585585558...

(vii) Rational number: 3 and Irrational number: 3.101101110...

(viii) Rational number: 0.00011 and Irrational number: 0.0001131331333...

(ix) Rational number: 1 and Irrational number: 1.909009000...

(x) Rational number: 6.3753 and Irrational number: 6.375414114111...

(i) 7

Draw a number line and mark 7 on it.

(ii) 7.2

Draw a number line. Here, 7.2 will be situated between 7 and 8 so on redrawing the number line between 7 and 8 we can easily find 7.2.

(iii)

Draw a number line. Here, -1.5 will be situated between -2 and -1 so on redrawing the number line between -2 and -1 we can easily find -1.5.

(iv)

Draw a number line. Here, -2.4 will be situated between -3 and -2 so on redrawing the number line between -3 and -2 we can easily find -2.4.

(i) √5

We can write 5 as the sum of the squares of two natural numbers: 5 = 1 + 4

⇒ 5 = 1² + 2²

On the number line, take OA = 2 units.

Draw BA = 1 units, perpendicular to OA. Join OB.

By Pythagoras theorem, OB = √5 Using a compass with centre O and radius OB, draw an arc which intersects the number line at the point C. Then, C corresponds to √5.

(ii) √10

We can write 10 as the sum of the squares of two natural numbers: 10 = 1 + 9

⇒ 10 = 1² + 3²

On the number line, take OA = 3 units.

Draw BA = 1 units, perpendicular to OA. Join OB.

By Pythagoras theorem, OB = √10 Using a compass with centre O and radius OB, draw an arc which intersects the number line at the point C. Then, C corresponds to √10.

(ii). √17

We can write 17 as the sum of the squares of two natural numbers: 17 = 1 + 16

⇒ 17 = 1² + 4²

On the number line, take OA = 4 units.

Draw BA = 1 units, perpendicular to OA. Join OB.

By Pythagoras theorem, OB = √17 Using a compass with centre O and radius OB, draw an arc which intersects the number line at the point C. Then, C corresponds to √17.

√4.5

Mark the distance 4.5 units from a fixed point A on a given line to obtain a point B such that AB = 4.5 units.

From B, mark a distance of 1 unit and mark the new point as C.

Find the mid-point of AC and mark that point as O. Draw a semicircle with centre O and radius OC.

Draw a line perpendicular to AC passing through B and intersecting the semicircle at D. Then, BD = √4.5.

Draw an arc with centre B and radius BD, meeting AC produced at E, then BE = BD = √4.5 units.

√5.6

Mark the distance 5.6 units from a fixed point A on a given line to obtain a point B such that AB = 5.6 units.

From B, mark a distance of 1 unit and mark the new point as C.

Find the mid-point of AC and mark that point as O. Draw a semicircle with centre O and radius OC.

Draw a line perpendicular to AC passing through B and intersecting the semicircle at D. Then, BD = √5.6.

Draw an arc with centre B and radius BD, meeting AC produced at E, then BE = BD = √5.6 units.

√8.1

Mark the distance 8.1 units from a fixed point A on a given line to obtain a point B such that AB = 8.1 units.

From B, mark a distance of 1 unit and mark the new point as C.

Find the mid-point of AC and mark that point as O. Draw a semicircle with centre O and radius OC.

Draw a line perpendicular to AC passing through B and intersecting the semicircle at D. Then, BD = √8.1.

Draw an arc with centre B and radius BD, meeting AC produced at E, then BE = BD = √8.1 units.

√2.3

Mark the distance 2.3 units from a fixed point A on a given line to obtain a point B such that AB = 2.3 units.

From B, mark a distance of 1 unit and mark the new point as C.

Find the mid-point of AC and mark that point as O. Draw a semicircle with centre O and radius OC.

Draw a line perpendicular to AC passing through B and intersecting the semicircle at D. Then, BD = √2.3.

Draw an arc with centre B and radius BD, meeting AC produced at E, then BE = BD = √2.3 units.

0.2

Now,

0.888…

Let 𝑥 = 0.888 …

⇒ 𝑥 = 0.8 ……………. (1)

Multiplying both sides by 10, we get

10 𝑥 = 8.8 ……………. (2)

Subtracting equation (1) from equation (2), we get

10 𝑥 − 𝑥 = 8.8 − 0.8

⇒ 9𝑥 = 8.0

5. 2

Let 𝑥 = 5.2 ……………. (1)

Multiplying both sides by 10, we get

10 𝑥 = 52.2 …………… (2)

Subtracting equation (1) from equation (2), we get

10 𝑥 − 𝑥 = 52.2 − 5.2

⇒ 9𝑥 = 47

0. 001

Let 𝑥 = 0.001 ……………. (1)

Multiplying both sides by 1000, we get

1000 𝑥 = 1.001 …………… (2)

Subtracting equation (1) from equation (2), we get

1000𝑥 − 𝑥 = 1.001 − 0.001

⇒ 999𝑥 = 1

0.2555…

Let 𝑥 = 0.2555 …

⇒ x = 0.25 ……………. (1)

Multiplying both sides by 10, we get

10 x = 2.5 ……………. (2)

Multiplying both sides by 100, we get

100 x = 25.5 …………. (3)

Subtracting equation (2) from equation (3), we get

100 x-10x = 25.5 - 2.5

⇒ 90𝑥 = 23

0.134

Let 𝑥 = 0.134 ………….…. (1)

Multiplying both sides by 10, we get

10 𝑥 = 1.34 ………………. (2)

Multiplying both sides by 1000, we get

1000 𝑥 = 134.34 …………. (3)

Subtracting equation (2) from equation (3), we get

1000 𝑥 − 10𝑥 = 134.34 − 1.34

⇒ 990𝑥 = 133

.00323232...

Let 𝑥 = 0.00323232 …

⇒ x = 0.0032 ………….…. (1)

Multiplying both sides by 100, we get 100

100x = 0.32 ……………. (2)

Multiplying both sides by 10000, we get

10000 x = 32.32 …………. (3)

Subtracting equation (2) from equation (3), we get

10000 x-100x = 32.32 - 0.32

⇒ 9900𝑥 = 32

.404040....

Let 𝑥 = 0.404040 …

⇒ 𝑥 = 0. 40 ………..….…. (1)

Multiplying both sides by 100, we get

100 𝑥 = 40.40 ……….…. (2)

Subtracting equation (1) from equation (2), we get

100 𝑥 − 𝑥 = 40.40 − 0.40

⇒ 99𝑥 = 40

Let 𝑥 = 0.142857142857 …

⇒ x = 0.142857 ……..….…. (1)

Multiplying both sides by 1000000, we get

1000000 x = 142857.142857 ……….…. (2)

Subtracting equation (1) from equation (2), we get

1000000 x-x = 142857.142857 - 0.142857

⇒ 999999x = 142857

Hence, proved.

√45 − 3√20 + 4√5

= √3 × √3 × √5 − 3√2 ×√ 2 × √5 + 4√5

= 3√5 − 6√5 + 4√5

= √5

(iv)

(√3 − √2)²

= (√3)² + (√2)² − 2(√3)(√2)

{Using (a-b)² = a²+b²-2ab}

= 3 + 2 − 2√6

= 5 − 2√6

= 3 - 8× 6 + 15× 2 + 15

= 3 – 48 + 30 + 15

= - 45 + 45

= 0

and

and

⇒ 11 - 6√3 = a - 6√3

⇒ a = 11

Given that 𝑎 = 2 + √3,

∴ We have

Now

(i)

(ii)

(iii)

(iv)

(v)

{Using a^x. a^y = a^x+y}

{Using a^x. a^y = a^x+y}

{Using a^x. a^y = a^x+y}

Let x = 0.6

Multiply by 10 on both sides,

10x = 6

x = 6/10

x = 3/5

So, the p/q form of 0.6 = 3/5

Let y =

Multiply by 10 on both sides,

10y =

10y – y = -

= 7.7777777……. – 0.7777777…………..

9y = 7

y = 7/9

So the p/q form of = 7/9

Let z =

Multiply by 10 on both sides,

10z =

10z – z = -

= 4.7777777…………. – 0.47777777……………………

9z = 4.2999

z ≈ 4.3/9

z = 43/90

So the p/q form of

= 43/90

=

Cross Multiplying to make the denominators same

Cross Multiplying again to make the denominators same

Cross Multiplying to make the denominators same

Here the denominators form the expansion as

(a + b) × (a – b) = (a² – b²)

Here a = 3√3

b = 2√2

a² = (3√3)²

= 27

b² = (2√2)²

= 8

We know the standard expansion as,

(a + b)² = a² + 2ab + b²

Let a = a

b = 1/a

= 9 – 2

= 7

(a + b) ² = a² + 2ab + b²

Also x = 1 / y or y = 1/x

Let a = x

b = y

(x + y) ² = x² + 2xy + y²

But we know y = 1/x

Here the denominators form the expansion as

(a + b) × (a – b) = (a² – b²)

Here a = √3

b = √2

a² = (√3)²

= 3

b² = (√2)²

= 2

= 10² – 2

= 100-2

= 98

By Law indices (a^m) ⁿ = a^mn

= 1/2

By Law indices (a^m) ⁿ = a^mn

= 144 + 64 + 6

= 214